# Linear Algebra: Basis Question

Let $b_1 = \left(\begin{matrix}-1 \\ -3\end{matrix}\right)$ and $b_2 = \left(\begin{matrix} -3 \\ -10\end{matrix}\right)$. The set $B = \{b_1, b_2\}$ is a basis for $\mathbb{R}^2$. Let $T: \mathbb{R}^2 \to \mathbb{R}^2$ is a linear transformation such that $T(b_1) = 4b_1 + 4b_2$ and $T(b_2) = 4b_1 + 5b_2$. Then the matrix of $T$ relative to the basis $B$ is

$$[T]_B = \left(\begin{matrix} \color{red}{4} && \color{red}{4} \\ \color{red}{4} && \color{red}{5} \end{matrix}\right),$$

and the matrix of $T$ relative to the standard basis $E$ for $\mathbb{R}^2$ is

$$[T]_E = \left(\begin{matrix} \color{red}{?} && \color{red}{?} \\ \color{red}{?} && \color{red}{?}\end{matrix}\right).$$

I managed to figure out the first part. I think that the standard basis $E$ would be $e_1 = (1,0)$ and $e_2 = (0,1)$ but I don't know how to use this to solve the second half of the question.

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Find $a_1, a_2 \in \mathbf R$ such that $a_1b_1 + a_2b_2 = (1, 0)$. You need to make the second coordinate zero, i.e. arrange for $-3a_1 - 10a_2 = 0$. –  Dylan Moreland Mar 26 '12 at 15:24
en.wikipedia.org/wiki/Change_of_basis –  Rasmus Mar 26 '12 at 15:25
@DylanMoreland I'm not sure that there is an a1 and a2 that meet this requirement... –  StickFigs Mar 26 '12 at 15:39
@DylanMoreland Ok I got a1 = {-10,-10} and a2 ={3,3} but these are not the correct answers... –  StickFigs Mar 26 '12 at 15:52

Recall theorem :We know that $[T]_E=P[T]_BP^{-1}$ where $P=[\operatorname{id}]_{E,B}=[b_1 \,\, b_2]$. Now you can easily solve it.

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I get it now! Thank you! –  StickFigs Mar 26 '12 at 16:04

Recall what the matrix $[\, T\,]_B$ is:

If you write a vector $x\in\Bbb R^2$ in terms of the basis $B=\{\,b_1,b_2\,\}$ $$x=\alpha_1 b_1+\alpha_2 b_2,$$ then if you multiply $[\, T\,]_B$ by the coordinate vector $x_B=\bigl[{\alpha_1\atop\alpha_2}\bigr]$, you get the coordinate vector of $T(x)$ with respect to $B$. That is $$\tag{1} [T(x)]_B = [\,T\,]_B x_B.$$
Now the columns of the matrix $[\,T\,]_E$ where $E=\{e_1,e_2\}$ are the vectors $T(e_1)$ and $T(e_2)$. To find these vectors, we can use the matrix $[\, T\,]_B$. There are three steps involved here. Considering the vector $e_2$, we have to

1. Find the coordinates of $e_2$ with respect to the basis $B$.
2. Find the coordinates of $T(e_2)$ with respect to the basis $B$.
3. Find $T(e_2)$ expressed in the standard basis.

Step 1: For $e_2=(0,1)$, we first find the coordinates of $e_2$ in terms of the basis $B$. Towards this end, we have to solve the system $$\Bigl[{0\atop1}\Bigr]= \alpha_1 \Bigl[{-1\atop-3}\Bigr] +\alpha_2\Bigl[{-3\atop-10}\Bigr].$$ Doing so gives: $$\alpha_1=3,\quad \alpha_2=-1$$ The coordinate vector of $e_2$ with respect to $B$ is $\bigl[{3\atop-1}\bigr]$.

Note we could have done this differently: the coordinate vector $\bigl[{\alpha_1\atop\alpha_2}\bigr]$ of $x$ with respect to $B$ satisfies $[b_1 \ b_2] \bigl[{\alpha_1\atop\alpha_2}\bigr] =x$; so $\bigl[{\alpha_1\atop\alpha_2} \bigr]=[b_1 \ b_2]^{-1} x$. Thus we could have found $[b_1\ b_2]^{-1}$ and just multiplied this by $e_2$. This is actually preferable, since we can use the inverse when considering $e_1$ later.

Step 2: Using $(1)$ now, the coordinate vector of $T(e_2)$ with respect to $B$ is $$[\,T\,]_B (e_2)_B = \Bigl[ \matrix{4&4\cr 4&5 }\Bigr]\Bigl[{3\atop -1} \Bigr] = \Bigl[{8\atop 7} \Bigr].$$

Step 3: But note that $T(e_2)$ is not the vector $\bigl[{8\atop 7} \bigr]$; this vector gives the coordinates of $T(e_2)$ with respect to the basis $B$. In general, if $x_B$ is the coordinate vector of $x$ with respect to $B$, then $x=[b_1\ b_2] x_B$; so $$T(e_2)=[b_1\ b_2]\Bigl[{8\atop 7} \Bigr] =8\, b_1+7\, b_2= 8\Bigl[{-1\atop -3} \Bigr] +7 \Bigl[{-3\atop -10} \Bigr] =\Bigl[{-29\atop -94} \Bigr].$$

Thus, the second column of $[\,T\,]_E$ is $\bigl[{-29\atop -94} \bigr]$.

To find the first column of $[\,T\,]_E$, apply the same procedure to the vector $e_1$. The first step here would be to write $e_1$ in terms of the basis $B$. To do that, you need to solve the system $$\Bigl[{1\atop0}\Bigr]= \alpha_1 \Bigl[{-1\atop-3}\Bigr] +\alpha_2\Bigl[{-3\atop-10}\Bigr].$$ or compute: $$\Bigl[{\alpha_1\atop\alpha_2} \Bigr]=[b_1 \ b_2]^{-1} e_1.$$

In general, given $x$ written in terms of the standard basis, the coordinates of $x$ with respect to $B$ are given by $P^{-1} x$ where $P=[b_1\ b_2]$. Then the coordinates of $T(x)$ with respect to $B$ are $[\,T\,]_B P^{-1} x$. Then we have that $T(x)$ expressed in terms of the usual basis is $P [\,T\,]_B P^{-1} x$. So, $$[\,T\,]_E = P[\,T\,]_B P^{-1}.$$

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Well we need the values of $T(e_1), T(e_2)$ in terms of $e_1,e_2$.

So first we are going to have to write $b_1, b_2$ in terms of $e_1,e_2$.

If we do this and use it in our definition of $T$ then we see that:

$-T(e_1) - 3T(e_2) = -16e_1 -52e_2$

$-3T(e_1) - 10T(e_2) = -19e_1 -62e_2$

We can now solve these equations for $T(e_1), T(e_2)$ and get:

$T(e_1) = 103e_1 +334e_2$

$T(e_2) = -29e_1 - 94e_2$

From this you get the matrix of $T$ with respect to the standard basis.

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