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Wize University Linear Algebra Textbook > Vector Spaces and Subspaces

Column Space and Null Space (Range and Kernel)

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Column Space and Null Space

Given a matrix Am×nA_{m \times n}, there are some special subspaces we can define.

Column Space

The column space (or range, or image) of AA is the set of all possible vectors b⃗\vec b that result from multiplying AA by any vector x⃗∈Rn\vec x \in \reals^n.
Col(A)={Ax⃗  ∣  x⃗∈Rn }\boxed{\quad {\rm Col}(A) = \{ A\vec x \ \ |\ \ \vec x \in \reals^n\ \} \quad}
Why is it called the column space?
Think of Am×nA_{m \times n} as a matrix of nn column vectors: A=[∣  ∣  a⃗1…a⃗n∣  ∣  ]A = \left[ \begin{array}{rrr} | \ \ & & |\ \ \\ \vec a_1 & \dots & \vec a_n \\ | \ \ & & |\ \ \\ \end{array} \right]
Then for any vector x⃗=[x1⋮xn]\vec x = \begin{bmatrix} x_1\\ \vdots\\ x_n\\ \end{bmatrix} , we have:
Ax⃗=[∣  ∣  a⃗1…a⃗n∣  ∣  ][x1⋮xn]=x1a⃗1+…+xna⃗nA\vec x = \left[ \begin{array}{rrr} | \ \ & & |\ \ \\ \vec a_1 & \dots & \vec a_n \\ | \ \ & & |\ \ \\ \end{array} \right] \begin{bmatrix} x_1\\ \vdots\\ x_n\\ \end{bmatrix} = x_1 \vec a_1 + \ldots + x_n\vec a_n
This is simply a linear combination of the columns.
So, the column space can also be defined as the span of the columns of AA:
Col(A)=span({a⃗1,…,a⃗n})\boxed{\quad {\rm Col}(A) = \text{span}(\{ \vec a_1, \dots, \vec a_n \}) \quad}

Wize Tip
You can make sure you have the right number of vectors by checking the rank:
dim(Col(A))=rank(A)\text{dim}\big(\text{Col}(A)\big)=\text{rank}(A)

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Null Space

The null space (or kernel) is the set of vectors that, when left multiplied by Am×nA_{m \times n}, produce the zero vector:
Null(A)={ x⃗∈Rn ∣ Ax⃗=0⃗ }\boxed{\quad \text{Null}(A)=\{\ \vec x\in \mathbb{R}^n\ |\ A\vec x=\vec 0\ \} \quad}

Wize Concept
Recall that Ax⃗=0⃗A\vec x=\vec 0 is a homogeneous system:
Solve for x⃗\vec x by row-reducing AA, then setting each row of the system equal to 0.

Wize Tip
You can make sure you have the right number of vectors by checking the rank:
rank(A)+dim(Null(A))=n\text{rank}(A) + \text{dim}\big(\text{Null}(A)\big) = n

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Fundamental Subspaces

The row space of matrix AA is the span of its rows (or simply the column space of the transpose!).
Row(A)=Col(AT)\boxed{\quad {\rm Row}(A) = {\rm Col}(A^T) \quad}
The left null space of AA is the null space of the transpose, Null(AT){\rm Null}(A^T).
Together, these four subspaces are called the fundamental subspaces of the matrix AA:
  • column space, Col(A){\rm Col} (A): the result of multiplying Ax⃗A\vec x for any x⃗\vec x.
  • row space, Row(A){\rm Row} (A): the result of multiplying ATx⃗A^T \vec x for any x⃗\vec x.
  • null space, Null(A){\rm Null} (A): the vectors x⃗\vec x that result in Ax⃗=0⃗A \vec x = \vec 0 .
  • left null space, Null(AT){\rm Null} (A^T): the vectors x⃗\vec x that result in ATx⃗=0⃗A^T \vec x = \vec 0 .
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Example: Column Space and Null Space

Given A=[1122−20]A= \left[\begin{array}{rrr} 1&1&2\\ 2&-2&0 \end{array}\right], find Col(A)\text{Col}(A) and Null(A)\text{Null}(A).

Column Space

Col(A)\text{Col}(A) is the span of the columns of A.
We can write the columns as vectors: a⃗1=[12],a⃗2=[1−2],a⃗3=[20]\vec a_1 = \begin{bmatrix} 1\\ 2\\ \end{bmatrix} ,\quad \vec a_2 = \begin{bmatrix} 1\\ -2\\ \end{bmatrix} ,\quad \vec a_3 = \begin{bmatrix} 2\\ 0\\ \end{bmatrix} .
Notice that {a⃗1,a⃗2}\{\vec a_1,\vec a_2\} are two linearly independent vectors, so we already know that they form a generating set for R2\reals^2.
That is, any vector b⃗=[b1b2]∈R2\vec b = \begin{bmatrix} b_1\\ b_2\\ \end{bmatrix} \in \reals^2 can be written as a linear combination of the vectors a⃗1, a⃗2\vec a_1, \ \vec a_2.
Adding the third column, {a⃗1,a⃗2,a⃗3}\{\vec a_1,\vec a_2, \vec a_3\} is still a generating set for R2\reals^2 (although it is not linearly independent).
Therefore:
Col(A)=span({a⃗1,a⃗2,a⃗3})=span({a⃗1,a⃗2})=R2\text{Col}(A) = \text{span}(\{\vec a_1,\vec a_2, \vec a_3\}) = \text{span}(\{\vec a_1,\vec a_2\}) = \boxed{\reals^2}

Null Space

To find Null(A)\text{Null}(A), we must find the vectors x⃗=[xyz]\vec x = \begin{bmatrix} x\\ y\\ z\\ \end{bmatrix} such that Ax⃗=0⃗A\vec x=\vec 0.

Let's row reduce AA to solve this homogeneous linear system:
[11   22−20]R2−2R1⟶[11   20−4−4]−14R2⟶[1   1   2011]R1−R2⟶[1   0   1011]\begin{aligned} &\left[ \begin{array}{rrr} 1 & 1 & \ \ \ 2\\ 2 & -2 & 0\\ \end{array} \right] \begin{array}{l} \\ R_2 - 2R_1\\ \end{array}\\[1.8em] \longrightarrow& \left[ \begin{array}{rrr} 1 & 1 & \ \ \ 2\\ 0 & -4 & -4\\ \end{array} \right] \begin{array}{l} \\ -\dfrac{1}{4}R_2\\ \end{array}\\[1.8em] \longrightarrow& \left[ \begin{array}{rrr} 1 & \ \ \ 1 & \ \ \ 2\\ 0 & 1 & 1\\ \end{array} \right] \begin{array}{l} R_1-R_2\\ \\ \end{array}\\[1.8em] \longrightarrow& \left[ \begin{array}{rrr} 1 & \ \ \ 0 & \ \ \ 1\\ 0 & 1 & 1\\ \end{array} \right] \end{aligned}
Rewriting as a linear system with a parameter for the free variable, and setting each row equal to 0:
{1x+0y+1z=00x+1y+1z=0z=t  ⟹  {x=−1z=−ty=−1z=−tz=t  ⟹  [xyz]=t[−1−11]\begin{cases} 1x +0y +1z = 0\\ 0x + 1y + 1z = 0\\ z=t \end{cases} \quad\implies\quad \begin{cases} x = -1z = -t\\ y = -1z = -t\\ z=t \end{cases} \quad\implies\quad \begin{bmatrix} x\\ y\\ z\\ \end{bmatrix} =t \begin{bmatrix} -1\\ -1\\ 1\\ \end{bmatrix}
This means that matrix AA times any multiple of [−1−11]\begin{bmatrix} -1\\ -1\\ 1\\ \end{bmatrix}produces 0⃗\vec 0.
Wize Tip
The number of free variables/parameters is always equal to the number of vectors in the null space.

Therefore, dim(Null(A))=1{\rm dim(Null}(A)) = 1 and:
Null(A)=span{[−1−11]}\boxed{ \text{Null}(A) = {\rm span} \left\{ \begin{bmatrix} -1\\ -1\\ 1\\ \end{bmatrix} \right\} }
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Practice: Column Space and Null Space

A=[101022−132]A= \left[\begin{array}{rrr} 1&0&1\\ 0&2&2\\ -1&3&2 \end{array}\right]
Find Col(A)\text{Col}(A) and Null(A)\text{Null}(A).
Extra Practice
Practice: Null Space and Range
Suppose that T:R3→R3T:\mathbb{R}^3\rightarrow\mathbb{R}^3 is defined by T(x⃗):=Ax⃗T(\vec x):=A\vec x , where A=[a⃗1a⃗2a⃗3]=[112224−136]A=\left[\begin{array}{rrr}\vec a_1&\vec a_2&\vec a_3\end{array}\right]=\left[\begin{array}{rrr}1&1&2\\2&2&4\\-1&3&6\end{array}\right]
Find Range(T)Range(T) and Null(T)Null(T)
And write each as a span of as few vectors as possible.
Example: Rank-Nullity Theorem
Let M2×2M_{2\times2} be the vector space of 2×22\times 2 matrices with real number entries, and define the transformation
L:M2×2→M2×2L:M_{2\times2}\to M_{2\times2} by:
L(B)=ABL(B)=AB where A=[3131]A=\left[\begin{array}{rr}3&1\\3&1\end{array}\right]
Practice: Rank-Nullity Theorem
Let P2P_2 be the vector space of polynomials of degree at most 2, with the standard operations of polynomial addition and scalar multiplication:
P2(x)={p(x)=ax2+bx+c  ∣  a,b,c∈R}P_2(x)=\{p(x)=ax^2+bx+c\;|\;a,b,c\in\mathbb{R}\}
Let L:P2→P2L:P_2\to P_2 be the transformation defined by:
133 - FML 3 - 18.1W - e.g. 46_$\tkcth{mock 1 ✓}$
The matrix
  ⁣M ⁣‾  =[40101103−30−10−1−9−270102166−60−20−2−18−5] \M = \begin{bmatrix} 4 & 0 & 1 & 0 & 1 & 10 & 3 \\[5pt] -3 & 0 & -1 & 0 & -1 & -9 & -2 \\[5pt] 7 & 0 & 1 & 0 & 2 & 16 & 6 \\[5pt] -6 & 0 & -2 & 0 & -2 & -18 & -5 \end{bmatrix}
Has reduced-row-echelon form given by:
19.4F_Mid_Builder_$\tkcth{8.4.}\tkcf{45}$_
Let A‾=[ai1  ⋯  ain]\A = \big[ a_{i1} \; \cdots \; a_{in} \big], where {ai2, ai3, ai5}\{ a_{i2},\, a_{i3},\, a_{i5} \} is a basis for the column space of A‾\A. The the subset
U={y⃗ ∈ Rn : A‾y⃗=b⃗,b⃗ ∈ span{ai2, ai3, ai5}} U = \left\{ \vec{y} \, \in \, \mathbb{R}^{n} \, : \, \A \vec{y} = \vb,\quad \vb \, \in \, \text{span}\{ a_{i2},\, a_{i3},\, a_{i5}\} \right\}
is a subspace of Rn\mathbb{R}^{n}.
19.4F_Mid_Builder_$\tkcth{8.4.17}\tkcf{p2}$__$\key{F.Quiz}$
A basis for ker⁡  ⁣M ⁣‾  \ker\M (i.e. null   ⁣M ⁣‾  \text{null}\,\M) for the matrix
  ⁣M ⁣‾  =[1000001000001000001000001] \M = \begin{bmatrix} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ \end{bmatrix}
is given by the set of vectors (select all that apply):
19.4F_Mid_Builder_$\tkcth{8.4.17}\tkcf{p1}$_$\key{F.Quiz}$
A basis for Im   ⁣M ⁣‾  \text{Im}\, \M for the matrix
  ⁣M ⁣‾  =[1000001000001000001000001] \M = \begin{bmatrix} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ \end{bmatrix}
is given by the set of vectors (select all that apply):
19.4F_Mid_Builder_$\tkcth{8.4.}\tkcf{20}\tkcth{p2}$_$\tkco{Mock 3}$
A basis for the kernel of the matrix
  ⁣M ⁣‾  =[50−71014000105001]\M = \begin{bmatrix} 5 & 0 & -7 & 1 \\ 0 & 1 & 4 & 0 \\ 0 & 0 & 1 & 0 \\ 5 & 0 & 0 & 1 \\ \end{bmatrix}
is given by the set of vectors (select all that apply):
19.4F_Final_Builder_Ch_17.3_Least_Squares_$\tkco{eg3}$_$\key{Final}$_Builder_$\tkcth{17.3.}\tkcf{3}$_
The system given by:
{x−2y=−42x−4y=7 \left\{ \begin{array}{rr} x - 2y \quad =& -4 \\ 2x - 4y \quad =& 7 \end{array} \right.
can be represented as a single matrix equation A‾ x⃗=b⃗\A \, \vx = \vb, where A‾\A is the matrix of coefficients and b⃗\vb is the constant vector [−47]\colvec{-4}{7}.
19.4F_Mid_Builder_$\tkcth{8.4.19}\tkcf{p2}$_$\tkct{redo solution: *not* obvious =D}$
A basis for the kernel of the matrix
  ⁣M ⁣‾  =[030−5013000100001]\M = \begin{bmatrix} 0 & 3 & 0 & -5 \\ 0 & 1 & 3 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix}
is given by the set of vectors (select all that apply):
Dimension of Null Spaces
Prove that dim(null(AB))≥dim(null(B))dim(null(AB))\geq dim(null(B)) for any matrices A,BA,B for which ABAB is defined.
19.4F_WML_6_$\tkco{eg6}$_$\key{Final}$_Builder_$\tkcth{17.1.}\tkct{6}$_
Find a basis for the kernel of the matrix of coefficients for the system:
[102214112][x1x2x3]=[000] \begin{bmatrix} 1 & 0 & 2 \\ 2 & 1 & 4 \\ 1 & 1 & 2 \end{bmatrix} % \colvecth{x_1}{x_2}{x_3} % = % \colvecth{0}{0}{0}
Use your result to show that any vector in the kernel of the matrix is orthogonal to any vector in the space spanned by the rows of the matrix.
Practice Question: Bases of Column and Row Spaces
Practice Question: Bases of Column and Row Spaces
Find bases for each of row(A),col(A)row(A) ,col(A) when A=[01−22511220112−31]A=\left[\begin{array}{rrrr}0&1&-2&2&5\\1&1&2&2&0\\1&1&2&-3&1\end{array}\right]
$\tkco{move to mock ✓}$ 133 - FML 3 - 18.1W - e.g. 50 $\tkcth{copy - not update for now}$ $\tkcf{mock2}$
Given the matrix   ⁣M‾=[−2−44−212−4−4]\bcb{\ul{M} = \begin{bmatrix} -2 & -4 & 4 & -2 \\ 1 & 2 & -4 & -4 \end{bmatrix}}
$\tkco{move to mock ✓}$ $\tkct{needs vid / solution!}$ 133 - FML 3 - 18.1W - e.g. 50
Given the matrix   ⁣M‾=[−2−44−212−4−4]\bcb{\ul{M} = \begin{bmatrix} -2 & -4 & 4 & -2 \\ 1 & 2 & -4 & -4 \end{bmatrix}}, find a basis for the row-, column- and null-space of   ⁣M‾\bcb{\ul{M}}.
Practice Question: Rank-Nullity
Write (−2,2,0,−5)(-2,2,0,-5) as a sum r⃗+n⃗\vec r+\vec n , where r⃗∈row(A),n⃗∈null(A)\vec r\in row(A),\vec n\in null(A) , and where A:=[141−31111]A:=\left[\begin{array}{rrrr}1&4&1&-3\\1&1&1&1\end{array}\right] .
19.4F_133_8.2_Mock_F1_$\tkco{eg8}$_$\key{Final}$_Builder_$\tkcth{8.2.}\tkcf{8}$_
If dim(ker(A‾))=0\textrm{dim}(\textrm{ker}(\A)) = 0, for an n×nn \times n matrix A‾\A, then Im (A‾)\text{Im}\, (\A) is Rn\mathbb{R}^{n}, and every vector in Rn\mathbb{R}^{n} is perpendicular to the space spanned by the rows of A‾\A.
133 - FML 3 - 18.1W - e.g. 46_$\tkco{mock 1 ✓}$
The matrix
  ⁣M ⁣‾  =[40101103−30−10−1−9−270102166−60−20−2−18−5] \M = \begin{bmatrix} 4 & 0 & 1 & 0 & 1 & 10 & 3 \\[5pt] -3 & 0 & -1 & 0 & -1 & -9 & -2 \\[5pt] 7 & 0 & 1 & 0 & 2 & 16 & 6 \\[5pt] -6 & 0 & -2 & 0 & -2 & -18 & -5 \end{bmatrix}
Has row-echelon form given by:
Bases for Row and Column Spaces
Find bases for the row and column spaces of A=[123456789101112]A=\left[\begin{array}{rrr}1&2&3\\4&5&6\\7&8&9\\10&11&12\end{array}\right]
Let
A=(12−2014−3−6624−4)A = \left ( \begin{array}{ccc} 1 & 2 & -2\\ 0 & 1 & 4\\ -3 & -6 & 6 \\ 2 & 4 & -4 \end{array} \right )
be a real 4×34 \times 3 matrix. Find a basis for the kernel of AA , and for the image of AA . Using the dimensions of the kernel and image of AA , verify the rank-nullity theorem.
Bases of Row, Column, & Null Spaces
Suppose that A=[1−13−11−1−2231−11]A=\left[\begin{array}{rrr}1&-1&3\\-1&1&-1\\-2&2&3\\1&-1&1\end{array}\right] .
Find a basis for each of row(A),col(A)row(A),col(A) .
What are rank(A) & dim(null(A))rank(A)\text{ \& }dim(null(A)) ?
19.4F_Mid_Builder_$\tkcth{8.4.19}\tkcf{p1}$_$\tkct{redo solution}$
A basis for Im   ⁣M ⁣‾  \text{Im}\, \M for the matrix
  ⁣M ⁣‾  =[030−5013000100001]\M = \begin{bmatrix} 0 & 3 & 0 & -5 \\ 0 & 1 & 3 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix}
is given by the set of vectors (select all that apply):
Let L:R2⟶R3L:\mathbb{R}^2\longrightarrow \mathbb{R}^3 be a linear transformation
Prove that a basis for Im(L)Im(L) can not be a basis for R3\mathbb{R}^3
Basis and Dimension
Practice: Finding a Basis
Let A=[1−11]A=\left[\begin{array}{rrr} 1&-1&1\\ \end{array}\right] and consider the special subspace of R3\mathbb{R}^3 called the null space:
N={x⃗∈R3  ∣  Ax⃗=0⃗}N=\{\vec x\in\mathbb{R}^3\;|\;A\vec x=\vec{0}\}
133 - FML 3 - 18.1W - e.g. 37
Given A=[−42141−1140−309111−20]⟶RREF[10−30011000010000]\bcb{A = \begin{bmatrix} -4 & 2 & 14 & 1 \\ -1 & 1 & 4 & 0 \\ -3 & 0 & 9 &1 \\ 1 & 1 & -2 & 0 \end{bmatrix} \overset{\text{\footnotesize{RREF}}}{\longrightarrow} \begin{bmatrix} 1 & 0 & -3 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 0 &1 \\ 0 & 0 & 0 & 0 \end{bmatrix} }, determine the dimension of the row space, column space and null space of A\bcb{A}, respectively, and provide a basis for each.
Practice: Matrix Invertibility
Suppose AA is an n×nn\times n matrix whose rows span Rn\mathbb{R}^n
Show that for all choices of vectors x⃗1,x⃗2,...,x⃗k∈Rn\vec x_1,\vec x_2,...,\vec x_k\in\mathbb{R}^n :
span{x⃗1,x⃗2,...,x⃗k}=Rnspan\{\vec x_1,\vec x_2,...,\vec x_k\}=\mathbb{R}^n if and only if span{Ax⃗1,Ax⃗2,...,Ax⃗k}=Rnspan\{A\vec x_1,A\vec x_2,...,A\vec x_k\}=\mathbb{R}^n .
Let L:R3⟶R2L:\mathbb{R}^3\longrightarrow\mathbb{R}^2 be defined by:
L((x,y,z))=(x+2y+3z,−x+2y+5z)L\big((x,y,z)\big)=(x+2y+3z,-x+2y+5z)
(a) Prove that LL is a linear transformation
133- 05.4F - Final - PartI Question#6
Let A\bcb{A} be the matrix
A=[1−2144−121232−4021]\bcb{A=\begin{bmatrix} 1&-2&1&4&4\\ -1&2&1&2&3\\ 2&-4&0&2&1 \end{bmatrix}}
Let r\bcb{r} be the rank of A\bcb{A} and q\bcb{q} the dimension of the null space of A\bcb{A} (i.e. the nullity of A\bcb{A}. What are the values of r\bcb{r} and q\bcb{q}?
Basis and Dimension
Practice: Finding a Basis
Let A=[1−11]A=\left[\begin{array}{rrr} 1&-1&1\\ \end{array}\right] and consider the special subspace of R3\mathbb{R}^3 called the null space:
N={x⃗∈R3  ∣  Ax⃗=0⃗}N=\{\vec x\in\mathbb{R}^3\;|\;A\vec x=\vec{0}\}
Suppose that AAis a 5×75\times7matrix and d=dim⁡(null(A))d=\dim\left(null\left(A\right)\right). Which of the following is always true?
Which of the following is a one-dimensional subspace of R3\mathbb{R}^3?