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Practice Question: Bases of Column and Row Spaces

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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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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]


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}^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.
$\tkco{move to mock ✓}$ 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?