MATH 332 — Fall 2026

Matrix Spaces and Rank
Anton, Rorres, & Kaul §§4.8–4.9

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

Input: \(\reals^n\)

\(r=\operatorname{rank}(\mat{A})\)

Output: \(\reals^m\)

\(\operatorname{row}(\mat{A})\)
\(r\) independent directions

\(\xrightarrow{\mat{A}}\)

\(\operatorname{col}(\mat{A})\)
\(r\) independent directions

\(\perp\)

\(\perp\)

\(\operatorname{null}(\mat{A})\)
\(n-r\) directions
\(\mat{A}\vct{x}=\vct{0}\)

\(\operatorname{null}(\mat{A}^{\mathsf T})\)
\(m-r\) directions
\(\mat{A}^{\mathsf T}\vct{y}=\vct{0}\)

• Matrix Spaces and Rank
• Rank and Solution Sets
• Orthogonal Pairs
• Rank Factorization and the Left Null Space
• What the Structure Buys Us
• Big Ideas

Preview for a real m-by-n matrix of rank r. The input splits orthogonally into Row(A) and Null(A); the output splits into Col(A) and Null(A^T). The displayed arrow is A restricted to the row space: it maps that space one-to-one onto the column space. Both have dimension r. The n-r null-space directions collapse to zero; they are not mapped to the left null space. The output’s m-r left-null directions are perpendicular to every attainable output. Thus n=r+(n-r) and m=r+(m-r). The deck develops these claims after computing concrete bases and dimensions.

Matrix Spaces and Rank

• A matrix has four fundamental subspaces
• Rank and nullity measure different parts of a matrix
• One reduction exposes the structure
• Column and row bases come from different matrices
• Free variables build a basis for the null space

A matrix has four fundamental subspaces

For an \(m\times n\) matrix \(\mat{A}\):

Space	Definition	Lives in
\(\operatorname{col}(\mat{A})\)	span of the columns of \(\mat{A}\)	\(\reals^m\)
\(\operatorname{null}(\mat{A})\)	solutions of \(\mat{A}\vct{x}=\vct{0}\)	\(\reals^n\)
\(\operatorname{row}(\mat{A})\)	span of the rows of \(\mat{A}\)	\(\reals^n\)
\(\operatorname{null}(\mat{A}^{\mathsf T})\)	solutions of \(\mat{A}^{\mathsf T}\vct{y}=\vct{0}\)	\(\reals^m\)

Always identify the ambient space before comparing subspaces

Transposing exchanges rows and columns; view row-space vectors as columns

\(\operatorname{col}(\mat{A})=\operatorname{row}(\mat{A}^{\mathsf T})\), \(\quad\operatorname{row}(\mat{A})=\operatorname{col}(\mat{A}^{\mathsf T})\)

Rank and nullity measure different parts of a matrix

Recall: dimension is the number of vectors in a basis

For \(\mat{A}\in\reals^{m\times n}\), define

\[ \operatorname{rank}(\mat{A})=\dim(\operatorname{col}(\mat{A})), \qquad \operatorname{nullity}(\mat{A})=\dim(\operatorname{null}(\mat{A})) \]

Row reduction gives

• rank = number of pivot columns = dimension of the row space
• nullity = number of free variables

\[ 0\le\operatorname{rank}(\mat{A})\le\min(m,n) \]

Rank counts independent output directions; nullity counts lost input directions

Rank is defined here using column-space dimension. The row-rank/column-rank theorem says the dimensions agree. Pivot columns of the original matrix are a column-space basis; the nonzero RREF rows are a row-space basis with the same number of vectors. Each free variable contributes one independent null-space vector. We use real matrices throughout this deck.

One reduction exposes the structure

Use one example throughout, recalling pivots and free variables from Systems and Matrices:

\[ \mat{A}= \begin{bmatrix} 1&2&1&3\\ 2&4&0&4\\ 1&2&2&4 \end{bmatrix} \quad\longrightarrow\quad \operatorname{rref}(\mat{A})= \left[ \begin{array}{cccc} \bbox[3px,border:2px solid teal]{\begin{matrix}1\\0\\0\end{matrix}} &\begin{matrix}2\\0\\0\end{matrix} &\bbox[3px,border:2px solid teal]{\begin{matrix}0\\1\\0\end{matrix}} &\begin{matrix}2\\1\\0\end{matrix} \end{array} \right] \]

The RREF has pivots \(1\) and zeros elsewhere in each pivot column

Pivot columns: \(1,3\); free variables: \(x_2,x_4\)

The boxed columns are \(\vct{e}_1,\vct{e}_2\): the first two columns of \(\mat{I}_3\)

\[ \operatorname{rank}(\mat{A})=2 \]

Subtract twice the first row from the second and the first row from the third to get \((0,0,-2,-2)\) and \((0,0,1,1)\). Keep the latter as pivot row, eliminate the former, then subtract the pivot row from the first row. This gives the displayed reduced matrix. The teal outlines mark whole pivot columns, including their trailing zero entries. They are the first two columns of I_3; the same selected columns in the nonzero-row factor R form I_2.

Column and row bases come from different matrices

For the running \(3\times4\) matrix, the pivot columns are \(1\) and \(3\)

A basis for the column space uses pivot columns of the original matrix:

\[ \left\{ \begin{bmatrix}1\\2\\1\end{bmatrix}, \begin{bmatrix}1\\0\\2\end{bmatrix} \right\} \]

A basis for the row space uses nonzero rows of the reduced matrix:

\[ \left\{ \begin{bmatrix}1\\2\\0\\2\end{bmatrix}, \begin{bmatrix}0\\0\\1\\1\end{bmatrix} \right\} \]

Row reduction preserves the row space but generally changes the column space

Free variables build a basis for the null space

Row operations preserve solutions: \(\mat{A}\vct{x}=\vct{0}\iff\operatorname{rref}(\mat{A})\vct{x}=\vct{0}\)

For the running \(3\times4\) matrix, the free variables are \(x_2,x_4\):

\[ x_1=-2x_2-2x_4, \qquad x_3=-x_4 \]

Therefore

\[ \vct{x} =x_2\begin{bmatrix}-2\\1\\0\\0\end{bmatrix} +x_4\begin{bmatrix}-2\\0\\-1\\1\end{bmatrix} \]

and

\[ \dim(\operatorname{null}(\mat{A}))=2 \]

Each free variable contributes one independent null-space direction

Rank and Solution Sets

• Rank-nullity counts inputs
• Rank separates existence from uniqueness
• Rank-nullity predicts solution sets

Rank-nullity counts inputs

For an \(m\times n\) matrix \(\mat{A}\),

\[ \operatorname{rank}(\mat{A}) +\operatorname{nullity}(\mat{A})=n \]

Running \(3\times4\) example: two pivots and two free variables

\[ \underbrace{2}_{\text{rank}}+\underbrace{2}_{\text{nullity}} =\underbrace{4}_{\text{input variables}} \]

Interpretation:

• rank counts independent input directions that remain visible
• nullity counts independent input directions sent to zero

Rank-nullity accounts for every direction in the input space

Rank separates existence from uniqueness

For \(\mat{A}\in\reals^{m\times n}\) and \(\mat{A}\vct{x}=\vct{b}\):

Question	Condition
Does this \(\vct{b}\) have a solution?	\(\vct{b}\in\operatorname{col}(\mat{A})\)
Is a solution unique, when it exists?	\(\operatorname{null}(\mat{A})=\{\vct{0}\}\), or rank \(=n\)
Does every \(\vct{b}\in\reals^m\) have a solution?	\(\operatorname{col}(\mat{A})=\reals^m\), or rank \(=m\)

Two inputs produce the same output exactly when

\[ \mat{A}\vct{x}=\mat{A}\vct{x}' \quad\Longleftrightarrow\quad \vct{x}-\vct{x}'\in\operatorname{null}(\mat{A}) \]

For a square matrix, full rank gives existence and uniqueness for every right-hand side

Full column rank gives at most one solution; it does not guarantee existence for every right-hand side in a taller system. Full row rank guarantees every right-hand side is attainable; a wider system can still have many solutions. The square case reconnects these statements to the invertible-matrix results in Inverses. Whenever a solution x_p exists, all solutions are x_p + Null(A); choosing a null-space basis supplies the independent parameters.

Rank-nullity predicts solution sets

\(\exstar\) Compare two \(3\times5\) matrices: \(\operatorname{rank}(\mat{A})=2\) and \(\operatorname{rank}(\mat{B})=3\) Find each nullity Could either system \(\mat{A}\vct{x}=\vct{b}\) or \(\mat{B}\vct{x}=\vct{b}\) have exactly one solution? Which system is solvable for every \(\vct{b}\in\reals^3\)?

Use \(\mat{A}\) on the board and leave \(\mat{B}\) for students. The nullities are \(3\) and \(2\). Neither system can have exactly one solution when consistent, because each null space contains nonzero directions. Only \(\mat{B}\) has rank \(3\) and hence column space \(\reals^3\), so only \(\mat{B}\vct{x}=\vct{b}\) is solvable for every \(\vct{b}\in\reals^3\).

Orthogonal Pairs

• An orthogonal complement contains all perpendicular directions
• The four spaces pair through orthogonality
• Example: see both orthogonal pairs

An orthogonal complement contains all perpendicular directions

For a subspace \(W\subseteq\reals^d\), its orthogonal complement is

\[ W^\perp=\{\vct{z}\in\reals^d:\vct{z}^{\mathsf T}\vct{w}=0 \text{ for every }\vct{w}\in W\} \]

It is enough to check the dot product with each vector in a basis of \(W\)

Every \(\vct{x}\in\reals^d\) has a unique decomposition

\[ \vct{x}=\vct{w}+\vct{z},\qquad \vct{w}\in W,\quad \vct{z}\in W^\perp \]

\(\dim(W)+\dim(W^\perp)=d\): the two subspaces account for every direction

This is Euclidean orthogonality using the dot product from Euclidean Spaces. If z is perpendicular to the basis vectors, linearity makes it perpendicular to every combination. The Euclidean orthogonal-decomposition theorem gives existence. Uniqueness follows because W intersects W^perp only in zero: a vector perpendicular to itself has zero squared length.

The four spaces pair through orthogonality

For \(\mat{A}\in\reals^{m\times n}\) with \(r=\operatorname{rank}(\mat{A})\),

\[ \operatorname{row}(\mat{A})^\perp =\operatorname{null}(\mat{A}) \qquad\text{inside }\reals^n \]

\[ \operatorname{col}(\mat{A})^\perp =\operatorname{null}(\mat{A}^{\mathsf T}) \qquad\text{inside }\reals^m \]

The dimensions balance:

\[ \dim(\operatorname{row}(\mat{A}))=\dim(\operatorname{col}(\mat{A}))=r \]

\[ \dim(\operatorname{null}(\mat{A}))=n-r, \qquad \dim(\operatorname{null}(\mat{A}^{\mathsf T}))=m-r \]

Example: see both orthogonal pairs

Take the rank-one matrix

\[ \mat{A}=\begin{bmatrix}1&2&0\\2&4&0\end{bmatrix} \]

Inside \(\reals^3\):

\[ \operatorname{row}(\mat{A})=\operatorname{span}\{(1,2,0)^{\mathsf T}\}, \qquad \operatorname{null}(\mat{A})=\{(-2s,s,t)^{\mathsf T}:s,t\in\reals\} \]

Their dot product is \(-2s+2s=0\)

Inside \(\reals^2\):

\[ \operatorname{col}(\mat{A})=\operatorname{span}\{(1,2)^{\mathsf T}\}, \qquad \operatorname{null}(\mat{A}^{\mathsf T})=\operatorname{span}\{(-2,1)^{\mathsf T}\} \]

Their dot product is \(-2+2=0\)

Row–null orthogonality and column–left-null orthogonality use different ambient spaces

Orthogonality and ambient spaces

Let

\[ \mat{B}=\begin{bmatrix}1&0&1\\0&1&1\end{bmatrix} \]

\(\exstar\) Find the two orthogonal pairs Find a basis for \(\operatorname{null}(\mat{B})\); check its dot products with both rows What are \(\operatorname{col}(\mat{B})\) and \(\operatorname{null}(\mat{B}^{\mathsf T})\)? Check the dimension sums in \(\reals^3\) and \(\reals^2\) Does it make sense to take the dot product of a row-space vector with a column-space vector here?

The null space has basis {(-1,-1,1)^T}; its dot product with each row is zero. The column space is R^2 and the left null space is {0}. The dimension sums are 2+1=3 and 2+0=2. Row-space vectors have three coordinates and column-space vectors have two, so their dot product is not defined.

Rank Factorization and the Left Null Space

• Rank factorization keeps the independent columns and rows
• Check the factorization column by column
• Find the left null space

Return to our running \(3\times4\) example:

\[ \mat{A}=\begin{bmatrix} 1&2&1&3\\ 2&4&0&4\\ 1&2&2&4 \end{bmatrix} \quad\longrightarrow\quad \operatorname{rref}(\mat{A})=\left[ \begin{array}{cccc} \bbox[3px,border:2px solid teal]{\begin{matrix}1\\0\\0\end{matrix}} &\begin{matrix}2\\0\\0\end{matrix} &\bbox[3px,border:2px solid teal]{\begin{matrix}0\\1\\0\end{matrix}} &\begin{matrix}2\\1\\0\end{matrix} \end{array} \right] \]

The RREF has pivot columns \(1\) and \(3\)

The companion notebook computes the four spaces, compatibility tests, and rank factorization

Rank factorization keeps the independent columns and rows

For the running \(3\times4\) matrix, take original columns \(1,3\) for \(\mat{C}\) and the two nonzero RREF rows for \(\mat{R}\):

\[ \mat{C}= \begin{bmatrix} 1&1\\ 2&0\\ 1&2 \end{bmatrix}, \qquad \mat{R}= \left[ \begin{array}{cccc} \bbox[3px,border:2px solid teal]{\begin{matrix}1\\0\end{matrix}} &\begin{matrix}2\\0\end{matrix} &\bbox[3px,border:2px solid teal]{\begin{matrix}0\\1\end{matrix}} &\begin{matrix}2\\1\end{matrix} \end{array} \right] \]

Row operations preserve column relations; the boxed columns form \(\mat{I}_2\), making \(\mat{R}\) the coefficient matrix:

\[ \mat{A}=\mat{C}\mat{R}, \qquad \operatorname{rref}(\mat{A})=\begin{bmatrix}\mat{R}\\\mat{0}\end{bmatrix} \]

The columns of \(\mat{C}\) form a basis for \(\operatorname{col}(\mat{A})\)

The rows of \(\mat{R}\) form a basis for \(\operatorname{row}(\mat{A})\)

Row operations preserve column relations

Let \(\mat{B}\) be obtained from \(\mat{A}\in\reals^{m\times n}\) by reversible row operations

For any coefficient vector \(\vct{c}\in\reals^n\), the homogeneous solutions stay the same:

\[ \mat{A}\vct{c}=\vct{0} \quad\Longleftrightarrow\quad \mat{B}\vct{c}=\vct{0} \]

Write \(\vct{A}_j,\vct{B}_j\) for column \(j\) of each matrix; matrix–vector multiplication gives

\[ \sum_{j=1}^n c_j\vct{A}_j=\vct{0} \quad\Longleftrightarrow\quad \sum_{j=1}^n c_j\vct{B}_j=\vct{0} \]

Move any column relation to one side to apply this zero test

Every column relation survives elimination with the same coefficients

Each elementary row operation preserves the entire solution set of a homogeneous system, and its inverse gives the reverse implication. Since A c is the sum of c_j A_j, the same coefficients c describe exactly the same zero relations in A and B. In particular, move A_j minus a proposed combination of selected columns to the left. The corresponding relation is zero before reduction if and only if it is zero after reduction. This argument applies even when column j is one of the selected columns.

Why the rank-factorization recipe works

For \(\mat{A}\in\reals^{m\times n}\) of rank \(r\), let \(\mat{B}=\operatorname{rref}(\mat{A})\)

List pivot indices left to right: \(p_1<\cdots<p_r\)

Use \(\mat{C}=[\vct{A}_{p_1}\ \cdots\ \vct{A}_{p_r}]\in\reals^{m\times r}\); collect the nonzero rows in \(\mat{R}\in\reals^{r\times n}\)

Let \(\vct{d}_j\in\reals^r\) be column \(j\) of \(\mat{R}\); \(\mat{B}\) has \(m-r\) trailing zero rows

RREF pivot columns have a single \(1\), so

\[ \vct{B}_j=\begin{bmatrix}\vct{d}_j\\\vct{0}\end{bmatrix} =\sum_{\ell=1}^r (\vct{d}_j)_\ell\vct{B}_{p_\ell} \]

Undo the row operations, preserving those coefficients:

\[ \vct{A}_j=\sum_{\ell=1}^r(\vct{d}_j)_\ell\vct{A}_{p_\ell} =\mat{C}\vct{d}_j \]

This holds for every column \(j\), so \(\mat{A}=\mat{C}\mat{R}\)

B contains the r nonzero rows of R, followed by m-r zero rows. Its selected pivot column p_l is [e_l;0], where e_l is the lth standard basis vector in R^r. Thus any reduced column is the combination of these pivot columns with coefficients given by its top r entries. The preceding slide proves that reversing row reduction restores A’s columns with those same coefficients. Collect d_j as the columns of R to obtain A=C R.

C has r independent columns because their reduced versions are independent unit columns and row reduction preserves dependence relations. The r rows of R are independent because their pivot-column block is I_r. Row operations preserve the row space, so those rows span Row(A). Consequently Col(C)=Col(A), Row(R)=Row(A), and both factors have rank r.

An equivalent matrix proof uses the invertible row-operation product M: M A=B=rref(A)=[R;0] and M C=[I_r;0]. Then M(C R)=[I_r;0] R=[R;0]=M A. Multiply by M^{-1}. For rank zero A is the zero matrix; the empty-column C and empty-row R give the zero product. If m=r, the displayed bottom zero block is empty.

Echelon form finds bases; RREF supplies these coefficients

For our running \(3\times4\) matrix \(\mat{A}\), an echelon form with unit pivots is

\[ \mat{E}=\begin{bmatrix} 1&2&1&3\\ 0&0&1&1\\ 0&0&0&0 \end{bmatrix} \xrightarrow{R_1\leftarrow R_1-R_2} \operatorname{rref}(\mat{A})=\left[ \begin{array}{cccc} \bbox[3px,border:2px solid teal]{\begin{matrix}1\\0\\0\end{matrix}} &\begin{matrix}2\\0\\0\end{matrix} &\bbox[3px,border:2px solid teal]{\begin{matrix}0\\1\\0\end{matrix}} &\begin{matrix}2\\1\\0\end{matrix} \end{array} \right] \]

• Either form identifies original columns \(1,3\) as a column-space basis
• The nonzero rows of either form give a row-space basis
• Remove the zero row from the boxed pivot columns to obtain \(\mat{I}_2\) in \(\mat{R}\)

For the direct formula \(\mat{A}=\mat{C}\mat{R}\), use the nonzero RREF rows

Echelon form is enough for both pivot-column extraction and a nonzero-row basis. Back substitution also finds null-space bases. RREF is not necessary for the existence of rank factorization. It is necessary for the particular shortcut used here: choose C as the original pivot columns and take R directly from the reduced rows. E’s pivot columns in its first two rows form the matrix [[1,1],[0,1]], whereas R’s form I_2. Thus the nonzero rows E_r are not the coefficient matrix for these selected original columns. One can instead use E_p^{-1} E_r, where E_p is that invertible pivot-column block; clearing above the pivots performs exactly that normalization here.

Check the factorization column by column

Recall the independent columns and their coefficient matrix:

\[ \mat{C}=[\vct{C}_1\ \vct{C}_2] =\begin{bmatrix}1&1\\2&0\\1&2\end{bmatrix}, \qquad \mat{R}=\left[ \begin{array}{cccc} \bbox[3px,border:2px solid teal]{\begin{matrix}1\\0\end{matrix}} &\begin{matrix}2\\0\end{matrix} &\bbox[3px,border:2px solid teal]{\begin{matrix}0\\1\end{matrix}} &\begin{matrix}2\\1\end{matrix} \end{array} \right] \]

The columns of \(\mat{R}\) give the coefficients, just as in the matrix product by columns in Systems and Matrices:

\[ \begin{aligned} \vct{A}_1&=\vct{C}_1 & \vct{A}_2&=2\vct{C}_1\\ \vct{A}_3&=\vct{C}_2 & \vct{A}_4&=2\vct{C}_1+\vct{C}_2 \end{aligned} \]

Thus \(\mat{C}\mat{R}\) rebuilds all four columns of \(\mat{A}\)

For the ordered column-space basis \(\mathcal{C}=\{\vct{C}_1,\vct{C}_2\}\),

\[ [\mat{A}\vct{x}]_{\mathcal{C}}=\mat{R}\vct{x} \]

\(\mat{R}\vct{x}\) gives the coordinates; \(\mat{C}\) rebuilds the output

Construct rank factorizations

\(\exstar\) For each matrix \(\mat{M}\) below, find \(\mat{C}\) and \(\mat{R}\) with \(\mat{M}=\mat{C}\mat{R}\) \(\displaystyle\mat{B}=\begin{bmatrix}1&2&0\\0&0&1\\1&2&1\end{bmatrix}\) \(\displaystyle\mat{D}=\begin{bmatrix}1&0&2&3\\2&1&4&7\end{bmatrix}\)

For each factorization, identify bases for the row and column spaces

For \(\mat{B}\), use columns \(1\) and \(3\): \(\mat{C}=\begin{bmatrix}1&0\\0&1\\1&1\end{bmatrix}\) and \(\mat{R}=\begin{bmatrix}1&2&0\\0&0&1\end{bmatrix}\). For \(\mat{D}\), use columns \(1\) and \(2\): \(\mat{C}=\begin{bmatrix}1&0\\2&1\end{bmatrix}\) and \(\mat{R}=\begin{bmatrix}1&0&2&3\\0&1&0&1\end{bmatrix}\). The columns of each \(\mat{C}\) and rows of each \(\mat{R}\) give the requested bases.

Find the left null space

For the running \(3\times4\) matrix, the column-space basis is stored in

\[ \mat{C}=\begin{bmatrix}1&1\\2&0\\1&2\end{bmatrix} \]

Every column of \(\mat{A}\) is a combination of these two columns

Thus \(\mat{A}^{\mathsf T}\vct{y}=\vct{0}\iff\mat{C}^{\mathsf T}\vct{y}=\vct{0}\):

\[ y_1+2y_2+y_3=0, \qquad y_1+2y_3=0 \]

Taking \(y_3=2\) gives \(\vct{y}=(-4,1,2)^{\mathsf T}\)

\[ \operatorname{null}(\mat{A}^{\mathsf T}) =\operatorname{span}\{(-4,1,2)^{\mathsf T}\} \]

This vector is perpendicular to both columns of \(\mat{C}\)

Left-null vectors test compatibility of equations

If \(\mat{A}^{\mathsf T}\vct{y}=\vct{0}\) and \(\mat{A}\vct{x}=\vct{b}\), then

\[ \vct{y}^{\mathsf T}\vct{b} =\vct{y}^{\mathsf T}\mat{A}\vct{x}=0 \]

A right-hand side is attainable exactly when it is perpendicular to every left-null vector

In our example, \(\operatorname{null}(\mat{A}^{\mathsf T})\) has basis \(\{(-4,1,2)^{\mathsf T}\}\), so consistency is equivalent to

\[ -4b_1+b_2+2b_3=0 \]

\(\exstar\) Use this test for \(\vct{b}=(0,1,0)^{\mathsf T}\) and \(\vct{b}=(1,2,1)^{\mathsf T}\)

For (0,1,0)^T, y^T b=1, so there is no solution. For (1,2,1)^T, the condition holds, and b is the first column of A, so x=(1,0,0,0)^T solves it. The sufficiency comes from Col(A)=Null(A^T)perp. The same y also describes a linear dependence among the equations; a consistent right-hand side must obey that relation. This is a concrete solvability certificate, not just another dimension count.

One matrix supports several basis questions

Return to the rank-factorization exercise:

\[ \mat{B}=\begin{bmatrix}1&2&0\\0&0&1\\1&2&1\end{bmatrix}, \qquad \mat{C}=\begin{bmatrix}1&0\\0&1\\1&1\end{bmatrix} \quad\text{(original columns $1,3$)} \]

\(\exstar\) Complete the picture of its four fundamental subspaces Find bases for \(\operatorname{null}(\mat{B})\) and \(\operatorname{null}(\mat{B}^{\mathsf T})\) Check that your left-null vector is perpendicular to both columns of \(\mat{C}\) Compare the four dimensions with rank–nullity

For \(\mat{B}\), a null-space basis is \(\{(-2,1,0)^{\mathsf T}\}\), and a left-null basis is \(\{(-1,-1,1)^{\mathsf T}\}\). Each is orthogonal to the appropriate row or column basis. The dimensions are \(2,1,2,1\) for column, null, row, and left null, respectively.

What the Structure Buys Us

Preview: Anton §§8.1 and 8.4

• A basis turns an abstract problem into a matrix problem
• Example: differentiation is a matrix
• A well-chosen basis can simplify an operator

A basis turns an abstract problem into a matrix problem

Let \(V\) and \(W\) be finite-dimensional real vector spaces

A linear transformation \(T:V\to W\) preserves linear combinations:

\[ T(c\vct{u}+d\vct{v})=cT(\vct{u})+dT(\vct{v}) \]

Choose ordered bases \(\mathcal{B}=\{\vct{b}_1,\ldots,\vct{b}_n\}\) for the input \(V\) and \(\mathcal{C}\) for the output \(W\): \(\dim(V)=n\), \(\dim(W)=m\)

Coordinates and matrix columns use the basis vectors in the listed order

Recall coordinates: \([\vct{v}]_{\mathcal{B}}=(c_1,\ldots,c_n)^{\mathsf T}\) means \(\vct{v}=\sum_{j=1}^n c_j\vct{b}_j\)

The \(m\times n\) coordinate matrix \(\mat{M}\) has column \(j\) equal to \([T(\vct{b}_j)]_{\mathcal{C}}\) and satisfies

\[ [T(\vct{v})]_{\mathcal{C}} =\mat{M}[\vct{v}]_{\mathcal{B}} \]

A linear transformation is determined by its action on a basis

Example: differentiation is a matrix

Let \(D:\mathbb{P}_2\to\mathbb{P}_1\) send \(p\) to \(p'\)

Use \(\mathcal{B}=\{1,t,t^2\}\) for the input and \(\mathcal{C}=\{1,t\}\) for the output

\[ D(1)=0, \qquad D(t)=1, \qquad D(t^2)=2t \]

Their output coordinate vectors become the columns:

\[ \mat{M}=\begin{bmatrix}0&1&0\\0&0&2\end{bmatrix}, \qquad [D(p)]_{\mathcal{C}}=\mat{M}[p]_{\mathcal{B}} \]

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

\[ \mat{M}\begin{bmatrix}2\\-3\\5\end{bmatrix} =\begin{bmatrix}-3\\10\end{bmatrix} \quad\Longleftrightarrow\quad p'=-3+10t \]

Apply the map to each input basis vector; put its output coordinates in a column

Evaluate a polynomial with a matrix

Let \(T:\mathbb{P}_2\to\reals^2\) be

\[ T(p)=\begin{bmatrix}p(0)\\p(1)\end{bmatrix} \]

Use \(\mathcal{B}=\{1,t,t^2\}\) and standard output coordinates

\(\exstar\) Construct and interpret the matrix of \(T\) Find \(T(1)\), \(T(t)\), and \(T(t^2)\); make them the columns of \(\mat{M}\) Use \(\mat{M}\) to evaluate \(p=2-3t+5t^2\) at both points Find a nonzero polynomial sent to zero; explain why two values cannot identify every quadratic

The columns are (1,1)^T, (0,1)^T, (0,1)^T, so M=[[1,0,0],[1,1,1]]. For p=2-3t+5t^2, the output is (2,4)^T. Its null space consists of coefficient vectors (0,-s,s)^T: the corresponding polynomials are s(t^2-t). The rank is two and nullity one. Adding any multiple of t^2-t leaves both sampled values unchanged. T preserves linear combinations by pointwise evaluation.

A well-chosen basis can simplify an operator

For \(\mat{A}:\reals^n\to\reals^n\), use the same basis for input and output

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

\[ \mat{P}=[\vct{b}_1\ \cdots\ \vct{b}_n], \qquad \vct{x}=\mat{P}[\vct{x}]_{\mathcal{B}} \]

The matrix in \(\mathcal{B}\) coordinates is \(\mat{P}^{-1}\mat{A}\mat{P}\)

Eigenvalues asks whether a basis can make it diagonal:

\[ \mat{P}^{-1}\mat{A}\mat{P} = \begin{bmatrix} \lambda_1&&0\\ &\ddots&\\ 0&&\lambda_n \end{bmatrix} \]

Each basis vector must satisfy \(\mat{A}\vct{b}_j=\lambda_j\vct{b}_j\): its span is preserved

Eigenvectors are basis candidates adapted to the action of a matrix

Big Ideas

• Reduction reveals bases for row, column, and null spaces
• Rank separates attainable outputs from lost input directions
• Row/null and column/left-null spaces form orthogonal pairs
• Rank factorization separates independent columns from their combinations
• Left-null vectors give compatibility conditions for the equations
• Bases turn finite-dimensional linear maps into matrices

How Far We Have Come

• Systems and Matrices — reduce \([\,\mat{A}\mid\vct{b}\,]\) to determine consistency, pivots, and free parameters
• Matrix actions — PLU reuses elimination; products compose, inverses undo; determinants detect collapse
• Geometry and bases — dot products, projection, and 3D cross products describe geometry; subspaces admit linear combinations and bases give unique coordinates

• Rank factorization — \(\mat{A}=\mat{C}\mat{R}\); \(\mat{C}\)’s columns and \(\mat{R}\)’s rows are column/row-space bases; each has \(r=\operatorname{rank}(\mat{A})\) members
• Four spaces — column space gives attainable outputs; null space gives lost inputs; row/null and column/left-null are orthogonal pairs
• Compatibility — consistent iff \(\vct{b}\in\operatorname{col}(\mat{A})\), equivalently \(\vct{y}^{\mathsf T}\vct{b}=0\) for every \(\vct{y}\in\operatorname{null}(\mat{A}^{\mathsf T})\)

For consistent \(\mat{A}\vct{x}=\vct{b}\) with \(\mat{A}\in\reals^{m\times n}\): all solutions are \(\vct{x}=\vct{x}_p+\mat{N}\vct{t}\), \(\vct{t}\in\reals^{n-r}\); \(\vct{x}_p\) is one solution and \(\mat{N}\)’s columns are a null-space basis

This is a cumulative synthesis, rather than a chronology of deck titles. Retain the original question Ax=b, the reversible computations and reusable factors, the geometric meaning of actions, and the progressively richer descriptions of vectors. N’s columns are a basis of Null(A), so t supplies exactly n-r independent free parameters. A column-space basis C describes which right-hand sides are attainable; a null-space basis N describes all inputs that give the same attainable right-hand side. The condition can also be tested by y^T b=0 for every left-null vector y. Full row rank r=m makes every b attainable, and full column rank r=n gives uniqueness when a solution exists. Projections, point differences, and cross-product normals and areas retain the Euclidean geometric tools from the earlier deck.

What Comes Next

Which directions does a matrix preserve, and when can eigenvectors form a basis?

Next: Eigenvalues and Eigenvectors