High School
SAT
ACT
University
MCAT
LSAT
DAT

Let L:R^3R^2 be defined by: L((x,y,z))=(x+2y+3z,-x+2y+5z) (a) Prove that L is a…

Related Topics

Wize University Linear Algebra Textbook > Vector Spaces and Subspaces

Column Space and Null Space (Range and Kernel)

3 Activities

Wize University Linear Algebra Textbook > Linear Transformations

Linear Transformations

5 Activities

checklist
Mark Yourself Question
  1. Grab a piece of paper and try this problem yourself.
  2. When you're done, check the "I have answered this question" box below.
  3. View the solution and report whether you got it right or wrong.
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

(b) Find a basis for ker(L)ker(L) and state the dimension of ker(L)ker(L)

(c) Find a basis for Im(L)Im(L) and state the dimension of Im(L)Im(L)
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}}.
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 .
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?
More Linear Transformations Questions:
Consider the linear transformation T:R2→R3T:\mathbb{R}^2 \rightarrow \mathbb{R}^3 such that T[12]=[123]T\begin{bmatrix} 1\\2 \end{bmatrix}=\begin{bmatrix} 1\\2\\3 \end{bmatrix} and T[11]=[−2−12]T\begin{bmatrix} 1\\1 \end{bmatrix}=\begin{bmatrix} -2\\-1\\2 \end{bmatrix}
Then T[1−1]T\begin{bmatrix} 1\\-1 \end{bmatrix} is equal to
True or False: Let T:R4→R2T:\mathbb{R}^4\to\mathbb{R}^2 defined by T(x1,x2,x3,x4)=(x1+2x2,x3−x4)T(x_1,x_2,x_3,x_4)=(x_1+2x_2,x_3-x_4) , then TT is a linear transformation and the matrix of TT is given by
T=[1200001−1]T=\begin{bmatrix}1&2&0&0\\0&0&1&-1\end{bmatrix}
Determine if a transformation is linear
Let W={(x,y,z)  ∣  x,y,z∈R+}W=\{(x,y,z)\;|\;x,y,z\in\mathbb{R}^+\} be the vector with defined operations:
(x1,y1,z2)+(x2,y2,z2)=(x1x2,  y1y2,  z1z2)(x_1,y_1,z_2)+(x_2,y_2,z_2)=(x_1x_2,\;y_1y_2,\;z_1z_2)
k⋅(x,y,z)=(xk,  yk,  zk)k\cdot(x,y,z)=(x^k,\;y^k,\;z^k)
Matrix of a Linear Transformation
Consider the map T:R3→R3x⃗→proj(1,1,0)x⃗+2proj(1,0,−1)x⃗−proj(0,0,1)x⃗\begin{array}{l}T:\mathbb{R}^3\rightarrow\mathbb{R}^3\\\vec x\rightarrow proj_{(1,1,0)}\vec x+2proj_{(1,0,-1)}\vec x-proj_{(0,0,1)}\vec x\end{array}
a) Show that T is a linear transformation
b) Compute the matrix of T, and its inverse matrix.
Find the matrix of a given transformation
(a) Let L:R3⟶R4L:\mathbb{R}^3\longrightarrow\mathbb{R}^4 be defined by:
L((x,y,z))=(2x−y,2y+3z,−3x+5z,x−y+z)L\big((x,y,z)\big)=(2x-y,2y+3z,-3x+5z,x-y+z)
Find the matrix AA that represents LL
Supposse T[01]=[112],  T[11]=[21−1],  B=[010],  and  C=[001]T\begin{bmatrix} 0\\1 \end{bmatrix}=\begin{bmatrix} 1\\1\\2 \end{bmatrix},\; T\begin{bmatrix} 1\\1 \end{bmatrix}=\begin{bmatrix} 2\\1\\-1 \end{bmatrix},\; B=\begin{bmatrix} 0\\1\\0 \end{bmatrix},\; and\; C=\begin{bmatrix} 0\\0\\1 \end{bmatrix}
Then:
Finding Non-linear Transformation
Which of the following transformations are linear?
a)[xy]→[x−y2x+y+60]a)\begin{bmatrix} x\\y \end{bmatrix}→\begin{bmatrix} x-y\\ 2x+y+6\\ 0 \end{bmatrix}
Let T be a linear transformation. Suppose that
T([101])=[−20−1],T([011])=[032],T([010])=[0−40]T\left(\begin{bmatrix} 1\\0\\1 \end{bmatrix}\right)=\begin{bmatrix} -2\\0\\-1 \end{bmatrix},T\left(\begin{bmatrix} 0\\1\\1 \end{bmatrix}\right)=\begin{bmatrix} 0\\3\\2 \end{bmatrix},T\left(\begin{bmatrix} 0\\1\\0 \end{bmatrix}\right)=\begin{bmatrix} 0\\-4\\0 \end{bmatrix}
a) Find T([100])a)\ Find\ T\left(\begin{bmatrix} 1\\0\\0 \end{bmatrix}\right)
Find a linear transformation T from R2 to R2, if
T([1−4])=[−36]T\left(\begin{bmatrix} 1\\-4 \end{bmatrix}\right)=\begin{bmatrix} -3\\6 \end{bmatrix}
and it transforms points on liney=0 y=0 to new line y=2xy=2x and vice versa:
Let u⃗=[−314]\vec{u} = \begin{bmatrix} -3 \\ 1 \\ 4 \end{bmatrix} and let T:R3→R3T : \mathbb{R}^3 \rightarrow \mathbb{R}^3 be the linear transformation defined by T(v⃗)=proj⁡u⃗(v⃗)T(\vec{v}) = \operatorname{proj}_{_{\vec{u}}} (\vec{v}). Find the matrix of TT.
Let T:R2→R2T:\mathbb{R}^2\to\mathbb{R}^2 be a linear transformation with T[3−2]=[62]T\begin{bmatrix}3\\-2\end{bmatrix}=\begin{bmatrix}6\\2\end{bmatrix} and T[−54]=[−40]T\begin{bmatrix}-5\\4\end{bmatrix}=\begin{bmatrix}-4\\0\end{bmatrix}
(a) Find the matrix of TT, that is, find a matrix AA such that Tv⃗=Av⃗T\vec{v}=A\vec{v} for all v⃗∈R2\vec{v}\in\mathbb{R}^2.
(b) Is TT invertible? If so, find the matrix of T−1T^{-1}. If not, prove why not.
The unit square in R2 is the set of all points (x; y) with 0 ≤ 𝑥 ≤ 2 and 0 ≤ 𝑦 ≤ 2. Suppose each point in
the unit square undergoes a linear transformation that has the given matrix representation
A=[1221].Draw a labelled sketch of the transformed points.A=\begin{bmatrix} 1&2\\2&1 \end{bmatrix}\text{.Draw a labelled sketch of the transformed points.}
Find the matrix A representing the linear transformation T: R3 to R3 such that it satisfies:
𝑻(𝑒1) = 𝑒1 + 2𝑒2 + 3𝑒3, 𝑻(𝑒1 + 𝑒2) = 𝑒1 + 2𝑒2, 𝑻(𝑒1 + 𝑒2 + 𝑒3) = −2𝑒2 + 𝑒3.
Suppose T is a linear transformation such that
T[113]=[2−1],T[202]=[32],T[424]=[−14]T\begin{bmatrix} 1\\1\\3 \end{bmatrix}=\begin{bmatrix} 2\\-1\\ \end{bmatrix},T\begin{bmatrix} 2\\0\\2 \end{bmatrix}=\begin{bmatrix} 3\\2 \end{bmatrix},T\begin{bmatrix} 4\\2\\4 \end{bmatrix}=\begin{bmatrix} -1\\4 \end{bmatrix}
Find the matrix of T.
Let u⃗=[−311]\vec{u} = \begin{bmatrix} -3 \\ 1 \\ 1 \end{bmatrix} and let T:R3→R3T : \mathbb{R}^3\rightarrow\mathbb{R}^3 be the linear transformation defined by T(v⃗)=proj⁡u⃗(v⃗)T(\vec{v})=\operatorname{proj}_{_{\vec{u}}}(\vec{v}). Find the matrix of TT.
Find the matrix for the linear transformation S :R3 to R3 given by
S([xyz])=([−y+zx+y+z−x+2y])S\left(\begin{bmatrix} x\\y\\z \end{bmatrix}\right)=\left(\begin{bmatrix} -y+z\\ x+y+z\\ -x+2y \end{bmatrix}\right)
Suppose a⃗\vec{a}is a fixed vector in ℝ3 and consider the transformation 𝑇:ℝ3 → ℝ3 defined by 𝑇((x)⃗=a⃗ ×x.⃗𝑇(\vec{\left(x\right)}=\vec{a}\ \times\vec{x.} Prove that T is a linear transformation and find the matrix of T.
If T is a linear transformation such that T[−12]=[31]T\begin{bmatrix}-1\\2\end{bmatrix}=\begin{bmatrix}3\\1\end{bmatrix} and T[13]=[−12]T\begin{bmatrix}1\\3\end{bmatrix}=\begin{bmatrix}-1\\2\end{bmatrix} , find T[4−3]T\begin{bmatrix}4\\-3\end{bmatrix}
Suppose T is a linear transformation and
T[25]=[123],T[38]=[−101]T\begin{bmatrix} 2\\5 \end{bmatrix}=\begin{bmatrix} 1\\2\\3 \end{bmatrix},T\begin{bmatrix} 3\\8 \end{bmatrix}=\begin{bmatrix} -1\\0\\1 \end{bmatrix}
Find the matrix of this linear transformation.
Consider the transformation T defined by 𝑇(𝑥1,𝑥2,𝑥3,x4)=(x1+2x2,𝑥3−𝑥4)𝑇(𝑥_1,𝑥_2,𝑥_3,x_4)=(x_1+2x_2,𝑥_3−𝑥_4). Prove T is a linear transformation and find the matrix of T.
Suppose a⃗\vec{a} is a fixed vector in ℝn and consider the transformation 𝑇:ℝn → ℝ defined by 𝑇 (x⃗)=a.⃗  x.⃗  𝑇\ \left(\vec{x}\right)=\vec{a.}\ \ \vec{x.}\ \ Prove that T is a linear transformation and find the matrix of T.
Suppose T is a linear transformation and
T[25]=[123],T[38]=[−101]T\begin{bmatrix} 2\\5 \end{bmatrix}=\begin{bmatrix} 1\\2\\3 \end{bmatrix},T\begin{bmatrix} 3\\8 \end{bmatrix}=\begin{bmatrix} -1\\0\\1 \end{bmatrix}
Find the matrix of this linear transformation.
For a linear transformation W from R2 to R2, if 𝑊
W([20])=[32],W([12])=[−14],find W([−44]).W \left(\begin{bmatrix} 2\\0 \end{bmatrix}\right)=\begin{bmatrix} 3\\2 \end{bmatrix},W\left(\begin{bmatrix} 1\\2 \end{bmatrix}\right)=\begin{bmatrix} -1\\4 \end{bmatrix},find\ W\left(\begin{bmatrix} -4\\4 \end{bmatrix}\right).
True or False: Suppose a ⃗\vec{a\ } is a fixed vector in Rn\mathbb{R}^n and T:Rn→RT:\mathbb{R}^n\to\mathbb{R} defined by T(x⃗)=a⃗⋅x⃗T(\vec{x})=\vec{a}\cdot\vec{x} then TT is a linear transformation and its matrix is [a1⋯an]\begin{bmatrix}a_1&\cdots&a_n\end{bmatrix}
Suppose T:R2→R3T:\mathbb{R}^2\to\mathbb{R}^3 is a linear transformation and
T[25]=[123],T[38]=[−101]T\begin{bmatrix}2\\5\end{bmatrix}=\begin{bmatrix}1\\2\\3\end{bmatrix},\quad T\begin{bmatrix}3\\8\end{bmatrix}=\begin{bmatrix}-1\\0\\1\end{bmatrix}
Find the matrix of this linear transformation.
Example: Finding the Linear Transformation Matrix
Find the corresponding matrix M of the linear transformation 𝑓(𝑥, 𝑦, 𝑧) = (𝑥 − 𝑦, 3𝑦, 3𝑥 + 2𝑧).
Finding Non-linear Transformation
Which of the following transformations are linear?
Is the given transformation linear?
Let VV be the vector space {(x,y)  ∣  x,y∈R+}\{(x,y)\;|\;x,y\in\mathbb{R}^+\} with the operations:
(x1,y1)+(x2,y2)=(x1x2,y1y2)(x_1,y_1)+(x_2,y_2)=(x_1x_2,y_1y_2)
k⋅(x,y)=(xk,yk)k\cdot(x,y)=(x^k,y^k)
Practice Question: Linear and Non-Linear Maps
Determine whether the following maps are linear or not.
1. L:R→RL: \mathbb{R} \rightarrow \mathbb{R} , defined by L(x)=cos⁡(x)L(x) = \cos(x) .
2. L:R→RL : \mathbb{R} \rightarrow \mathbb{R} , defined by L(x)=3x−2L(x) = 3x -2
19.4F_Mid_Builder_$\tkcth{8.4.}\tkcf{15}$_$\tkco{Mock 3}$.
The following is known about the system of linear equations given by   ⁣M ⁣‾   x⃗=y⃗\M \, \vx = \vec{y}:
There is a unique solution for any y⃗ ∈ R5\vec{y} \, \in \, \mathbb{R}^{5}.
If one interchanges the 3rd and 4th rows and then adds twice the third row to the first before finally multiplying the 2nd and 4th rows by 3,   ⁣M ⁣‾  \M will be in its RREF.
Find the matrix of a given transformation
(a) Let L:R3⟶R4L:\mathbb{R}^3\longrightarrow\mathbb{R}^4 be defined by:
L((x,y,z))=(2x−y,2y+3z,−3x+5z,x−y+z)L\big((x,y,z)\big)=(2x-y,2y+3z,-3x+5z,x-y+z)
Find the matrix AA that represents LL
19.4F_Mid_Builder_$\tkcth{8.4.}\tkcf{14}$_
Given the transformation T:R2 → R2T: \mathbb{R}^2 \, \to \, \mathbb{R}^2 where T(v⃗)=x⃗T(\vv) = \vx and T(w⃗)=y⃗T(\vw) = \vec{y}, where
v⃗=[14]\vv = \colvec{1}{4}, w⃗=[311]\vw = \colvec{3}{11}, x⃗=[02]\vx = \colvec{0}{2} and y⃗=[−30]\vec{y} = \colvec{-3}{0},
find the matrix representation   ⁣M ⁣‾  TB\M_{T_B} of the transformation TT in the basis B={[28], [−3−11]}B = \left\{ \colvec{2}{8},\, \colvec{-3}{-11} \right\} .
Let L:R2→R3L : \mathbb{R}^2 \rightarrow \mathbb{R}^{3} be the map defined by
L(x,y)=(2x,3y,x+y)L(x, y) = (2x, 3y, x + y)
Is this map a linear transformation? If so, what is the kernel and image of this function?
133 - FML 3 - 18.1W - e.g. 34
If T[10]=[4−4]\bcb{T\begin{bmatrix} 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 4 \\ -4 \end{bmatrix}} and T[01]=[3−1]\bcb{T\begin{bmatrix} 0 \\ 1 \end{bmatrix} = \begin{bmatrix} 3 \\ -1 \end{bmatrix}}, find the matrix representation of the transformation in the space spanned by the basis vectors u⃗1=(1,1)\bcb{\vec{u}_1 = (1,1)} and u⃗2=(1,−1)\bcb{\vec{u}_2 = (1,-1)}.
  ⁣M ⁣‾  T{u⃗1,u⃗2}=\M_{T_{\{\vu_1, \vu_2\}}} =
7
1
-5
-3
19.4F_Mid_Builder_$\tkcth{8.5.}\tkco{5}$__$\tkco{Mock 3}$
Given the basis for R2\mathbb{R}^2
v⃗=[1−1]\vv = \colvec{1}{-1} and w⃗=[2−3]\vw = \colvec{2}{-3}.
If T(v⃗)=2v⃗+3w⃗T(\vv) = 2\vv + 3\vw and T(w⃗)=v⃗+3w⃗T(\vw) = \vv + 3\vw, determine the matrix representation of the transformation TT.
Let L:R3⟶RL:\mathbb{R}^3\longrightarrow\mathbb{R} be defined by:
L((x,y,z))=3x−y+2zL\big((x,y,z)\big)=3x-y+2z
(a) Show that v→=(2,−10,−8)\overrightarrow{v}=(2,-10,-8) is in ker(L)ker(L)
What is the matrix for the linear map
L(x,y,z)=(x+y+z,2x−3z)L(x, y, z) = (x + y + z, 2x - 3z)
with respect to the standard basis?
Let LL be a linear transformation defined by:
L(x→)=Ax→L(\overrightarrow{x})=A\overrightarrow{x}
where A=[2−31−14−3]A=\left[\begin{array}{rrr} 2&-3&1\\ -1&4&-3 \end{array}\right]
Properties
Practice: Similar Matrices
Consider the matrix A=[−1421]A= \left[ \begin{array}{rrr} -1 & 4\\ 2 & 1 \end{array} \right].
Concept Clarifier
If TT and SS are linear transformations induced by the matrices
[−2024],[1−201]\begin{bmatrix}-2&0\\2&4\end{bmatrix},\quad \begin{bmatrix}1&-2\\0&1\end{bmatrix}
respectively, then find (S∘T)(x⃗)(S \circ T)(\vec{x}) where x⃗=[15]\vec{x}=\begin{bmatrix}1\\5\end{bmatrix}
133 - FML 3 - 18.1W e.g. 2.3
Determine whether the following transformation is linear:{y1= 3(x1+x2)y2=−x1\boldsymbol{ \begin{cases} y_1 &=~ 3(x_1 + x_2) \\ y_2 & = -x_1 \end{cases} }
133 - FML 3 - 18.1W e.g. 2.3
Determine whether the following transformation is linear:{y1= 1y2=1y3= 1\boldsymbol{ \begin{cases} y_1 &=~ 1 \\ y_2 & = 1 \\ y_3 &=~ 1 \end{cases} }
133 - FML 3 - 18.1W e.g. 2.2
Determine whether the following transformation is linear:{y1= x1y2=x1x2\boldsymbol{ \begin{cases} y_1 &=~ x_1 \\ y_2 & = x_1x_2 \end{cases} }
133 - FML 3 - 18.1W e.g. 2.1
Determine whether the following transformation is linear:{y1= x3y2= x3y3= x3\boldsymbol{ \begin{cases} y_1 &=~ x_3 \\ y_2 & = ~x_3 \\ y_3 &=~ x_3 \end{cases} }
Linear Transformations
Practice: Identifying Non-Linear Transformations
Select all of the following transformations that are non-linear.
Linear Transformations
Practice: Linear Transformations
Suppose T:R2→R3T:\mathbb{R}^2\to\mathbb{R}^3 is a linear transformation where
T[25]=[123],T[38]=[−101]T\begin{bmatrix}2\\5\end{bmatrix}=\begin{bmatrix}1\\2\\3\end{bmatrix},\quad T\begin{bmatrix}3\\8\end{bmatrix}=\begin{bmatrix}-1\\0\\1\end{bmatrix}.
Define the vector space V={(x,y,z)  ∣  x,y,z>0}V=\{(x,y,z)\;|\;x,y,z>0\} with the operations:
(x1,y1,z1)+(x2,y2,z2)=(x1x2,y1y2,z1+z2)(x_1,y_1,z_1)+(x_2,y_2,z_2)=(x_1x_2,y_1y_2,\textcolor{red}{z_1+z_2})
k⋅(x,y,z)=(xk,yk,kz)k\cdot(x,y,z)=(x^k,y^k,\textcolor{red}{kz})
Let LL be the linear transformation such that L(v→)=Av→L(\overrightarrow{v})=A\overrightarrow{v} where:
A=[111−10201021−1]A=\left[\begin{array}{rrr} 1&1&1\\[0.5em] -1&0&2\\[0.5em] 0&1&0\\[0.5em] 2&1&-1\\ \end{array}\right]
Find the rule that defines LL including the domain and co-domain vector spaces
Introduction to Transformations

Practice: Linear Transformations
Suppose T:R2→R3T:\mathbb{R}^2\to\mathbb{R}^3 is a linear transformation and
T[25]=[123],T[38]=[−101]T\begin{bmatrix}2\\5\end{bmatrix}=\begin{bmatrix}1\\2\\3\end{bmatrix},\quad T\begin{bmatrix}3\\8\end{bmatrix}=\begin{bmatrix}-1\\0\\1\end{bmatrix}
Find the matrix of this linear transformation.
Linear Transformations

Given that S:R2→R2S:\mathbb{R}^2 \to \mathbb{R}^2and T:R2→R2T:\mathbb{R}^2 \to \mathbb{R}^2such that
S[35]=[1−1],S[3−1]=[25]S\begin{bmatrix}3\\5\end{bmatrix}=\begin{bmatrix}1\\-1\end{bmatrix}, \quad S\begin{bmatrix}3\\-1\end{bmatrix}=\begin{bmatrix}2\\5\end{bmatrix}and T[2−2]=[10],T[11]=[25]T\begin{bmatrix}2\\-2\end{bmatrix}=\begin{bmatrix}1\\0\end{bmatrix}, \quad T\begin{bmatrix}1\\1\end{bmatrix}=\begin{bmatrix}2\\5\end{bmatrix}
Find S∘T[3−1]S \circ T\begin{bmatrix}3\\-1\end{bmatrix}
Matrix Multiplication
Find a 3 × 3 matrix M such that
  ⁣M ⁣‾  [   0−2   0]=[−1   1−1] ⁣,  ⁣M ⁣‾  [0018]=[   0−2   3] ⁣,and  ⁣M ⁣‾  [1300]=[−2   2−1]\bcb{ \M\begin{bmatrix} \ \ \ 0\\-2\\\ \ \ 0 \end{bmatrix}= \begin{bmatrix} -1\\\ \ \ 1\\-1 \end{bmatrix}\!, \hspace{0.7cm} \M\begin{bmatrix} 0\\0\\\tfrac{1}{8} \end{bmatrix}= \begin{bmatrix} \ \ \ 0\\-2\\\ \ \ 3 \end{bmatrix}\!, \hspace{0.5cm} \text{\small{and}}\hspace{0.5cm} \M\begin{bmatrix} \tfrac{1}{3}\\0\\0 \end{bmatrix}= \begin{bmatrix} -2\\\ \ \ 2\\-1 \end{bmatrix} }
Matrix of a Projection
a) Find the matrix AA so that for x⃗∈R3\vec x \in \mathbb{R}^3 the transformation mapping x⃗→proj(−13,1,12)x⃗\vec x\rightarrow proj_{(\frac{-1}{3},1,\frac{1}{2})}\vec x is the same as the transformation mapping x⃗→Ax⃗\vec x\rightarrow A\vec x
b) What is rank(A)rank(A) ?
Matrix Multiplication
e.g. Find a 3 × 3 matrix M such that
 ⁣M[   0−2   0]=[−1   1−1] ⁣, ⁣M[000]=[   0−2   3] ⁣,and M[1300]=[−2   2−1]\!M\begin{bmatrix} \ \ \ 0\\-2\\\ \ \ 0 \end{bmatrix}= \begin{bmatrix} -1\\\ \ \ 1\\-1 \end{bmatrix}\!, \hspace{0.7cm} \!M\begin{bmatrix} 0\\0\\0 \end{bmatrix}= \begin{bmatrix} \ \ \ 0\\-2\\\ \ \ 3 \end{bmatrix}\!, \hspace{0.7cm} \text{\small{and}}\ M\begin{bmatrix} \frac{1}{3}\\0\\0 \end{bmatrix}= \begin{bmatrix} -2\\\ \ \ 2\\-1 \end{bmatrix}
Given the following matrix, what can you conclude?
A=[10−2151]A = \begin{bmatrix} 1 & 0 \\ -2 & 1\\ 5 & 1 \end{bmatrix}
Projection onto a subspace
Suppose u⃗,v⃗∈Rn\vec u,\vec v\in\mathbb{R}^n satisfy u⃗⋅v⃗=0,∣∣u⃗∣∣=∣∣v⃗∣∣=1\vec u\cdot\vec v=0, ||\vec u||=||\vec v||=1, and define U:=span{u⃗,v⃗}U := span\{\vec u,\vec v\} .
Using the definition that projUx⃗=A(ATA)−1ATx⃗proj_U\vec x=A(A^TA)^{-1}A^T\vec x , where A=[u⃗v⃗]A=\left[\begin{array}{rr}\vec u&\vec v\end{array}\right] , show that for all x⃗∈Rn\vec x\in\mathbb{R}^n we have that projUx⃗=proju⃗x⃗+projv⃗x⃗proj_U\vec x=proj_{\vec u}\vec x+proj_{\vec v}\vec x .