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Given: \[ T\left(\begin{bmatrix}-1 \\ -2\end{bmatrix}\right) = \begin{bmatrix}17 \\ 11\end{bmatrix} \text{ and } T\left(\begin{bmatrix}2 \\ 3\end{bmatrix}\right) = \begin{bmatrix}-30 \\ -17\end{bmatrix} \] Find a matrix such that: \[ T(v) = \begin{bmatrix}? & ? \\ ? & ? \end{bmatrix}\cdot v \]

I have no idea where to start... Any help will be much appreciate it.

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It isn't at all clear what's going on, here. What is your notation supposed to mean? – Cameron Buie Nov 9 '12 at 7:00
I have no idea what you are asking. – copper.hat Nov 9 '12 at 7:21
It helps if you look at your post after it is displayed. You should have seen that it wasn't formatted as you thought. A look at the source helped, I TeXified it. – martini Nov 9 '12 at 8:37
up vote 0 down vote accepted

The matrix $A=\begin{pmatrix}17&-30\\11&-17\end{pmatrix}$ defines your linear map $T$, but using the basis $B=[b_1,b_2]=[\binom{-1}{-2},\binom23]$ (rather than the canonical basis $E=[e_1,e_2]=\binom10,\binom01]$) at departure, while using the canonical basis $E$ at arrival: this means that $$T(xb_1+yb_2)=A\cdot\binom xy$$ (you can check that this is what is given, for $(x,y)=(1,0)$ and for $(x,y)=(0,1)$; the general case follows by linearity). I would personally write this (but the notation is not widely used) as $A={}_C\!\operatorname{Mat}_B(T)$; the idea of this notation is that the basis at departure is written at the right where the vector argument "enters", and the basis at arrival is written on the left where the result "leaves" the matrix multiplication. You want to know $M={}_C\!\operatorname{Mat}_C(T)$, the matrix when using the canonical bases all the time.

This means you want to take a vector $\binom xy$ expressed in $C$, first transform it into coordinates on $B$ (find $x',y'$ such that $\binom xy=x'b_1+y'b_2$) and then compute $A\cdot\binom{x'}{y'}$. The opposite coordinate transformation is easy: given $x',y'$ one has $\binom xy=x'b_1+y'b_2=\begin{pmatrix}-1&2\\-2&3\end{pmatrix}\cdot\binom{x'}{y'}$. But then we must have $\binom{x'}{y'}=\begin{pmatrix}-1&2\\-2&3\end{pmatrix}^{-1}\cdot\binom xy$, and one can compute $$ \begin{pmatrix}-1&2\\-2&3\end{pmatrix}^{-1} = \begin{pmatrix}3&-2\\2&-1\end{pmatrix}. $$ Now the compound operation of first doing the coordinate transform from $\binom xy$ to $\binom{x'}{y'}$ and then applying the linear map $T$ through its matrix $A$ has as matrix $M$ the matrix product $$ M = A \cdot\begin{pmatrix}-1&2\\-2&3\end{pmatrix}^{-1} =\begin{pmatrix}17&-30\\11&-17\end{pmatrix}\cdot \begin{pmatrix}3&-2\\2&-1\end{pmatrix} =\begin{pmatrix}-9&-4\\-1&-5\end{pmatrix}. $$

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Hint: $B=\{\begin{pmatrix}-1\\ -2 \end{pmatrix},\begin{pmatrix}2\\ 3 \end{pmatrix}\}$ is a basis for $\mathbb{R}^{2}$. Can you write $[T]_{B}$ ?

Can you change basis from $B$ to $B'=\{\begin{pmatrix}1\\ 0 \end{pmatrix},\begin{pmatrix}0\\ 1 \end{pmatrix}\}$ to get $[T]_{B'}$which is the desired result ?

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This is needlessly complicated. Images are already expressed in the canonical basis at arrival, so one only needs to change basis at departure, which amounts to multiplying $A=\binom{17~~-30}{11~~-17}$ on the right by the inverse of $M_B=\binom{-1~~2}{-2~~3}$. First left-multiplying $A$ by that inverse to get $[T]_B$ and then conjugating by $M_B$ is a roundabout way of doing that right-multiplication. – Marc van Leeuwen Nov 9 '12 at 13:49

The answer is $\begin{bmatrix} T(e_1) & T(e_2) \end{bmatrix}$ where $e_1 = \begin{bmatrix} 1 \\ 0 \end{bmatrix}$ and $e_2 = \begin{bmatrix} 0 \\ 1 \end{bmatrix}$. We just need to figure out what $T(e_1)$ and $T(e_2)$ are.

Let $v_1 = \begin{bmatrix} -1 \\ -2 \end{bmatrix}$ and $v_2 = \begin{bmatrix} 2 \\ 3 \end{bmatrix}$. Notice that

\begin{equation} e_1 = 3 v_1 + 2 v_2 \end{equation} and \begin{equation} e_2 = -2 v_1 - v_2. \end{equation}

These equations allow us to compute $T(e_1)$ and $T(e_2)$.

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Let's write $$\begin{bmatrix} a & b \\ c & d\end{bmatrix}$$ for the matrix we want to find. Then the two given equations read \begin{align*} -a - 2b &= 17\\ -c - 2d &= 11\\[3mm] 2a + 3b &= -30\\ 2c + 3d &= -17 \end{align*} We have to solve this system, so let's first look at the equations for $a$ and $b$ \begin{align*} -a - 2b &= 17\\ 2a + 3b &= -30 \end{align*} Adding the first two times to the second gives \begin{align*} -a - 2b &= 17\\ -b &= 4 \end{align*} Hence $b = -4$ and $a = -9$. The second system reads \begin{align*} -c - 2d &= 11\\ 2c + 3d &= -17 \end{align*} Adding again, we have \begin{align*} -c - 2d &= 11\\ -d &= 5 \end{align*} Hence $d = -5$ and $c = -1$. So $$ \begin{bmatrix} -9 & -4 \\ -1 & -5\end{bmatrix} $$ is the matrix we looked for.

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