Vector Spaces,
Subspaces, and Span
Anton, Rorres, & Kaul §§4.1–4.3
Quiz 3 on 10/15
October 9, 2026
As Systems and Matrices showed, elimination reveals free variables in a consistent system \(\mat{A}\vct{x}=\vct{b}\)
Choose one particular solution \(\vct{x}_p\), so \(\mat{A}\vct{x}_p=\vct{b}\)
As Euclidean Spaces showed, all solutions have the form
\[ \{\text{all solutions}\} =\vct{x}_p+\{\vct{z}:\mat{A}\vct{z}=\vct{0}\} \]
The homogeneous solutions can be
This deck names the abstract structure shared by collections with these linear operations
A real vector space is a nonempty set \(V\) whose elements are called vectors, with addition and multiplication by elements of \(\reals\), called scalars:
\[ \vct{u}+\vct{v}\in V, \qquad c\vct{v}\in V \qquad (\vct{u},\vct{v}\in V,\ c\in\reals) \]
These operations obey the vector-space axioms on the next two slides
We specify the set and operations so polynomials, functions, and other objects can be vectors, not only columns in \(\reals^n\)
For all \(\vct{u},\vct{v},\vct{w}\in V\),
\[ \begin{aligned} \vct{u}+\vct{v}&\in V &&\text{closure}\\ \vct{u}+\vct{v}&=\vct{v}+\vct{u} &&\text{commutativity}\\ (\vct{u}+\vct{v})+\vct{w}&=\vct{u}+(\vct{v}+\vct{w}) &&\text{associativity}\\ \vct{v}+\vct{0}&=\vct{v} &&\text{zero vector}\\ \vct{v}+(-\vct{v})&=\vct{0} &&\text{additive inverse} \end{aligned} \]
The symbol \(\vct{0}\) means the zero element of this particular space
For all \(c,d\in\reals\) and \(\vct{u},\vct{v}\in V\),
\[ \begin{aligned} c\vct{v}&\in V &&\text{closure}\\ c(\vct{u}+\vct{v})&=c\vct{u}+c\vct{v}\\ (c+d)\vct{v}&=c\vct{v}+d\vct{v}\\ c(d\vct{v})&=(cd)\vct{v}\\ 1\vct{v}&=\vct{v} \end{aligned} \]
Linear algebra uses only addition, scalar multiplication, and consequences of these rules
Every vector space satisfies
\[ 0\vct{v}=\vct{0}, \qquad c\vct{0}=\vct{0}, \qquad (-1)\vct{v}=-\vct{v} \]
Let \(\vct{w}=0\vct{v}\). Since \(0+0=0\) for scalars, distributivity gives
\[ \vct{w}=(0+0)\vct{v}=\vct{w}+\vct{w} \]
The additive-inverse axiom supplies \(-\vct{w}\). Adding it to both sides gives \(\vct{0}=\vct{w}=0\vct{v}\)
Once the axioms hold, familiar algebraic manipulations work for every kind of vector
Addition and scalar multiplication are defined object by object
Different objects can obey the same linear rules
Most of this course uses real scalars, but the same axioms work over any field
In the two-element field
\[ \mathbb{F}_2=\{0,1\}, \qquad 1+1=0 \]
Addition in \(\mathbb{F}_2\) is exclusive OR (XOR) on bits, familiar from computer science
A vector in \(\mathbb{F}_2^n\) is a binary string; vector addition applies XOR coordinatewise
For example,
\[ [1,0,1,1]^{\mathsf T}+[1,1,0,1]^{\mathsf T}=[0,1,1,0]^{\mathsf T} \]
Changing the scalar field changes which linear combinations are possible
An additive inverse adds to give \(\vct{0}\); in \(\mathbb{F}_2\), \(1+1=0\)
\(\exstar\) Find the additive inverse of \(\vct{u}=[1,0,1,1]^{\mathsf T}\) in \(\mathbb{F}_2^4\) Repeat for \(\vct{w}=[0,1,1,0]^{\mathsf T}\) Does the answer generalize to every vector in \(\mathbb{F}_2^n\)? Explain using XOR
Write \(\mathbb{P}_k\) for real polynomials in one variable \(t\) of degree at most \(k\), including the zero polynomial
For example,
\[ \mathbb{P}_2=\{a_0+a_1t+a_2t^2:a_0,a_1,a_2\in\reals\} \]
For \(p,q\in\mathbb{P}_2\) and \(c\in\reals\), use ordinary polynomial operations:
\[ (p+q)(t)=p(t)+q(t), \qquad (cp)(t)=c\,p(t) \]
The zero vector is the zero polynomial
\[ 0(t)=0 \]
A polynomial is one vector; its coefficients provide one possible coordinate description
For integers \(m\ge 0\) and \(n\ge 1\), \(\mathbb{P}_m(\reals^n)\) contains real polynomials in \(x_1,\ldots,x_n\) of total degree at most \(m\), including the zero polynomial
A monomial’s total degree is the sum of its exponents:
\[ x_1^{\alpha_1}\cdots x_n^{\alpha_n} \quad\text{has total degree}\quad \alpha_1+\cdots+\alpha_n, \qquad \alpha_j\in\{0,1,2,\ldots\} \]
A nonzero polynomial’s total degree is the largest total degree of its monomials with nonzero coefficients
\[ \mathbb{P}_2(\reals^2) =\{a+bx+cy+dx^2+exy+fy^2:a,b,c,d,e,f\in\reals\} \]
\(x^2y\in\mathbb{P}_3(\reals^2)\), but \(x^2y^2\notin\mathbb{P}_3(\reals^2)\)
\(m\) bounds the total degree; \(n\) counts the variables
Our one-variable notation is \(\mathbb{P}_m=\mathbb{P}_m(\reals)\)
In \(\mathbb{P}_2(\reals^2)\), take
\[ p(x,y)=1+x+xy, \qquad q(x,y)=2-y+x^2 \]
Combine matching monomials:
\[\begin{align*} 2p-q &=2(1+x+xy)-(2-y+x^2)\\ &=2x+y-x^2+2xy \end{align*}\]
Every remaining monomial has total degree at most two
Addition and scalar multiplication change coefficients; they do not introduce new monomials
\(\exstar\) Work in \(\mathbb{P}_2(\reals^2)\): each monomial’s exponents sum to at most two Which belong: \(x^2+xy+y^2\), \(x^2y\), \(7\), and \(0\)? For \(p=1+x+xy\) and \(q=2-y+x^2\), compute \(p+3q\) Must the product of two members stay in this space? Give an example
Let \(C[a,b]\) contain all continuous real-valued functions on \([a,b]\)
\[ (f+g)(t)=f(t)+g(t), \qquad (cf)(t)=c\,f(t) \]
The vector-space operations happen pointwise
\[ (2f-3g)(t)=2f(t)-3g(t) \]
No finite list of functions can generate all of \(C[a,b]\) by linear combinations, for \(a<b\)
A function is one vector, even though its values vary over an entire interval
The solutions of \(f''+f=0\) on \(\reals\) are functions
\[ f(t)=a\cos t+b\sin t, \qquad a,b\in\reals \]
For solutions \(f,g\) and real constants \(c,d\),
\[ (cf+dg)''+(cf+dg)=c(f''+f)+d(g''+g)=0 \]
The same closure holds for vector-valued solutions of \(\vct{y}'=\mat{A}\vct{y}\) with a fixed matrix \(\mat{A}\)
Solutions of a homogeneous linear DE or system form a vector space of functions
\(\exstar\) Test whether linear combinations of solutions are still solutions For \(h''-h=0\), verify that \(e^t\) and \(e^{-t}\) solve it and that every \(ae^t+be^{-t}\) does too For \(\mat{A}=\operatorname{diag}(1,-1)\), verify that \([e^t,0]^{\mathsf T}\) and \([0,e^{-t}]^{\mathsf T}\) solve \(\vct{y}'=\mat{A}\vct{y}\) and so does every linear combination
For \(h''-h=1\), the constant function \(h_p=-1\) is one solution
The solutions of \(z''-z=0\) are \(z=ae^t+be^{-t}\), so every solution is
\[ h(t)=-1+ae^t+be^{-t}, \qquad a,b\in\reals \]
The zero function does not solve \(h''-h=1\)
When a nonhomogeneous linear DE has a solution, its solution set is a particular solution plus the homogeneous solution space
Use one particular solution plus the homogeneous solutions
\(\exstar\) Find one particular solution and describe the full solution set \(g''+g=2\) \(y'-y=3\)
The corresponding homogeneous solutions are \(a\cos t+b\sin t\) and \(ce^t\), respectively
For each set, explain why it is not a vector space under the usual operations
Use \(\sqrt{-1}\) for the imaginary unit, leaving \(i\) available as an index
On \([0,1]\), use the Fourier modes
\[ \phi_k(t)=e^{2\pi\sqrt{-1}\,kt}, \qquad k\in\mathbb{Z}, \qquad \phi_k(t+1)=\phi_k(t) \]
Start with their finite sums over complex scalars:
\[ \mathcal{T} =\left\{\sum_{k=-N}^{N}c_k\phi_k:N=0,1,2,\ldots,\ c_k\in\mathbb{C}\right\} \]
Each vector in \(\mathcal{T}\) is a complex-valued function on \([0,1]\) with a 1-periodic extension
Adding two finite sums or scaling by a complex number stays in \(\mathcal{T}\)
Complex exponentials are vectors; finite Fourier sums form a vector space
Under ordinary operations, the zero vector is \([0,0]^{\mathsf T}\)
\[ L=\left\{ \begin{bmatrix}1\\2\end{bmatrix} +t\begin{bmatrix}3\\1\end{bmatrix}:t\in\reals \right\} \]
This line does not contain \(\vct{0}\), so it cannot be a vector space under the usual operations
Its direction set
\[ W=\left\{t\begin{bmatrix}3\\1\end{bmatrix}:t\in\reals\right\} \]
is a vector space
An affine set is a point plus a direction subspace
It is itself a subspace exactly when it contains the origin
On \(V=\{(x,y):x,y\in\reals\}\), suppose
\[ (x,y)\mathbin{\boxplus}(u,v)=(x+u-1,y+v-1) \]
The usual pair \((0,0)\) is not the zero vector for this addition
\(\exstar\) Find the element \(\vct{z}\) satisfying \((x,y)\mathbin{\boxplus}\vct{z}=(x,y)\) for every \((x,y)\) Why must scalar multiplication also be specified before deciding whether these operations define a vector space?
The companion notebook explores polynomial constraints, closure, and affine sets
A subset \(W\subseteq V\) is a subspace when \(W\) is itself a vector space using the addition and scalar multiplication already defined on \(V\)
Examples in \(\reals^n\) for \(n\ge 2\):
Examples in \(\mathbb{P}_3\) (polynomials in \(t\) of degree at most three):
A subspace uses the same operations as its ambient space
A nonempty subset \(W\subseteq V\) is a subspace if
\[ c\vct{u}+d\vct{v}\in W \]
for every \(\vct{u},\vct{v}\in W\) and every \(c,d\in\reals\)
This one condition gives
Subspaces are exactly the subsets closed under linear combinations
For each subset of \(\reals^2\), ask first whether \(\vct{0}\) belongs
\[ W_1=\left\{\begin{bmatrix}x\\y\end{bmatrix}:x+2y=0\right\} \]
\[ W_2=\left\{\begin{bmatrix}x\\y\end{bmatrix}:x+2y=1\right\} \]
\(W_1\) passes the zero test and is closed under linear combinations
\(W_2\) fails immediately because \(\vct{0}\notin W_2\)
Homogeneous linear conditions define subspaces; shifted conditions usually define affine sets
The union of the coordinate axes contains the origin
\[ W=\left\{\begin{bmatrix}x\\y\end{bmatrix}:xy=0\right\} \]
Every scalar multiple of a vector in \(W\) stays on its axis
\(\exstar\) Does \(W\) form a subspace of \(\reals^2\)? Choose one nonzero vector on each axis Does their sum belong to \(W\)?
The zero-vector test can reject a set; closure is still needed to accept it
Recall the matrix families in Matrix Transformations
Use ordinary matrix addition and real scalar multiplication in \(\reals^{n\times n}\)
The upper-triangular matrices satisfy
\[ U=\{\mat{A}\in\reals^{n\times n}:a_{ij}=0\text{ whenever }i>j\} \]
For \(n=3\), their form is
\[ \begin{bmatrix}a&b&c\\0&d&e\\0&0&f\end{bmatrix}, \qquad a,b,c,d,e,f\in\reals \]
The zero matrix belongs to \(U\); for \(\mat{A},\mat{B}\in U\) and \(r,s\in\reals\),
\[ (r\mat{A}+s\mat{B})_{ij}=ra_{ij}+sb_{ij}=0\qquad(i>j) \]
Upper-triangular matrices form a vector space: linear combinations preserve the zeros below the diagonal
For fixed \(n\ge1\), a permutation matrix has exactly one \(1\) in each row and column and \(0\) everywhere else
In \(\reals^{2\times2}\), the only permutation matrices are
\[ \mat{I}=\begin{bmatrix}1&0\\0&1\end{bmatrix}, \qquad \mat{P}=\begin{bmatrix}0&1\\1&0\end{bmatrix} \]
The zero matrix is not a permutation matrix
Addition and scalar multiplication also fail closure:
\[ \mat{I}+\mat{P}=\begin{bmatrix}1&1\\1&1\end{bmatrix}, \qquad 2\mat{I}=\begin{bmatrix}2&0\\0&2\end{bmatrix} \]
Permutation matrices are closed under matrix multiplication, but fail the vector-space tests for addition and scalar multiplication
Use ordinary addition and real scalar multiplication in \(\reals^{n\times n}\), with fixed \(n\ge2\)
\(\exstar\) Decide whether each family is a subspace Diagonal matrices: \(a_{ij}=0\) when \(i\ne j\) Lower-triangular matrices: \(a_{ij}=0\) when \(i<j\) Symmetric matrices: \(\mat{A}^{\mathsf T}=\mat{A}\) Tridiagonal matrices: \(a_{ij}=0\) when \(|i-j|>1\) Invertible matrices: a matrix inverse exists Orthogonal matrices: \(\mat{A}^{\mathsf T}\mat{A}=\mat{I}\)
For each yes, prove closure under linear combinations; for each no, give a failed test or counterexample
The null space of an \(m\times n\) matrix \(\mat{A}\) is
\[ \operatorname{null}(\mat{A}) =\{\vct{x}\in\reals^n:\mat{A}\vct{x}=\vct{0}\} \]
If \(\vct{x},\vct{y}\in\operatorname{null}(\mat{A})\), then
\[ \mat{A}(c\vct{x}+d\vct{y}) =c\mat{A}\vct{x}+d\mat{A}\vct{y} =\vct{0} \]
If \(\mat{A}\vct{x}_p=\vct{b}\), the full solution set is \(\vct{x}_p+\operatorname{null}(\mat{A})\)
For \(\vct{b}\ne\vct{0}\), this affine set does not contain the zero vector
This gives a name to the homogeneous solutions found by elimination in Systems and Matrices
The null space records every input direction that a matrix sends to zero
For vectors \(\vct{v}_1,\ldots,\vct{v}_k\) in a real vector space \(V\),
\[ \operatorname{span}\{\vct{v}_1,\ldots,\vct{v}_k\} =\left\{c_1\vct{v}_1+\cdots+c_k\vct{v}_k:c_i\in\reals\right\} \]
The span
Span answers a reachability question: which vectors can these generators build?
For \(\vct{v}_j\in\reals^n\), collect the chosen vectors as columns:
\[ \mat{V}=\begin{bmatrix}\vct{v}_1&\cdots&\vct{v}_k\end{bmatrix}\in\reals^{n\times k}, \qquad \vct{c}=\begin{bmatrix}c_1&\cdots&c_k\end{bmatrix}^{\mathsf T}\in\reals^k \]
\[ \operatorname{span}\{\vct{v}_1,\ldots,\vct{v}_k\} =\{\mat{V}\vct{c}:\vct{c}\in\reals^k\}, \qquad \mat{V}\vct{c}=\sum_{j=1}^k c_j\vct{v}_j \]
This is matrix–vector multiplication in Systems and Matrices
The span is the set of all outputs \(\mat{V}\vct{c}\) as the coefficient vector \(\vct{c}\) varies
Take \(\vct{v}_1=[1,0,1]^{\mathsf T}\) and \(\vct{v}_2=[0,1,1]^{\mathsf T}\)
\[ \mat{V}\vct{c} =\begin{bmatrix}1&0\\0&1\\1&1\end{bmatrix} \begin{bmatrix}c_1\\c_2\end{bmatrix} =\begin{bmatrix}c_1\\c_2\\c_1+c_2\end{bmatrix} \]
Thus
\[ \operatorname{span}\{\vct{v}_1,\vct{v}_2\} =\{[x,y,z]^{\mathsf T}:z=x+y\} \]
For example, \([2,-1,1]^{\mathsf T}=2\vct{v}_1-\vct{v}_2\) belongs; \([0,0,1]^{\mathsf T}\) does not
A spanning description and an equation can describe the same subspace
Keep \(\vct{v}_1=[1,0,1]^{\mathsf T}\) and \(\vct{v}_2=[0,1,1]^{\mathsf T}\); let \(\vct{v}_3=[1,1,2]^{\mathsf T}\)
\(\exstar\) Use coefficients or the plane equation \(z=x+y\) Is \([3,2,5]^{\mathsf T}\) in the span? If so, find coefficients Is \([3,2,4]^{\mathsf T}\) in the span? If so, find coefficients Does adding \(\vct{v}_3\) enlarge the span? Explain Find two different coefficient lists in \(\vct{v}_3=c_1\vct{v}_1+c_2\vct{v}_2+c_3\vct{v}_3\); zero coefficients are allowed
Let
\[ p_1(t)=1+t, \qquad p_2(t)=t+t^2, \qquad p_3(t)=1-t^2 \]
To span \(\mathbb{P}_2\), the equation must be solvable for every real \(a_0,a_1,a_2\)
\[ c_1p_1+c_2p_2+c_3p_3=a_0+a_1t+a_2t^2 \]
Matching coefficients gives \(\mat{A}\vct{c}=\vct{a}\):
\[ \underbrace{\begin{bmatrix} 1&0&1\\ 1&1&0\\ 0&1&-1 \end{bmatrix}}_{\mat{A}} \begin{bmatrix}c_1\\c_2\\c_3\end{bmatrix} = \begin{bmatrix}a_0\\a_1\\a_2\end{bmatrix} \]
Keep \(p_1=1+t\), \(p_2=t+t^2\), and \(p_3=1-t^2\)
The columns of \(\mat{A}\) collect their coefficients of \(1,t,t^2\), so
\[ \vct{A}_3=\vct{A}_1-\vct{A}_2 \quad\implies\quad \det(\mat{A})=0 \]
Thus \(\mat{A}\) is singular: \(\mat{A}\vct{c}=\vct{a}\) cannot be solved for every coefficient vector \(\vct{a}\in\reals^3\)
The same relation holds between the polynomials:
\[ p_3=p_1-p_2 \]
In every resulting \(a_0+a_1t+a_2t^2\), \(a_1=a_0+a_2\)
The constant polynomial \(1\) has \((a_0,a_1,a_2)=(1,0,0)\) and fails this condition
These three polynomials do not span \(\mathbb{P}_2\): their singular coefficient matrix cannot produce every coefficient vector
Keep \(p_1=1+t\) and \(p_2=t+t^2\), and let \(W=\operatorname{span}\{p_1,p_2\}\)
\(\exstar\) Test membership by solving \(c_1p_1+c_2p_2=p\) for real \(c_1,c_2\) Write \(2+3t+t^2\) as a combination of \(p_1,p_2\), if possible Decide whether \(1\) belongs to \(W\) Give a coefficient test for \(a_0+a_1t+a_2t^2\) to belong to \(W\)
In \(\mathbb{P}_3\) (degree at most three), consider \(U=\{p:p(0)=0\}\)
The zero polynomial lies in \(U\); for \(p,q\in U\) and \(c,d\in\reals\),
\[ (cp+dq)(0)=cp(0)+dq(0)=0 \]
Also, \(p(t)=a_1t+a_2t^2+a_3t^3\), so
\[ U=\operatorname{span}\{t,t^2,t^3\} \]
Use the defining condition to prove closure, then use coefficients to find a spanning set
Let
\[ W=\{p\in\mathbb{P}_3:p(1)=0\} \]
\(\exstar\) Show that \(W\) is a subspace of \(\mathbb{P}_3\) Start with \(p,q\in W\) and \(c,d\in\reals\) Evaluate \((cp+dq)(1)\) Find three polynomials that span \(W\) Can fewer than three polynomials span \(W\)? Why?
Use continuous real-valued functions on \([0,1]\), with pointwise addition and real scalar multiplication
\(\exstar\) Which sets are subspaces of \(C[0,1]\)? \(W_0=\{f:f(0)=0\}\) \(W_1=\{f:f(0)=1\}\) \(W_2=\{f:f(0)=f(1)\}\)
Check the zero function and closure under \(cf+dg\)
For \(\vct{v}_j\in\reals^n\) and \(\mat{V}=[\vct{v}_1\ \cdots\ \vct{v}_k]\): \(\vct{b}\) is in their span iff the coefficient system \(\mat{V}\vct{c}=\vct{b}\) is consistent
How can we remove redundant generators and describe each vector uniquely?
First see how spans, bases, and coordinates connect a transformed square, polynomial differentiation, regression, and PCA
© 2026 Fred J. Hickernell · Illinois Tech · assisted by ChatGPT and Codex · Vector Spaces and Span · MATH 332 — Fall 2026 Website · \(\exstar\) = exercise