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Practice Question: Rank-Nullity

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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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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] .

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.
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}}.
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?