MATH 332 — Fall 2026

Bases, Dimension,
and Coordinates
Anton, Rorres, & Kaul §§4.4–4.7

Fred J. Hickernell

October 9, 2026

Course Map

Course decks

• Why Linear Algebra?
• Systems and Matrices
• Inverses
• Matrix Transformations
• Determinants
• Euclidean Spaces
• Vector Spaces and Span
• Bases and Coordinates
• Matrix Spaces and Rank
• Eigenvalues
• Inner Products
• Special Topics

In this deck

Basis (coordinates use the listed order): \(\mathcal{B}=\{1,t,t^2\}\)

Coordinates in \(\mathcal{B}\)
\([p]_{\mathcal{B}}=\begin{bmatrix}2\\-3\\5\end{bmatrix}\)

\(\longrightarrow\)

Vector in \(\mathbb{P}_2\)
\(p(t)=2\cdot1-3\cdot t+5\cdot t^2\)

• Why Span, Bases, and Coordinates?
• Independence and Bases
• Bases and Unique Representations
• Coordinates in a Chosen Basis
• Dimension and Coordinates
• Arithmetic in Coordinates
• Changing Coordinate Systems
• Big Ideas

The polynomial is the vector in this example; the column contains its coordinates in the basis {1,t,t^2}, in the listed order. We use braces for bases; for coordinates, the displayed order matters. Its first entry multiplies 1, its second multiplies t, and its third multiplies t^2. In R^n with the standard basis, a vector and its coordinate column look identical. Here the different kinds of objects make the distinction visible: coefficients together with the chosen basis determine the polynomial.

Why Span, Bases, and Coordinates?

Which vectors can we build, and how can we describe each one uniquely?

• From building vectors to giving them addresses
• A square becomes a parallelogram
• Coordinates turn transformations into matrix multiplication
• Regression chooses the closest vector we can build
• PCA chooses coordinates adapted to the data

From building vectors to giving them addresses

Choose vectors \(\vct{b}_1,\ldots,\vct{b}_k\) in a space \(V\)

Concept	Question it answers	Payoff
Linear combination	What does \(c_1\vct{b}_1+\cdots+c_k\vct{b}_k\) build?	One vector from chosen ingredients
Span	What can all choices of coefficients reach?	The attainable subspace
Linear independence	Can two coefficient lists build the same vector?	No ambiguity when independent
Basis \(\{\vct{b}_1,\ldots,\vct{b}_k\}\)	Do we reach every vector of \(V\), uniquely?	A spanning, independent set
Coordinates \(c_1,\ldots,c_k\)	Which coefficients build this vector?	Its address in an ordered basis

Spanning gives existence; independence gives uniqueness

This is a motivation preview; the next sections develop the definitions and proofs. A list always spans its own span; being a basis for a specified V additionally requires that this span equal V. If two coefficient lists build the same vector, subtracting their combinations produces a nontrivial zero combination. Conversely a nontrivial zero combination changes coefficients without changing the vector. A vector and its coordinate list are different objects, even when both are displayed as columns.

A square becomes a parallelogram

An invertible map transforms the standard basis into a new basis

\[\begin{gather*} \mat{A}=\begin{bmatrix}2&1\\0&1\end{bmatrix},\qquad \vct{b}_1=\mat{A}\vct{e}_1=\begin{bmatrix}2\\0\end{bmatrix},\qquad \vct{b}_2=\mat{A}\vct{e}_2=\begin{bmatrix}1\\1\end{bmatrix} \end{gather*}\]

\[ \vct{c}=\begin{bmatrix}1/4\\3/4\end{bmatrix} \quad\longmapsto\quad \vct{v}=\mat{A}\vct{c}=\tfrac14\vct{b}_1+\tfrac34\vct{b}_2 =\begin{bmatrix}5/4\\3/4\end{bmatrix} \]

Recall the basis-image interpretation from Matrix Transformations. Both panels use the same scale in both coordinate directions. Grid lines and the highlighted point are transformed by the same A as the edge vectors. The square and parallelogram are bounded patches, not subspaces. Restricting the weights to [0,1] fills these patches; arbitrary real weights fill the plane. A singular map may collapse a direction, so it need not preserve a basis.

Unique addresses need independent directions

Keep \(\vct{b}_1=[2,0]^{\mathsf T}\) and \(\vct{b}_2=[1,1]^{\mathsf T}\)

For the ordered basis \(\mathcal{B}=\{\vct{b}_1,\vct{b}_2\}\),

\[ \vct{v}=\begin{bmatrix}5/4\\3/4\end{bmatrix} \quad\text{has coordinates}\quad [\vct{v}]_{\mathcal{B}}=\begin{bmatrix}1/4\\3/4\end{bmatrix} \]

• \(0\le c_1,c_2\le1\) fills the parallelogram; \(c_1,c_2\in\reals\) fills \(\reals^2\)
• \(\det[\vct{b}_1\ \vct{b}_2]=2\ne0\): every vector has one address
• \(\vct{b}_1^{\mathsf T}\vct{b}_2=2\ne0\): a basis need not be orthogonal

\(\exstar\) Add \(\vct{b}_3=\vct{b}_1+\vct{b}_2\) Does the span grow? Find two different lists in \(\vct{b}_3=c_1\vct{b}_1+c_2\vct{b}_2+c_3\vct{b}_3\); zero coefficients are allowed

The span stays R^2. The lists [1,1,0]^T and [0,0,1]^T both build b_3; the nontrivial combination b_1+b_2-b_3=0 certifies dependence. This ambiguity is precisely what independence removes. The coefficient bounds describe a patch only; the two edge directions form a basis of the whole plane, not a vector-space basis of the bounded parallelogram. No orthogonality is needed.

Coordinates turn transformations into matrix multiplication

Differentiation is a linear transformation \(D:\mathbb{P}_3\to\mathbb{P}_2\), \(D(p)=p'\)

\[ \text{Input basis: }\mathcal{B}=\{1,t,t^2,t^3\},\qquad \text{output basis: }\mathcal{C}=\{1,t,t^2\} \]

For \(p(t)=2-3t+5t^2+4t^3\), transform its coordinates:

\[ \underbrace{\begin{bmatrix}0&1&0&0\\0&0&2&0\\0&0&0&3\end{bmatrix}}_{\mat{M}} \underbrace{\begin{bmatrix}2\\-3\\5\\4\end{bmatrix}}_{[p]_{\mathcal{B}}} =\underbrace{\begin{bmatrix}-3\\10\\12\end{bmatrix}}_{[D(p)]_{\mathcal{C}}} \]

Reconstruct only when needed: \(D(p)(t)=-3+10t+12t^2\)

For a linear map \(T\) and fixed ordered bases: \([T(\vct{v})]_{\mathcal{C}}=\mat{M}[\vct{v}]_{\mathcal{B}}\)

Build \(\mat{M}\) once; compute with coordinates — Matrix Spaces and Rank

This is the computational payoff of coordinates, not merely a naming system. P_3 consists of polynomials of degree at most three, including zero; P_2 has degree at most two. Differentiation preserves linear combinations. Differentiate the ordered input basis: D(1)=0, D(t)=1, D(t^2)=2t, D(t^3)=3t2. Their coordinate columns in C form the 3-by-4 matrix M. The constant coordinate is lost under differentiation; invertibility is not required. The matrix is attached to the transformation AND the chosen input and output bases; changing bases generally changes it. For any linear map between finite-dimensional spaces, linearity lets us transform basis vectors once and then act on coordinate columns. The next deck proves the formula and works a lower-degree differentiation example, P_2 to P_1. This same principle works for functions in a chosen finite-dimensional function space, matrices, and other abstract vectors. For a general infinite-dimensional space, a finite coordinate column may represent only a selected subspace or approximation; no such restriction is needed for these polynomial spaces.

Regression chooses the closest vector we can build

\(\mat{A}\in\reals^{m\times n}\): \(m\) observations, \(n\) feature columns \(\vct{A}_1,\ldots,\vct{A}_n\); \(\vct{b}\in\reals^m\): observed responses

\[ \widehat{\vct{b}}=\mat{A}\widehat{\vct{x}} =\widehat{x}_1\vct{A}_1+\cdots+\widehat{x}_n\vct{A}_n \]

• Span: \(S=\operatorname{span}\{\vct{A}_1,\ldots,\vct{A}_n\}\) contains every possible fitted response
• Closest fit: choose \(\widehat{\vct{b}}\in S\) minimizing \(\|\vct{b}-\widehat{\vct{b}}\|^2\)
• Independent columns: \(\operatorname{rank}(\mat{A})=n\), called full column rank
• Basis and coordinates: the ordered columns form a basis of \(S\); \(\widehat{\vct{x}}\) gives the fitted vector’s coordinates

The closest fitted vector is unique; its coefficients are unique exactly when the columns are independent

Later: Inner Products — least squares

Ordinary unweighted least squares uses Euclidean distance in response space R^m. A response vector collects all m observed responses; a feature column collects one feature’s values at those same observations. The columns span the column space, which may be much smaller than R^m. Full column rank requires m >= n. Full row rank alone does not give unique coefficients. Dependent columns still yield a unique Euclidean projection but allow different coefficient lists for it. We preview rank as the number of independent columns; Matrix Spaces and Rank develops it formally. Leave normal equations and projection formulas to Inner Products.

Example: two features give coordinates for three fitted responses

Fit a line \(b\approx x_1+x_2t\) to \((t,b)=(0,1),(1,0),(2,2)\)

\[ \mat{A}=\begin{bmatrix}1&0\\1&1\\1&2\end{bmatrix},\qquad \vct{b}=\begin{bmatrix}1\\0\\2\end{bmatrix},\qquad \widehat{\vct{x}}=\begin{bmatrix}1/2\\1/2\end{bmatrix} \]

\[ \widehat{\vct{b}}=\tfrac12\begin{bmatrix}1\\1\\1\end{bmatrix} +\tfrac12\begin{bmatrix}0\\1\\2\end{bmatrix} =\begin{bmatrix}1/2\\1\\3/2\end{bmatrix},\qquad \vct{b}-\widehat{\vct{b}}=\begin{bmatrix}1/2\\-1\\1/2\end{bmatrix} \]

The independent feature columns span a plane in \(\reals^3\)

The residual is perpendicular to both columns: this is the closest fit

\([1/2,1/2]^{\mathsf T}\) locates the fitted response in that plane; it does not represent the observed response outside the plane

The feature columns have dot product 3, so this basis is not orthogonal. The residual r=[1/2,-1,1/2]^T has zero dot product with each column. For any candidate A x, Pythagoras gives ||b-A x||^2 = ||r||^2 + ||bhat-A x||^2, minimized at A x=bhat. Independence makes xhat unique. The squared minimum is 3/2. This is the later Inner Products line-fitting example with predictor values shifted by one: its intercept changes from 1 to 1/2 while its fitted responses stay the same. Geometry here takes place in the three-dimensional space of responses, not the plotted (t,b) plane.

PCA chooses coordinates adapted to the data

Principal component analysis (PCA): center the data, then choose orthogonal unit directions with the greatest remaining variation

\[ \vct{q}_1=\frac1{\sqrt2}\begin{bmatrix}1\\1\end{bmatrix},\qquad \vct{q}_2=\frac1{\sqrt2}\begin{bmatrix}-1\\1\end{bmatrix} \]

In this basis, one centered observation is

\[ \vct{z}=3\vct{q}_1+\tfrac12\vct{q}_2 \]

Its coordinate vector \([3,1/2]^{\mathsf T}\) contains its scores

Full basis: exact coordinates

Keep \(\vct{q}_1\): \(\vct{z}\approx3\vct{q}_1\) in its span

Later: Special Topics — low-rank approximation

Preview, not a PCA algorithm lesson. Subtract the sample mean before describing observations as vectors from the origin. Feature scaling is a modeling choice. PCA chooses orthonormal directions successively maximizing variance, with each new direction perpendicular to the preceding ones. A full orthonormal basis represents every centered observation exactly; the retained k directions are a basis only of their k-dimensional span. The scores are the coordinates of the projection in that retained basis. Add the mean back to reconstruct an observation in the original feature coordinates. Tied eigenvalues permit different equally valid choices of directions; uniqueness is relative to the chosen ordered basis. The synthetic figure has score pairs plus/minus (3,1/2), (2,-3/4), (1,0), and (0,1/2). Their cross moment is zero and the first variance is larger, so the displayed q_1 and q_2 really are principal directions. Its highlighted z uses the first pair.

[Sources] Andrew Ng, Stanford CS229, Principal components analysis, centering, variance-maximizing directions, and projected coordinates.

Independence and Bases

• Independence means the zero vector has one construction
• A homogeneous coefficient system detects independence
• Dependence identifies a redundant generator

Independence means the zero vector has one construction

Vectors \(\vct{v}_1,\ldots,\vct{v}_k\) are linearly independent when

\[ c_1\vct{v}_1+\cdots+c_k\vct{v}_k=\vct{0} \]

forces

\[ c_1=\cdots=c_k=0 \]

They are linearly dependent when some nonzero coefficient vector produces the zero vector

Independence means no generator can be synthesized from the others

Example: independent Fourier modes

Are \(1\) and \(e^{2\pi\sqrt{-1}\,t}\) independent over \(\mathbb{C}\) on \([0,1]\)?

Suppose \(a+be^{2\pi\sqrt{-1}\,t}=0\) for every \(t\in[0,1]\)

\[ \begin{array}{rcl} t=0&:&a+b=0\\ t=\tfrac12&:&a-b=0 \end{array} \quad\implies\quad a=b=0 \]

Only the trivial combination gives the zero function, so the two modes are independent

Are two Fourier modes independent?

\(\exstar\) Are \(e^{2\pi\sqrt{-1}\,t}\) and \(e^{-2\pi\sqrt{-1}\,t}\) linearly independent over \(\mathbb{C}\) on \([0,1]\)? Set their linear combination equal to zero for every \(t\); evaluate at \(t=0\) and \(t=\tfrac14\) Repeat for \(1\) and \(e^{4\pi\sqrt{-1}\,t}\), choosing two useful values of \(t\)

For \(e^{2\pi\sqrt{-1}\,t},e^{-2\pi\sqrt{-1}\,t}\): at \(t=0\), \(a+b=0\); at \(t=\tfrac14\), \(\sqrt{-1}\,a-\sqrt{-1}\,b=0\), so \(a=b=0\). For \(1,e^{4\pi\sqrt{-1}\,t}\): at \(t=0\), \(a+b=0\); at \(t=\tfrac14\), \(a-b=0\), so again \(a=b=0\).

A homogeneous coefficient system detects independence

For coordinate vectors, place the candidates into columns:

\[ \mat{V}= \begin{bmatrix} \vct{v}_1&\cdots&\vct{v}_k \end{bmatrix} \]

Then

\[ c_1\vct{v}_1+\cdots+c_k\vct{v}_k=\vct{0} \quad\Longleftrightarrow\quad \mat{V}\vct{c}=\vct{0} \]

Only \(\vct{c}=\vct{0}\) solves this system exactly when every column has a pivot

The familiar homogeneous system answers the abstract independence question

Dependence identifies a redundant generator

Suppose

\[ c_1\vct{v}_1+\cdots+c_k\vct{v}_k=\vct{0} \]

and \(c_j\ne0\). Then

\[ \vct{v}_j =-\sum_{i\ne j}\frac{c_i}{c_j}\vct{v}_i \]

Removing \(\vct{v}_j\) does not change the span

A dependence relation is a certificate of redundancy

Bases and Unique Representations

• A basis balances reach and economy
• Elimination extracts a basis from a spanning set
• Bases need not be finite

A basis balances reach and economy

A finite set

\[ \mathcal{B}=\{\vct{b}_1,\ldots,\vct{b}_n\} \]

is a basis for \(V\) when

1. \(\mathcal{B}\) spans \(V\)
2. \(\mathcal{B}\) is linearly independent

For coordinates, use the basis vectors in the order listed

Spanning guarantees existence of coefficients

Independence guarantees uniqueness of coefficients

Every vector has one coefficient list in a chosen ordered basis

Independence makes coordinates unique

Let \(\mathcal{B}=\{\vct{b}_1,\ldots,\vct{b}_n\}\) span \(V\) and be independent

Spanning supplies coefficients; suppose one vector has two coefficient lists

\[ \vct{v}=\sum_{j=1}^n c_j\vct{b}_j =\sum_{j=1}^n d_j\vct{b}_j \]

Subtract the two representations

\[ \sum_{j=1}^n(c_j-d_j)\vct{b}_j=\vct{0} \quad\implies\quad c_j-d_j=0\quad\text{for every }j \]

Linear independence forces the two coefficient lists to agree

A space can have many bases

Coordinates are unique once the ordered basis is fixed

Elimination extracts a basis from a spanning set

The retained vectors must span the same space and be independent

For vectors in \(\reals^m\):

1. place the spanning vectors into the columns of a matrix
2. row reduce
3. identify the pivot columns
4. keep the corresponding columns of the original matrix

Row operations preserve dependence relations among columns but change the columns themselves

Use the reduced matrix to locate pivots; take the basis vectors from the original matrix

Example: remove redundant generators

Let \(W\) be the span of the columns of \(\mat{V}\)

\[ \mat{V}=[\vct{v}_1\ \vct{v}_2\ \vct{v}_3\ \vct{v}_4] =\begin{bmatrix}1&0&1&2\\0&1&1&0\\1&1&2&2\end{bmatrix} \;\xrightarrow{R_3\leftarrow R_3-R_1-R_2}\; \begin{bmatrix}1&0&1&2\\0&1&1&0\\0&0&0&0\end{bmatrix} \]

The first two columns are pivot columns

\[ \vct{v}_3=\vct{v}_1+\vct{v}_2, \qquad \vct{v}_4=2\vct{v}_1 \]

Remove \(\vct{v}_3\) and \(\vct{v}_4\) without changing the span

\[ W=\operatorname{span}\{\vct{v}_1,\vct{v}_2,\vct{v}_3,\vct{v}_4\} =\operatorname{span}\{\vct{v}_1,\vct{v}_2\} \]

\(\mathcal{B}=\{\vct{v}_1,\vct{v}_2\}\) still spans \(W\)

\(a\vct{v}_1+b\vct{v}_2=\vct{0}\) forces \(a=b=0\): independence completes the basis proof

Subtract the first two rows from the third to obtain the displayed echelon form. The nonpivot columns reveal the two dependence relations. The retained columns span the original space because both removed vectors can be rebuilt from them. They are independent: the first and second components of \(a\vct{v}_1+b\vct{v}_2=\vct{0}\) force \(a=b=0\). Thus four generators become a two-vector basis without losing any vector in W. The space is the plane \(z=x+y\), connecting back to the first deck’s Course Map.

Different bases can describe the same span

Use the preceding generators of the plane \(W=\{[x,y,z]^{\mathsf T}:z=x+y\}\)

\[ \vct{v}_1=\begin{bmatrix}1\\0\\1\end{bmatrix},\quad \vct{v}_2=\begin{bmatrix}0\\1\\1\end{bmatrix},\quad \vct{v}_3=\begin{bmatrix}1\\1\\2\end{bmatrix},\quad \vct{v}_4=\begin{bmatrix}2\\0\\2\end{bmatrix} \]

\(\exstar\) A basis must span \(W\) and be independent Show that \(\{\vct{v}_2,\vct{v}_3\}\) is another basis for \(W\) Explain why \(\{\vct{v}_1,\vct{v}_4\}\) is not a basis for \(W\)

The first two components of a v2+b v3=0 force b=0, then a=0, so v2 and v3 are independent. Since v1=v3-v2 and v4=2(v3-v2), they span W. The pair v1,v4 is dependent because v4=2v1; it spans only a line in W. A basis need not be unique. Independence makes a vector’s coefficients unique after an ordered basis has been chosen.

A frame can keep redundant generators

For a subspace of \(\reals^n\), a finite spanning list is a frame

Independence is optional; for \(W=\{[x,y,z]^{\mathsf T}:z=x+y\}\), use

\[ \vct{v}_1=[1,0,1]^{\mathsf T},\quad \vct{v}_2=[0,1,1]^{\mathsf T}, \quad \vct{v}_3=\vct{v}_1+\vct{v}_2,\quad \vct{v}_4=2\vct{v}_1 \]

\[ \vct{v}_3 =0\vct{v}_1+0\vct{v}_2+1\vct{v}_3+0\vct{v}_4 =1\vct{v}_1+1\vct{v}_2+0\vct{v}_3+0\vct{v}_4 \]

A redundant frame admits more than one coefficient list for the same vector

Here any one of the four vectors can be lost without losing the span

Keep this as a brief preview, not a development of frame theory. In a finite-dimensional inner product space, a finite spanning list satisfies the frame bounds and is a frame. In infinite dimensions, spanning alone does not suffice. The word frame does not require redundancy: a basis is also a frame. For a redundant frame, a reconstruction rule can select one coefficient list (for example the minimum-norm list), even though the representation equation alone has many solutions. Inner products and norms come later in the course. For the displayed example, loss of v1 leaves v2 and v3, loss of v2 leaves v1 and v3, and loss of v3 or v4 leaves v1 and v2; each remaining list spans W. Reference: https://faculty.sites.iastate.edu/esweber/files/inline-files/book_final_0.pdf

Bases need not be finite

An algebraic (Hamel) basis uses only finite linear combinations

• All polynomials have a countably infinite basis \(\{1,t,t^2,\ldots\}\)
• Finite Fourier sums on \([0,1]\), over \(\mathbb{C}\), have a countable basis \(\{e^{2\pi\sqrt{-1}\,kt}:k\in\mathbb{Z}\}\)
• \(C(\reals)\), all continuous real-valued functions on the real line, has an uncountable algebraic basis

Algebraic bases give exact finite representations

Countable orthonormal bases in \(L^2[a,b]\) allow infinite expansions

This is only a taste of infinite-dimensional spaces. The basis here is algebraic: every vector is a finite combination. The family \(\{e^{\lambda t}:\lambda\in\reals\}\) is uncountably linearly independent in \(C(\reals)\), so any algebraic basis there must be uncountable. In Fourier theory, countably many complex exponentials can also represent functions through convergent infinite series; that is a different, analytic sense of basis. The orthonormal expansion converges in the L2 norm, not necessarily uniformly or pointwise everywhere. L2[a,b] itself also has uncountable algebraic dimension. To see the uncountable independent family in C(R) directly, take distinct real lambdas and differentiate a finite relation among their exponentials zero through k-1 times at t=0. The resulting Vandermonde system is invertible. This family is independent, not asserted to be a basis. C(R) is not being identified with a subspace of L2(R): its functions need not be square integrable. Frames and infinite algebraic bases are enrichment beyond Anton Chapter 4; infinite Fourier series preview Section 6.6. Reference: https://math.mit.edu/~rbm/18.103-S04/node3.html

Coordinates in a Chosen Basis

• A basis gives every vector unique coordinates
• Same vector, different coordinate columns
• Polynomial coefficients are coordinates in a chosen basis
• Example: a basis of symmetric matrices
• Example: test a polynomial basis

A basis gives every vector unique coordinates

A basis \(\mathcal{B}=\{\vct{b}_1,\ldots,\vct{b}_n\}\) spans \(V\) and is independent

For \(\vct{v}\in V\), spanning gives existence and independence gives uniqueness:

\[ \vct{v}=c_1\vct{b}_1+\cdots+c_n\vct{b}_n \]

The coordinate vector of \(\vct{v}\) relative to \(\mathcal{B}\) is

\[ [\vct{v}]_{\mathcal{B}} =\begin{bmatrix}c_1\\\vdots\\c_n\end{bmatrix} \]

For coordinates, the listed order matters: entry \(j\) multiplies \(\vct{b}_j\)

We have been using the standard basis

In \(\reals^2\), the standard basis is

\[ \mathcal{E}=\left\{\vct{e}_1=\begin{bmatrix}1\\0\end{bmatrix}, \vct{e}_2=\begin{bmatrix}0\\1\end{bmatrix}\right\} \]

Writing \(\vct{x}=\begin{bmatrix}5\\1\end{bmatrix}\) has meant

\[ \vct{x}=5\vct{e}_1+1\vct{e}_2, \qquad [\vct{x}]_{\mathcal{E}}=\begin{bmatrix}5\\1\end{bmatrix} \]

In \(\reals^n\), use \(\mathcal{E}=\{\vct{e}_1,\ldots,\vct{e}_n\}\), where \(\vct{e}_j\) has a 1 in entry \(j\) and zeros elsewhere

Our familiar vector column is its coordinate vector in the standard basis

Same vector, different coordinate columns

For \(\vct{x}=[5,1]^{\mathsf T}\), compare the standard basis \(\mathcal{E}=\{[1,0]^{\mathsf T},[0,1]^{\mathsf T}\}\) with

\[ \mathcal{B}=\left\{\vct{b}_1=\begin{bmatrix}1\\1\end{bmatrix}, \vct{b}_2=\begin{bmatrix}1\\-1\end{bmatrix}\right\} \]

The same vector has two descriptions:

\[ \vct{x}=5\vct{e}_1+1\vct{e}_2=3\vct{b}_1+2\vct{b}_2 \]

\[ [\vct{x}]_{\mathcal{E}}=\begin{bmatrix}5\\1\end{bmatrix}, \qquad [\vct{x}]_{\mathcal{B}}=\begin{bmatrix}3\\2\end{bmatrix} \]

The brackets mean “coordinates of this vector in this basis”

The entries 3 and 2 are weights on \(\vct{b}_1\) and \(\vct{b}_2\), not the standard horizontal and vertical components

The vector stays at \((5,1)\); changing the basis changes its description

Read the basis label before the numbers

Use \(\mathcal{E}=\{[1,0]^{\mathsf T},[0,1]^{\mathsf T}\}\) and \(\mathcal{B}=\{[1,1]^{\mathsf T},[1,-1]^{\mathsf T}\}\)

\(\exstar\) Interpret each coordinate column If \([\vct{u}]_{\mathcal{B}}=[1,0]^{\mathsf T}\), find \(\vct{u}\) in standard coordinates If \([\vct{v}]_{\mathcal{E}}=[1,0]^{\mathsf T}\), find \(\vct{v}\). Is it the same vector as \(\vct{u}\)? Reverse the listed order of \(\mathcal{B}\) to get \(\mathcal{B}'\). Find \([\vct{x}]_{\mathcal{B}'}\) for \(\vct{x}=[5,1]^{\mathsf T}\)

The first column means u = 1 b1 + 0 b2 = [1,1]^T. The second means v = 1 e1 + 0 e2 = [1,0]^T, a different vector despite identical coordinate entries. With B’ = {b2,b1} in the reversed order (the same underlying set), x = 2 b2 + 3 b1, so its coordinates are [2,3]^T. The jth coordinate multiplies the jth basis vector; their order matters.

Polynomial coefficients are coordinates in a chosen basis

For the standard polynomial basis

\[ \mathcal{E}=\{1,t,t^2\} \]

the polynomial \(p(t)=2-3t+5t^2\) has coordinates

\[ [p]_{\mathcal{E}} =\begin{bmatrix}2\\-3\\5\end{bmatrix} \]

For another basis, the same polynomial has another coordinate vector

Choosing a basis converts abstract vectors into coordinate columns

Example: a basis of symmetric matrices

For real symmetric \(2\times2\) matrices (\(\mat{A}^{\mathsf T}=\mat{A}\)), use

\[ \mat{B}_1=\begin{bmatrix}1&0\\0&0\end{bmatrix}, \quad \mat{B}_2=\begin{bmatrix}0&1\\1&0\end{bmatrix}, \quad \mat{B}_3=\begin{bmatrix}0&0\\0&1\end{bmatrix} \]

Every symmetric matrix has exactly one representation

\[ \begin{bmatrix}a&b\\b&d\end{bmatrix} =a\mat{B}_1+b\mat{B}_2+d\mat{B}_3 \]

Thus \(\mathcal{B}=\{\mat{B}_1,\mat{B}_2,\mat{B}_3\}\) spans the space and is independent

\[ \mat{A}=\begin{bmatrix}2&-1\\-1&3\end{bmatrix} \quad\implies\quad [\mat{A}]_{\mathcal{B}}=\begin{bmatrix}2\\-1\\3\end{bmatrix} \]

The vector is a matrix; its coordinate vector is a column of coefficients

Bases for other matrix spaces

\(\exstar\) For real \(2\times2\) matrices, build bases from the freely chosen entries Find an ordered basis for all upper-triangular \(2\times2\) matrices In your basis, find the coordinates of \(\begin{bmatrix}2&-1\\0&3\end{bmatrix}\) Find a basis for the skew-symmetric matrices: \(\mat{A}^{\mathsf T}=-\mat{A}\) Explain why your lists both span and are independent

Use E_11, E_12, E_22, where E_ij has a single 1 in entry (i,j), for the upper-triangular space; coordinates are [2,-1,3]^T in that order. A real skew-symmetric 2x2 matrix is [[0,b],[-b,0]], so the single matrix [[0,1],[-1,0]] is a basis. Matching entries proves spanning and independence. No dimension terminology is needed yet.

Example: test a polynomial basis

In \(\mathbb{P}_2\), the polynomials in \(t\) of degree at most two, take \(\mathcal{E}=\{1,t,1+t^2\}\)

Its coefficient columns relative to \(\{1,t,t^2\}\) form

\[ \begin{bmatrix}1&0&1\\0&1&0\\0&0&1\end{bmatrix} \]

Three pivots: every coefficient vector is reached uniquely, so \(\mathcal{E}\) spans \(\mathbb{P}_2\) and is independent

For \(p(t)=2+4t+3t^2\),

\[ p=-1(1)+4t+3(1+t^2), \qquad [p]_{\mathcal{E}}=[-1,4,3]^{\mathsf T} \]

A basis question combines two tests

In \(\mathbb{P}_2\), consider

\[ \mathcal{B}=\{1+t,\ t+t^2,\ 1+t^2\} \]

\(\exstar\) Does \(\mathcal{B}\) span all of \(\mathbb{P}_2\) and is it independent? Form the matrix of coefficients relative to \(\{1,t,t^2\}\) Test its columns for independence If it is a basis, find \([2+4t]_{\mathcal{B}}\)

The coefficient matrix is \(\begin{bmatrix}1&0&1\\1&1&0\\0&1&1\end{bmatrix}\), whose determinant is \(2\). Thus the list is a basis. Solving gives coordinates \((3,1,-1)\) for \(2+4t\).

Two more polynomial basis tests

In \(\mathbb{P}_2\), consider

\[ \mathcal{C}=\{1+t,\ t+t^2,\ 1+2t+t^2\}, \qquad \mathcal{D}=\{1+t,\ t+t^2,\ 1\} \]

\(\exstar\) Use spanning and independence to decide which list is a basis of \(\mathbb{P}_2\) For the basis, find the coordinates of \(2+4t\)

Use the previous basis slide as the board example. \(\mathcal{C}\) is dependent: its third polynomial is the sum of its first two. \(\mathcal{D}\) is a basis; \(2+4t=4(1+t)+0(t+t^2)-2(1)\), so the coordinates are \((4,0,-2)\).

Dimension and Coordinates

• Dimension counts vectors in any basis
• Dimension turns two tests into one
• Dimension depends on the scalar field
• Example: count monomials to find dimension

Dimension counts vectors in any basis

A space is finite-dimensional if it has a finite basis

Every basis then has the same number of vectors, the dimension:

\[ \dim(V)=n \]

Over \(\reals\): columns, one-variable polynomials of degree at most \(k\), and matrices

\[ \dim(\reals^n)=n, \qquad \dim(\mathbb{P}_k)=k+1, \qquad \dim(\reals^{m\times n})=mn \]

Our earlier \(U=\{p\in\mathbb{P}_3:p(0)=0\}\) has basis \(\{t,t^2,t^3\}\), so \(\dim(U)=3\)

The zero space has the empty basis, so \(\dim(\{\vct{0}\})=0\)

Dimension counts independent directions, not the number of vectors in the space

Dimension turns two tests into one

Suppose \(\dim(V)=n\) and a list contains exactly \(n\) vectors in \(V\)

Then either one of these facts proves the other:

\[ \text{linearly independent} \quad\Longleftrightarrow\quad \text{spans }V \]

More generally,

• more than \(n\) vectors must be dependent
• fewer than \(n\) vectors cannot span \(V\)
• every independent list can be extended to a basis
• every spanning list can be reduced to a basis

Knowing the dimension can cut a basis proof in half

Use dimension before computing

Since \(\dim(\mathbb{P}_3)=4\):

\(\exstar\) What can the dimension alone tell you? Can three polynomials span \(\mathbb{P}_3\)? Can five polynomials be linearly independent in \(\mathbb{P}_3\)? If four polynomials are independent, what else follows?

Three cannot span a four-dimensional space. Five must be dependent. Four independent polynomials form a basis and therefore span \(\mathbb{P}_3\).

Dimension depends on the scalar field

The same set \(\mathbb{C}\) can be viewed as two different vector spaces

Scalars	A basis	Dimension
\(\mathbb{C}\)	\(\{1\}\)	\(\dim_{\mathbb{C}}(\mathbb{C})=1\)
\(\reals\)	\(\{1,\sqrt{-1}\}\)	\(\dim_{\reals}(\mathbb{C})=2\)

Over \(\mathbb{C}\), any \(z\) is the scalar multiple \(z\cdot1\)

Over \(\reals\), write \(z=a+b\sqrt{-1}\) with two real coefficients

\(\exstar\) Find the dimensions of \(\mathbb{C}^2\) over \(\mathbb{C}\) and over \(\reals\)

This extends the earlier scalar-field discussion and previews Anton Section 5.3 rather than adding a Chapter 4 reading requirement. The dimensions of C^2 are 2 over C and 4 over R. A real basis is {(1,0),(sqrt(-1),0),(0,1),(0,sqrt(-1))}. Independence and spanning must always refer to the chosen scalar field. When the field is not shown in our later real examples, it is R.

Example: count monomials to find dimension

In \(\mathbb{P}_2(\reals^2)\), total degree is at most two: add the \(x\) and \(y\) exponents

\[ \underbrace{1}_{0}, \qquad \underbrace{x,\ y}_{1}, \qquad \underbrace{x^2,\ xy,\ y^2}_{2} \]

Every polynomial in the space is a unique linear combination of these six monomials

They form a basis, so

\[ \dim(\mathbb{P}_2(\reals^2))=1+2+3=6 \]

Count freely chosen coefficients after identifying a basis

Dimension beyond coordinate columns

\(\exstar\) Find a basis and count its vectors; total degree adds the \(x\) and \(y\) exponents \(\mathbb{P}_3(\reals^2)\): list monomials by total degree All upper-triangular \(3\times3\) matrices All symmetric \(3\times3\) matrices Could five symmetric \(3\times3\) matrices span their entire space?

For P_3(R^2), a basis is {1,x,y,x^2,xy,y2,x^3,x2y,xy^2,y3}, so the dimension is 1+2+3+4=10. Upper-triangular 3x3 matrices have six free entries; the six corresponding single-entry matrices form a basis. Symmetric 3x3 matrices also have six free entries: three diagonal and three off-diagonal pairs. Use three diagonal single-entry matrices and E_ij+E_ji for i<j. Matching entries proves independence. Five cannot span this six-dimensional space.

Arithmetic in Coordinates

• The coordinate map preserves linear combinations
• Example: combine coordinate vectors

The coordinate map preserves linear combinations

For a basis \(\mathcal{B}\) of an \(n\)-dimensional real vector space, define

\[ \vct{v}\longmapsto[\vct{v}]_{\mathcal{B}}\in\reals^n \]

The column \([\vct{v}]_{\mathcal{B}}\) lists coefficients in this ordered basis, so

\[ [c\vct{u}+d\vct{v}]_{\mathcal{B}} =c[\vct{u}]_{\mathcal{B}}+d[\vct{v}]_{\mathcal{B}} \]

The coordinate map is one-to-one, onto, and preserves linear combinations

Such a map is a linear isomorphism

It preserves linear combinations; ordinary coordinate lengths may change

Every \(n\)-dimensional real vector space has the same linear structure as \(\reals^n\)

This is a statement about linear structure. The ordinary Euclidean length of a coordinate column need not equal the original vector’s length. For the basis {[1,1]^T,[1,-1]T}, b1 has length sqrt(2), while its coordinate column [1,0]^T has Euclidean length 1. General coordinates preserve algebra; geometry requires the inner product in those coordinates.

Example: combine coordinate vectors

Use the polynomial basis \(\mathcal{B}=\{1,1+t,t^2\}\)

\[ p=2+3t+t^2, \quad [p]_{\mathcal{B}}=[-1,3,1]^{\mathsf T} \]

\[ q=1-t+2t^2, \quad [q]_{\mathcal{B}}=[2,-1,2]^{\mathsf T} \]

Compute in coordinates:

\[ [2p-q]_{\mathcal{B}} =2[-1,3,1]^{\mathsf T}-[2,-1,2]^{\mathsf T} =[-4,7,0]^{\mathsf T} \]

Reconstruct: \(-4(1)+7(1+t)+0t^2=3+7t=2p-q\)

Coordinate arithmetic uses coefficients in the chosen basis

Reconstruct after coordinate arithmetic

Keep \(\mathcal{B}=\{1,1+t,t^2\}\) and suppose

\[ [p]_{\mathcal{B}}=[1,2,-1]^{\mathsf T}, \qquad [q]_{\mathcal{B}}=[-2,1,3]^{\mathsf T} \]

\(\exstar\) Coordinate entries multiply \(1\), \(1+t\), and \(t^2\) in that order Find \([p+2q]_{\mathcal{B}}\) without first expanding the polynomials Reconstruct \(p+2q\) as a polynomial in \(t\) Why is its constant coefficient different from the first coordinate?

The coordinate vector is [-3,4,5]^T, and the polynomial is -3+4(1+t)+5t^2=1+4t+5t2. Both 1 and 1+t contribute to the constant coefficient. Directly, p=3+2t-t^2 and q=-1+t+3t^2, giving the same result.

Changing Coordinate Systems

• A basis matrix converts coordinates to standard form
• Change of basis passes through the vector
• Transition columns describe the old basis in the new one
• Example: change between polynomial bases

The companion notebook explores basis extraction and coordinate changes

A basis matrix converts coordinates to standard form

For a basis \(\mathcal{B}=\{\vct{b}_1,\ldots,\vct{b}_n\}\) of \(\reals^n\), form

\[ \mat{P}_{\mathcal{B}} =\begin{bmatrix}\vct{b}_1&\cdots&\vct{b}_n\end{bmatrix} \]

The column \([\vct{x}]_{\mathcal{B}}\) lists coefficients multiplying these basis vectors

Independence makes \(\mat{P}_{\mathcal{B}}\) invertible:

\[ \vct{x}=\mat{P}_{\mathcal{B}}[\vct{x}]_{\mathcal{B}}, \qquad [\vct{x}]_{\mathcal{B}}=\mat{P}_{\mathcal{B}}^{-1}\vct{x} \]

The columns of a basis matrix tell what the new coordinate axes mean in standard coordinates

Change of basis passes through the vector

Let \(\mathcal{B}\) and \(\mathcal{C}\) be ordered bases of \(\reals^n\)

\(\mat{P}_{\mathcal{B}}\) and \(\mat{P}_{\mathcal{C}}\) contain their basis vectors as columns

\[ \vct{x}=\mat{P}_{\mathcal{B}}[\vct{x}]_{\mathcal{B}} =\mat{P}_{\mathcal{C}}[\vct{x}]_{\mathcal{C}} \]

To obtain \(\mathcal{C}\)-coordinates, multiply by \(\mat{P}_{\mathcal{C}}^{-1}\):

\[ [\vct{x}]_{\mathcal{C}} =\underbrace{\mat{P}_{\mathcal{C}}^{-1}\mat{P}_{\mathcal{B}}}_{ \text{transition from }\mathcal{B}\text{ to }\mathcal{C}} [\vct{x}]_{\mathcal{B}} \]

Changing coordinates changes the description, not the vector

Transition columns describe the old basis in the new one

Let \(\mathcal{B}=\{\vct{b}_1,\ldots,\vct{b}_n\}\) and \(\mathcal{C}\) be ordered bases of the same finite-dimensional real vector space

\[ \mat{Q}_{\mathcal{C}\leftarrow\mathcal{B}} =\begin{bmatrix}[\vct{b}_1]_{\mathcal{C}}&\cdots&[\vct{b}_n]_{\mathcal{C}}\end{bmatrix} \]

Column \([\vct{b}_j]_{\mathcal{C}}\) lists the coefficients that rebuild \(\vct{b}_j\) from \(\mathcal{C}\)

\[ [\vct{v}]_{\mathcal{C}} =\mat{Q}_{\mathcal{C}\leftarrow\mathcal{B}}[\vct{v}]_{\mathcal{B}} \]

For bases of \(\reals^n\), this is \(\mat{P}_{\mathcal{C}}^{-1}\mat{P}_{\mathcal{B}}\)

The arrow records the direction: input coordinates in \(\mathcal{B}\), output coordinates in \(\mathcal{C}\)

Example: change between polynomial bases

In \(\mathbb{P}_2\), choose

\[ \mathcal{B}=\{1,1+t,t^2\},\qquad \mathcal{C}=\{1,t,t+t^2\} \]

Columns give \(\mathcal{C}\)-coefficients of the \(\mathcal{B}\) polynomials; use \(t^2=(t+t^2)-t\):

\[ \mat{Q}_{\mathcal{C}\leftarrow\mathcal{B}} =\begin{bmatrix}1&1&0\\0&1&-1\\0&0&1\end{bmatrix} \]

For \(p=2-3t+5t^2\),

\[ [p]_{\mathcal{B}}=\begin{bmatrix}5\\-3\\5\end{bmatrix} \quad\xrightarrow{\mat{Q}_{\mathcal{C}\leftarrow\mathcal{B}}}\quad [p]_{\mathcal{C}}=\begin{bmatrix}2\\-8\\5\end{bmatrix} \]

\(2(1)-8t+5(t+t^2)=p\): new coordinates, same polynomial

Coordinates change with the basis

Keep \(\vct{x}=[5,1]^{\mathsf T}\) and compare

\[ \mathcal{C}=\bigl\{[1,0]^{\mathsf T},[1,1]^{\mathsf T}\bigr\}, \qquad \mathcal{D}=\bigl\{[2,1]^{\mathsf T},[1,1]^{\mathsf T}\bigr\} \]

\(\exstar\) Find \([\vct{x}]_{\mathcal{C}}\) and \([\vct{x}]_{\mathcal{D}}\) Reconstruct \(\vct{x}\) from each coordinate vector Use basis vectors as columns of \(\mat{P}_{\mathcal{C}}\) and \(\mat{P}_{\mathcal{D}}\); check the change from \(\mathcal{C}\) to \(\mathcal{D}\) with \(\mat{P}_{\mathcal{D}}^{-1}\mat{P}_{\mathcal{C}}\)

The preceding slide is the board example. Here \([\vct{x}]_{\mathcal{C}}=[4,1]^{\mathsf T}\) and \([\vct{x}]_{\mathcal{D}}=[4,-3]^{\mathsf T}\); both reconstruct \(\vct{x}=[5,1]^{\mathsf T}\). The transition matrix is \(\mat{P}_{\mathcal{D}}^{-1}\mat{P}_{\mathcal{C}} =\begin{bmatrix}1&0\\-1&1\end{bmatrix}\), which sends \([4,1]^{\mathsf T}\) to \([4,-3]^{\mathsf T}\).

Big Ideas

• Independence means every vector has at most one construction from the generators
• A basis combines independence and spanning to give unique coordinates
• Coordinates are coefficients in an ordered basis; the same vector can have different coordinate columns
• Dimension counts basis vectors and depends on the scalar field
• Coordinate maps preserve linear structure; transition matrices change descriptions

How Far We Have Come

• Systems and Matrices — \([\,\mat{A}\mid\vct{b}\,]\) tests consistency and finds free parameters; PLU saves elimination
• Matrix actions — products compose, inverses undo; determinants give signed scale and detect collapse
• Euclidean Spaces — consistent solutions: \(\vct{x}_p+\operatorname{null}(\mat{A})\); dot products, projection, and 3D cross products describe geometry

• Vector Spaces and Span — subspaces contain all linear combinations; vectors can be matrices, polynomials, or functions
• Bases and Coordinates — spanning gives coefficients; independence makes them unique; dimension counts basis vectors
• Coordinates — ordered-basis coordinates preserve linear combinations; changing the basis changes the column, not the vector

With ordered basis \(\mathcal{B}=\{\vct{b}_1,\ldots,\vct{b}_d\}\) and \(d=\dim(V)<\infty\): \(\vct{v}=\sum_{j=1}^d c_j\vct{b}_j\) has unique coordinates \([\vct{v}]_{\mathcal{B}}=[c_1,\ldots,c_d]^{\mathsf T}\)

What Comes Next

How do bases and dimension expose the structure of a matrix?

Next: Matrix Spaces and Rank