# Change of Basis, Linear Mapping, Coordinate Vector

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.

• $T[v]_e$ 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. – amd Mar 19 at 19:14
• 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. – amd Mar 19 at 19:17

$$T(v)=Tv=[Tv]_e^T\left(\begin{array}{c} -1 & 1& 2\\ -2 & 1& 4\\ -1 & 1& 4\\ \end{array}\right)$$ $$Tv=A[v]_e^T\left(\begin{array}{c} -1 & 1& 2\\ -2 & 1& 4\\ -1 & 1& 4\\ \end{array}\right)$$ I suggest you read the chapter of linear thranformation about any linear algebra textbook.

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 $$[T(v)]_e = A[v]_e:$$

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