Linear transformation with respect to basis problem

Question

Let $T: \mathbb{R}^2 \to \mathbb{R}^2$ such that $T(\begin{bmatrix} 3 \\ 1 \end{bmatrix}) = \begin{bmatrix} 1 \\2 \end{bmatrix}$ and $T(\begin{bmatrix} -1 \\ 0 \end{bmatrix}) = \begin{bmatrix} 1 \\1 \end{bmatrix}$. Find the matrix $A$ representing $T$.

I understand that to approach this problem, I have to view $\begin{bmatrix} 3 \\1 \end{bmatrix}$ and $\begin{bmatrix} -1 \\ 0 \end{bmatrix}$ as a basis, $B = \{ v_1, v_2 \}$, where $[v1 v2]$ is the transition matrix from $[x]B$ to $x$. How do I use $\begin{bmatrix} 1\\2 \end{bmatrix}$ and $\begin{bmatrix} 1 \\1 \end{bmatrix}$? I'm unclear on their connection to the basis vectors $v_1$ and $v_2$.

Answer 1

Relative to the bases $B$ and the standard basis, the matrix is:. $\begin{pmatrix}1&1\\2&1\end{pmatrix}$.

The change of basis matrix is: $\begin{pmatrix}3&-1\\1&0\end{pmatrix}$.

The latter changes basis from $B$ to the standard basis.

Thus you want: $\begin{pmatrix}1&1\\2&1\end{pmatrix}\begin{pmatrix}3&-1\\1&0\end{pmatrix}^{-1}$.

I will leave this calculation for you.

$\begin{pmatrix}1&1\\2&1\end{pmatrix}\begin{pmatrix}0&1\\-1&3\end{pmatrix}=\begin{pmatrix}-1&4\\-1&5\end{pmatrix}$

Answer 2

Let T be the transformation represented by $\begin{pmatrix}a & b \\ c&d\end{pmatrix}$

Now, we are given that $$T([3,1]) = [1,2], T([-1, 0]) = [1,1]$$

Applyin the transformation for the first equation

$$\begin{pmatrix}a & b \\ c&d\end{pmatrix}.\begin{pmatrix} 3 \\ 1 \end{pmatrix} = \begin{pmatrix}1\\2\end{pmatrix}$$

This will give you two equations in $a,b,c,d$. Similarly, you will get two equations from the other equation. Can you proceed?

This is not the simplest way of solving it, but is the most basic. Other techniques are all based on this. Faster methods are available, but I suggest going through this exercise at least once

Answer 3

Let $\alpha$ be the standard ordered basis for $\mathbb R^2$ and $\beta = \{(3,1),(-1,0)\}$. Then we have $$\begin{align} [\textsf{T}]_\alpha &= [\textsf{T} \circ \operatorname{id}]_\alpha \\ &= [\textsf{T}]_\beta^\alpha [\operatorname{id}]_\alpha^\beta \\ &= [\textsf{T}]_\beta^\alpha ([\operatorname{id}]_\beta^\alpha)^{-1}. \end{align}$$ Now, $[\textsf{T}]_\beta^\alpha$ is already given and $[\operatorname{id}]_\beta^\alpha$ is just the matrix whose columns are the vectors of $\beta$.

Answer 4

$\begin {bmatrix} 1\\0 \end{bmatrix} = - \begin {bmatrix} -1\\0 \end{bmatrix}$

$T(\begin {bmatrix} 1\\0 \end{bmatrix} ) = -T(\begin {bmatrix} -1\\0 \end{bmatrix}) = -\begin {bmatrix} 1\\1 \end{bmatrix} = \begin {bmatrix} -1\\-1 \end{bmatrix}$

$\begin {bmatrix} 0\\1 \end{bmatrix} = \begin {bmatrix} 3\\1 \end{bmatrix} + 3\begin {bmatrix} -1\\0 \end{bmatrix}$

$T(\begin {bmatrix} 1\\0 \end{bmatrix} ) = T(\begin {bmatrix} 3\\1 \end{bmatrix}) + 3T(\begin {bmatrix} -1\\0 \end{bmatrix}) = \begin {bmatrix} 1\\2 \end{bmatrix} + 3\begin {bmatrix} 1\\1 \end{bmatrix} = \begin {bmatrix} 4\\5 \end{bmatrix}$

$T = \begin {bmatrix} -1&4\\-1&5 \end{bmatrix}$