19.4F_Mid_Builder_$\tkcth{8.4.}\tkcf{45}$_

Column Space and Null Space (Range and Kernel)

3 Activities

Let A‾=[ai1  ⋯  ain]\A = \big[ a_{i1} \; \cdots \; a_{in} \big]A​=[ai1​⋯ain​], where {ai2, ai3, ai5}\{ a_{i2},\, a_{i3},\, a_{i5} \}{ai2​,ai3​,ai5​} is a basis for the column space of A‾\AA​. 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\}
U={y​∈Rn:A​y​=b,b∈span{ai2​,ai3​,ai5​}}

is a subspace of Rn\mathbb{R}^{n}Rn. 

N.B. ain\bcb{a_{i\bct{n}}}ain​ represents the nth\bct{n}^{th}nth column of the matrix A‾\AA​.


Select all that apply:

The subset contains the origin. 

The subset is closed under addition.

The subset is closed under scalar multiplication.

The subset is a subspace.

The subset is not a subspace.

None of the other options. 

I don't know

More Column Space and Null Space (Range and Kernel) Questions:

Practice: Null Space and Range

Suppose that T:R3→R3T:\mathbb{R}^3\rightarrow\mathbb{R}^3T:R3→R3 is defined by T(x⃗):=Ax⃗T(\vec x):=A\vec xT(x):=Ax , 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]A=[a1​​a2​​a3​​]=​12−1​123​246​​
Find Range(T)Range(T)Range(T) and Null(T)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}M2×2​ be the vector space of 2×22\times 22×2 matrices with real number entries, and define the transformation
 L:M2×2→M2×2L:M_{2\times2}\to M_{2\times2}L:M2×2​→M2×2​ by:
L(B)=ABL(B)=ABL(B)=AB where A=[3131]A=\left[\begin{array}{rr}3&1\\3&1\end{array}\right]A=[33​11​]

Practice: Rank-Nullity Theorem

Let P2P_2P2​ 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}\}P2​(x)={p(x)=ax2+bx+c∣a,b,c∈R}
Let L:P2→P2L:P_2\to P_2L:P2​→P2​ 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}
M​=​4−37−6​0000​1−11−2​0000​1−12−2​10−916−18​3−26−5​​
Has reduced-row-echelon form given by:

19.4F_Mid_Builder_$\tkcth{8.4.17}\tkcf{p2}$__$\key{F.Quiz}$

A basis for  ker⁡  ⁣M ⁣‾  \ker\MkerM​ (i.e. null   ⁣M ⁣‾  \text{null}\,\MnullM​) 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}
M​=​10000​01000​00100​00010​00001​​
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}\, \MImM​ 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}
M​=​10000​01000​00100​00010​00001​​
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}M​=​5005​0100​−7410​1001​​
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.
        {x−2y=2x−4y=​−47​
can be represented as a single matrix equation A‾ x⃗=b⃗\A \, \vx = \vbA​x=b, where A‾\AA​ is the matrix of coefficients and b⃗\vbb is the constant vector [−47]\colvec{-4}{7}[−47​].

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}M​=​0000​3100​0310​−5001​​
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))dim(null(AB))≥dim(null(B)) for any matrices A,BA,BA,B for which ABAB AB 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}
         ​121​011​242​​​x1​x2​x3​​​=​000​​
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)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]A=​011​111​−222​22−3​501​​

$\tkco{move to mock ✓}$ 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}}M​=[−21​−42​4−4​−2−4​]

$\tkco{move to mock ✓}$ $\tkct{needs vid / solution!}$ 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}}M​=[−21​−42​4−4​−2−4​], find a basis for the row-, column- and null-space of   ⁣M‾\bcb{\ul{M}}M​. 

Practice Question: Rank-Nullity

Write (−2,2,0,−5)(-2,2,0,-5)(−2,2,0,−5) as a sum r⃗+n⃗\vec r+\vec nr+n , where r⃗∈row(A),n⃗∈null(A)\vec r\in row(A),\vec n\in null(A)r∈row(A),n∈null(A) , and where A:=[141−31111]A:=\left[\begin{array}{rrrr}1&4&1&-3\\1&1&1&1\end{array}\right]A:=[11​41​11​−31​] .

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)) = 0dim(ker(A​))=0, for an n×nn \times nn×n matrix A‾\AA​, then Im (A‾)\text{Im}\, (\A)Im(A​) is Rn\mathbb{R}^{n}Rn, and every vector in Rn\mathbb{R}^{n}Rn is perpendicular to the space spanned by the rows of A‾\AA​.

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}
M​=​4−37−6​0000​1−11−2​0000​1−12−2​10−916−18​3−26−5​​
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]A=​14710​25811​36912​​

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 )A=​10−32​21−64​−246−4​​
be a real 4×34 \times 34×3 matrix. Find a basis for the kernel of AAA , and for the image of AAA . Using the dimensions of the kernel and image of AAA , 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]A=​1−1−21​−112−1​3−131​​ .  
Find a basis for each of row(A),col(A)row(A),col(A)row(A),col(A) .  
What are rank(A) & dim(null(A))rank(A)\text{ \& }dim(null(A))rank(A) & dim(null(A)) ?

19.4F_Mid_Builder_$\tkcth{8.4.19}\tkcf{p1}$_$\tkct{redo solution}$

A basis for Im   ⁣M ⁣‾  \text{Im}\, \MImM​ 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}M​=​0000​3100​0310​−5001​​
is given by the set of vectors (select all that apply): 

Let L:R2⟶R3L:\mathbb{R}^2\longrightarrow \mathbb{R}^3L:R2⟶R3 be a linear transformation
Prove that a basis for Im(L)Im(L)Im(L) can not be a basis for R3\mathbb{R}^3R3

Basis and Dimension

Practice: Finding a Basis
Let A=[1−11]A=\left[\begin{array}{rrr}
1&-1&1\\
\end{array}\right]A=[1​−1​1​] and consider the special subspace of R3\mathbb{R}^3R3 called the null space:
N={x⃗∈R3  ∣  Ax⃗=0⃗}N=\{\vec x\in\mathbb{R}^3\;|\;A\vec x=\vec{0}\}N={x∈R3∣Ax=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} }A=​−4−1−31​2101​1449−2​1010​​⟶RREF​​1000​0100​−3100​0010​​, determine the dimension of the row space, column space and null space of A\bcb{A}A, respectively, and provide a basis for each.

Practice: Matrix Invertibility

Suppose AAA is an n×nn\times nn×n matrix whose rows span Rn\mathbb{R}^nRn
Show that for all choices of vectors x⃗1,x⃗2,...,x⃗k∈Rn\vec x_1,\vec x_2,...,\vec x_k\in\mathbb{R}^nx1​,x2​,...,xk​∈Rn : 
span{x⃗1,x⃗2,...,x⃗k}=Rnspan\{\vec x_1,\vec x_2,...,\vec x_k\}=\mathbb{R}^nspan{x1​,x2​,...,xk​}=Rn  if and only if span{Ax⃗1,Ax⃗2,...,Ax⃗k}=Rnspan\{A\vec x_1,A\vec x_2,...,A\vec x_k\}=\mathbb{R}^nspan{Ax1​,Ax2​,...,Axk​}=Rn .

Let L:R3⟶R2L:\mathbb{R}^3\longrightarrow\mathbb{R}^2L:R3⟶R2 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)L((x,y,z))=(x+2y+3z,−x+2y+5z)
(a) Prove that LLL is a linear transformation

133- 05.4F - Final - PartI Question#6

Let A\bcb{A}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}}A=​1−12​−22−4​110​422​431​​
Let r\bcb{r}r be the rank of A\bcb{A}A and q\bcb{q}q the dimension of the null space of A\bcb{A}A (i.e. the nullity of A\bcb{A}A. What are the values of r\bcb{r}r and q\bcb{q}q?

Basis and Dimension

Practice: Finding a Basis
Let A=[1−11]A=\left[\begin{array}{rrr}
1&-1&1\\
\end{array}\right]A=[1​−1​1​] and consider the special subspace of R3\mathbb{R}^3R3 called the null space:
N={x⃗∈R3  ∣  Ax⃗=0⃗}N=\{\vec x\in\mathbb{R}^3\;|\;A\vec x=\vec{0}\}N={x∈R3∣Ax=0}

Suppose that AAAis a 5×75\times75×7matrix and d=dim⁡(null(A))d=\dim\left(null\left(A\right)\right)d=dim(null(A)). Which of the following is always true?

Which of the following is a one-dimensional subspace of R3\mathbb{R}^3R3?

19.4F_Mid_Builder_$\tkcth{8.4.}\tkcf{45}$_

Related Topics

Column Space and Null Space (Range and Kernel)

More Column Space and Null Space (Range and Kernel) Questions:

Practice: Null Space and Range

Example: Rank-Nullity Theorem

Practice: Rank-Nullity Theorem

133 - FML 3 - 18.1W - e.g. 46_$\tkcth{mock 1 ✓}$

19.4F_Mid_Builder_$\tkcth{8.4.17}\tkcf{p2}$__$\key{F.Quiz}$

19.4F_Mid_Builder_$\tkcth{8.4.17}\tkcf{p1}$_$\key{F.Quiz}$

19.4F_Mid_Builder_$\tkcth{8.4.}\tkcf{20}\tkcth{p2}$_$\tkco{Mock 3}$

19.4F_Final_Builder_Ch_17.3_Least_Squares_$\tkco{eg3}$_$\key{Final}$_Builder_$\tkcth{17.3.}\tkcf{3}$_

19.4F_Mid_Builder_$\tkcth{8.4.19}\tkcf{p2}$_$\tkct{redo solution: *not* obvious =D}$

Dimension of Null Spaces

19.4F_WML_6_$\tkco{eg6}$_$\key{Final}$_Builder_$\tkcth{17.1.}\tkct{6}$_

Practice Question: Bases of Column and Row Spaces

$\tkco{move to mock ✓}$ 133 - FML 3 - 18.1W - e.g. 50 $\tkcth{copy - not update for now}$ $\tkcf{mock2}$

$\tkco{move to mock ✓}$ $\tkct{needs vid / solution!}$ 133 - FML 3 - 18.1W - e.g. 50

Practice Question: Rank-Nullity

19.4F_133_8.2_Mock_F1_$\tkco{eg8}$_$\key{Final}$_Builder_$\tkcth{8.2.}\tkcf{8}$_

133 - FML 3 - 18.1W - e.g. 46_$\tkco{mock 1 ✓}$

Bases for Row and Column Spaces

Bases of Row, Column, & Null Spaces

19.4F_Mid_Builder_$\tkcth{8.4.19}\tkcf{p1}$_$\tkct{redo solution}$

Basis and Dimension

133 - FML 3 - 18.1W - e.g. 37

Practice: Matrix Invertibility

133- 05.4F - Final - PartI Question#6

Basis and Dimension

19.4F_Mid_Builder_$\tkcth{8.4.19}\tkcf{p2}$_$\tkct{redo solution: not obvious =D}$