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Okay this may be a bit long. Only the last question is an issue for me but I can't verify whether the earlier questions (which affect the last) are done correctly.

Given \begin{eqnarray*} T(x,y,z) = (x+2y-z,x+z,4x-4y+5z) \end{eqnarray*} for $T:\mathbb{R}^3 \rightarrow \mathbb{R}^3$.

1) Find the matrix of $T$ with respect to the standard basis $\{(1,0,0),(0,1,0),(0,0,1)\}$.

My answer: \begin{eqnarray*} (x,2y,-z) &=& x(1,0,0)+ 2y(0,1,0) -z(0,0,1)\\ (x,0,z) &=& x(1,0,0)+ 0(0,1,0) +z(0,0,1)\\ (4x,-4y,5z) &=& 4x(1,0,0)-4y(0,1,0) +5z(0,0,1)\\ \end{eqnarray*}

$T = \left( \begin{matrix} 1&2&-1\\ 1&0&1\\ 4&-4&5 \end{matrix} \right).$

2) Find the matrix $A$ of $T$ with respect to the ordered basis $e = \{(-1,1,2),(-2,1,4),(-1,1,4)\}$.

My answer: \begin{eqnarray*} (x,2y,-z) &=& a_1(-x,y,2z)+ b_1(-2x,y,4z) +c_1 (-x,y,4z)\\ (x,0,z) &=& a_2(-x,y,2z)+ b_2(-2x,y,4z) + c_2(-x,y,4z)\\ (4x,-4y,5z) &=& a_3(-x,y,2z)+ b_3(-2x,y,4z) +c_3(-x,y,4z). \end{eqnarray*} $A = \left( \begin{matrix} a_1&b_1&c_1\\ a_2&b_2&c_2\\ a_3&b_3&c_3 \end{matrix} \right)= \left( \begin{matrix} 4.5&-3&0.5\\ -0.5&-1&1.5\\ -10.5&0&6.5 \end{matrix} \right).$

3) Find the coordinate vector $[v]_e$ when $v = (1,0,2)$.

My answer: \begin{eqnarray*} (1,0,2) &=& a(-1,1,2) + b(-2,1,4)+ c(-1,1,4) \end{eqnarray*} $[v]_e = \left[ \begin{matrix} a\\b\\c \end{matrix} \right] = \left[ \begin{matrix} -1\\-1\\2 \end{matrix} \right].$

4) Find $T(v)$ and $[T(v)]_e$.

\begin{eqnarray*} T(v) &=& \left( \begin{matrix} 1&2&-1\\ 1&0&1\\ 4&-4&5 \end{matrix} \right) \left( \begin{matrix} 1\\0\\2 \end{matrix} \right)\\ &=& \left( \begin{matrix} -1\\3\\4 \end{matrix} \right).\\ \text{[}T(v)\text{]}_e &=& \left( \begin{matrix} 1&2&-1\\ 1&0&1\\ 4&-4&5 \end{matrix} \right) \left( \begin{matrix} -1\\-1\\2 \end{matrix} \right)\\ &=& \left( \begin{matrix} -5\\1\\10 \end{matrix} \right). \end{eqnarray*}

5. Verify that $[T(v)]_e = A[v]_e:$

\begin{eqnarray*} A[v]_e &=& \left( \begin{matrix} 4.5&-3&0.5\\ -0.5&-1&1.5\\ -10.5&0&6.5 \end{matrix} \right) \left( \begin{matrix} -1\\-1\\2 \end{matrix} \right)\\ &=& \left( \begin{matrix} -0.5\\4.5\\23.5 \end{matrix} \right)\\ &\neq& [T(v)]_e \end{eqnarray*}

I think I misunderstood the difference between $T[v]_e$ and $[T(v)]_e$ (I assumed they're the same) but I don't know what is the difference.

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  • $\begingroup$ $T[v]_e$ is nonsensical. See math.stackexchange.com/q/3577264/265466 for a lengthy explanation of why. It would help you keep things straight not to use the same name for the linear transformation $T$ and its matrix w/r the standard basis. $\endgroup$ – amd Mar 19 at 19:14
  • $\begingroup$ You should check your work. For instance, $T(-1,1,2)=(-1,1,2)$, so the first column of $A$ should be $(1,0,0)^T$, which is not what you have. $\endgroup$ – amd Mar 19 at 19:17
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$$T(v)=Tv=[Tv]_e^T\left(\begin{array}{c} -1 & 1& 2\\ -2 & 1& 4\\ -1 & 1& 4\\ \end{array}\right)$$ $$Tv=A[v]_e^T\left(\begin{array}{c} -1 & 1& 2\\ -2 & 1& 4\\ -1 & 1& 4\\ \end{array}\right)$$ I suggest you read the chapter of linear thranformation about any linear algebra textbook.

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Definition 5.1.3. Let $T : \mathbb{K}^n \rightarrow \mathbb{K}^m$ be a linear transformation and define the matrix $A \in \mathbb{M}^{m \times n}(\mathbb{K})$ to be the matrix whose $j$-th column is $T(ij)$, i.e.

\begin{eqnarray*} A = [T(i_1) \dots T(i_n)] \end{eqnarray*} where {i_1 \dots i_n} is the standard basis of $\mathbb{K}^n$. Then for every $x \in \mathbb{K}^n, x = x_1i_1 + \dots x_n i_n$, and \begin{eqnarray*} T(x) = T(x_1i_1 + \dots x_ni_n) = x_1T(i_1) + \dots x_nT(i_n) = Ax. \end{eqnarray*}

That is, the action of $T$ can be expressed in terms of the matrix $A$ and we call $A$ the matrix of the linear transformation $T$.

Lemma 5.2.1. If $e = \{e_1, \dots, e_n\}$ is a basis for $\mathbb{R}^n$ and $P = [e_1 \dots e_n]$ is the matrix with $e_1, \dots, e_n$ as its columns, then $[v]_e = P^{-1}v$ for all $v \in \mathbb{R}^n$.

Proof. $P[v]_e = [e_1 \dots e_n] \left[ \begin{matrix} x_1 \\ \vdots \\ x_n \end{matrix} \right] = x_1e_1 + \cdots + x_ne_n = v. $ Hence $[v]_e = P^{-1}v.$

Source of Definition and Lemma: Lecture Notes - Linear Algebra Spring 2020

Solution

Given \begin{eqnarray*} T(x,y,z) = (x+2y-z,x+z,4x-4y+5z) \end{eqnarray*} for $T:\mathbb{R}^3 \rightarrow \mathbb{R}^3$.

  1. With respect to the standard basis $\{(1,0,0),(0,1,0),(0,0,1)\}$, the matrix $A$ of $T$ can be found by using Definition 5.1.3 which states that in this case: \begin{eqnarray*} A = [T(i_1) \quad T(i_2) \quad T(i_3)], \end{eqnarray*}

where $i_1, i_2, i_3$ are the standard basis $\{(1,0,0),(0,1,0),(0,0,1)\}$ respectively.

\begin{eqnarray*} T(i_1) &=& T(1,0,0)\\ &=& (1 + 0 - 0, 1 + 0, 4 -0 + 0)^T\\ &=& (1,1,4)^T.\\ T(i_2) &=& T(0,1,0)\\ &=& (2,0,-4)^T.\\ T(i_2) &=& T(0,0,1)\\ &=& (-1,1,5)^T. \end{eqnarray*}

Thus, $A = \left( \begin{matrix} 1&2&-1\\ 1&0&1\\ 4&-4&5 \end{matrix} \right). $

  1. With respect to the ordered basis $e = \{(-1,1,2),(-2,1,4),(-1,1,4)\}$, the matrix $A$ of $T$ can also be found by using Definition 5.1.3.

Let $e_1, e_2, e_3$ represent the three vectors of $e = \{(-1,1,2),(-2,1,4),(-1,1,4)\}$ respectively.

\begin{eqnarray*} T(e_1) &=& T(-1,1,2)\\ &=& (-1,1,2)^T.\\ T(e_2) &=& T(-2,1,4)\\ &=& (-4,2,8)^T.\\ T(e_3) &=& T(-1,1,4)\\ &=& (-3,3,12)^T.\\ \end{eqnarray*}

Next, we'll express $T(e_1), T(e_2), T(e_3)$ as linear combinations of $e_1, e_2, e_3.$

\begin{eqnarray*} T(e_1) &=& 1(-1,1,2)^T + 0(-2,1,4)^T + 0(-1,1,4)^T\\ &=& 1e_1+ 0e_2+ 0e_3\\ \text{[}T(e_1)\text{]}_e &=& (1,0,0)^T.\\ T(e_2) &=& 0e_1+ 2e_2+ 0e_3.\\\text{[}T(e_2)\text{]}_e &=& (0,2,0)^T.\\ T(e_3) &=& 0e_1+ 0e_2+ 3e_3.\\\text{[}T(e_3)\text{]}_e &=& (0,0,3)^T.\\ \end{eqnarray*}

Thus, $ A = [\text{[}T(e_1)\text{]}_e \quad \text{[}T(e_2)\text{]}_e \quad \text{[}T(e_3)\text{]}_e] = \left( \begin{matrix} 1&0&0\\ 0&2&0\\ 0&0&3 \end{matrix} \right). $

  1. Now we'll find the coordinate vector $\text{[}v\text{]}_e$ when $v = (1,0,2)$ using Lemma 5.2.1.

Let $P$ represent the matrix with the ordered basis $\{e_1, e_2, e_3\}$ as its columns:

\begin{eqnarray*} P &=& \left( \begin{matrix} -1&-2&-1\\ 1&1&1\\ 2&4&4 \end{matrix} \right). \end{eqnarray*}

\begin{eqnarray*} [v]_e &=& P^{-1}v\\ &=& \left( \begin{matrix} -1&-2&-1\\ 1&1&1\\ 2&4&4 \end{matrix} \right)^{-1} \left( \begin{matrix} 1\\0\\2 \end{matrix} \right). \end{eqnarray*}

We can find the inverse of $P$ through the following:

i) Matrix of Minors, $M$, of $P = \left( \begin{matrix} 0&2&2\\ -4&-2&0\\ -1&0&1 \end{matrix} \right). $

ii) Matrix of Cofactors, $C$, of $M = \left( \begin{matrix} 0&-2&2\\ 4&-2&0\\ -1&0&1 \end{matrix} \right). $

iii) Adjugate of $C = \left( \begin{matrix} 0&4&-1\\ -2&-2&0\\ 2&0&1 \end{matrix} \right). $

iv) Determinant of $P = -1(0) + 2(2) -1(2) = 2.$

v) Inverse of $P, P^{-1}= \frac{1}{2}C = \frac{1}{2} \left( \begin{matrix} 0&4&-1\\ -2&-2&0\\ 2&0&1 \end{matrix} \right) = \left( \begin{matrix} 0&2&-0.5\\ -1&-1&0\\ 1&0&0.5 \end{matrix} \right). $

Thus, \begin{eqnarray*} [v]_e &=& P^{-1}v\\ &=& \left( \begin{matrix} 0&2&-0.5\\ -1&-1&0\\ 1&0&0.5 \end{matrix} \right) \left( \begin{matrix} 1\\0\\2 \end{matrix} \right)\\ &=& \left( \begin{matrix} -1\\-1\\2 \end{matrix} \right). \end{eqnarray*}

  1. Let us find $T(v)$ and $[T(v)]_e$.

\begin{eqnarray*} T(v) &=& T(1,0,2)\\ &=& (-1, 3, 14)^T. \end{eqnarray*}

Expressing $T(v)$ as a linear combination of $\{e_1, e_2, e_3\}$, we get: \begin{eqnarray*} T(v) &=& (-1,3,14)^T\\ &=& a(e_1) + b(e_2) + c(e_3)\\ &=& a(-1,1,2) + b(-2,1,4) + c(-1,1,4)\\ &=& -1(-1,1,2) -2 (-2,1,4) + 6(-1,1,4). \end{eqnarray*}

Hence, $[T(v)]_e = (-1,-2,6)^T$.

  1. Verifying that $[T(v)]_e = A[v]_e:$

\begin{eqnarray*} A[v]_e &=& \left( \begin{matrix} 1&0&0\\ 0&2&0\\ 0&0&3 \end{matrix} \right) \left( \begin{matrix} -1\\-1\\2 \end{matrix} \right)\\ &=& \left( \begin{matrix} -1\\-2\\6 \end{matrix} \right).\\ \\ \text{[}T(v)\text{]}_e &=& \left( \begin{matrix} -1\\-2\\6 \end{matrix} \right) \quad \text{(from part (4.))}.\\ \\ A[v]_e &=& \text{[}T(v)\text{]}_e\\ \text{[}T(v)\text{]}_e &=& A[v]_e. \end{eqnarray*}

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