0
$\begingroup$

I have a linear transformation $T: \mathbb{R}^2 \to \mathbb{R}^2$, and all I know about is that it sends $$ \begin{bmatrix} 3 \\ 2 \end{bmatrix} \mapsto \begin{bmatrix} 6 \\ 8 \end{bmatrix}, \; \begin{bmatrix} 1 \\ 2 \end{bmatrix} \mapsto \begin{bmatrix} 5 \\ 5 \end{bmatrix}. $$ One suggested method for solving this (finding $[T]$ relative to the standard basis) in the lecture notes I'm following is to set up the equation: $$ [T] \begin{bmatrix} 3 & 1 \\ 2 & 2 \end{bmatrix} = \begin{bmatrix} 6 & 5 \\ 8 & 5 \end{bmatrix}. $$ Even though this ends up working, I cannot understand why it does or how this encodes what the linear transformation does. I know that there are three sets of bases at play: \begin{align*} \mathcal{B}_1 &= \left\{\begin{bmatrix} 3 \\ 2 \end{bmatrix}, \begin{bmatrix} 1 \\ 2 \end{bmatrix}\right\} \\ \mathcal{B}_2 &= \left\{\begin{bmatrix} 6 \\ 8 \end{bmatrix}, \begin{bmatrix} 5 \\ 5 \end{bmatrix} \right\} \\ \mathcal{B}_3 &= \left\{\begin{bmatrix} 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 1 \end{bmatrix}\right\} \end{align*} I tried to think of the above equation in $[T]$ as a change of basis. The columns of $T$ apply the vectors in $\mathcal{B}_2$ and write them in terms of the standard basis. The second matrix writes the $\mathcal{B}_1$ vectors in terms of the standard basis. The third matrix writes the $\mathcal{B}_3$ vectors in terms of the standard basis. As a result, this doesn't look like change of basis, so I'm having some trouble figuring out exactly how this is contructed.

$\endgroup$
3
  • 1
    $\begingroup$ The linear transformation has the form $T(x,y) =(a_{11}x+a_{12}y,a_{21}x+a_{22}y)$, where the $a_{ij}$ are the entries of $[T]$. You are setting up a system of four equations by plugging in the inputs you have, and the corresponding outputs, and then solving. The fact that the outputs form a basis is coincidence, so you should ignore $B_2$. You could have both outputs equal, and still have a linear transformation. $\endgroup$ Oct 27, 2023 at 16:35
  • $\begingroup$ @ArturoMagidin Thank you, this is very helpful. The only thing I'm still confused on is how the system of four equations that results from plugging in the two known inputs $(x,y)$ to $T$ translates into the matrix equation. $\endgroup$ Oct 27, 2023 at 21:16
  • 1
    $\begingroup$ Multiplying the matrix $(a_{ij})$ by $(3,2)^T$ results in $(6,8)^T$. So you would compute $[T](3,2)^T = (6,8)^T$, to get equations for the $a_{ij}$. Multiplying the matrix of $T$ by $(1,2)^T$ gives $(5,5)^T$, so you would be computing $[T](1,2)^T = (5,5)^T$ to get the other equations for the $a_{ij}$. Rather than do those two computations, you can do them simultaneously in matrix form, since the first column of $\left(\begin{array}{cc}3&1\\2&2\end{array}\right)$ is what gives the first column of the answer, and the second column is what gives the second column of the answer. $\endgroup$ Oct 27, 2023 at 21:29

1 Answer 1

1
$\begingroup$

You want to figure out the entries of $[T]$, which is of the form $$\left(\begin{array}{cc} a_{11} & a_{12}\\ a_{21} & a_{22} \end{array}\right)$$ where $T\left(\begin{array}{c}1\\ 0\end{array}\right) = \left(\begin{array}{c}a_{11}\\ a_{21}\end{array}\right)$, and $T\left(\begin{array}{c}0\\ 1\end{array}\right) = \left(\begin{array}{c}a_{12}\\ a_{22}\end{array}\right)$.

Each of the equations you have gives you two equations in the $a_{ij}$: $$\begin{align*} \left(\begin{array}{c} 6\\ 8\end{array}\right) = T\left(\begin{array}{c}3\\ 2\end{array}\right) &= \left(\begin{array}{c} 3a_{11} + 2a_{12}\\ 3a_{21} + 2a_{22}\end{array}\right);\\ \left(\begin{array}{c}5\\ 5\end{array}\right) = T\left(\begin{array}{c}1\\ 2\end{array}\right) &= \left(\begin{array}{c} a_{11} + 2a_{12}\\ a_{21} + 2a_{22} \end{array}\right). \end{align*}$$

You then take the four equations and solve the system of four equations in four unknowns to get the values of the $a_{ij}$.

But you can do this in one go with a matrix equation. Because when you compute $AB$, the first column of $AB$ is just $A\mathbf{b}_1$, where $\mathbf{b}_1$ is the first column of $B$; and the second column is $A\mathbf{b}_2$, where $\mathbf{b}_2$ is the second column of $B$, etc. So we can write out the two vector equations at the same time as a matrix equation: $$\left(\begin{array}{cc} a_{11} & a_{12}\\ a_{21} & a_{22} \end{array}\right) \left(\begin{array}{cc} 3&1\\ 2&2\end{array}\right) = [T]\left(\begin{array}{c|c} \mathbf{v}_1 & \mathbf{v}_2\end{array}\right) = \left(\begin{array}{c|c} T\mathbf{v}_1 & T\mathbf{v}_2\end{array}\right) = \left(\begin{array}{cc} 6 & 5\\ 8 & 5 \end{array}\right),$$ where $\mathbf{v}_1 = \left(\begin{array}{c}3\\ 2\end{array}\right)$ and $\mathbf{v}_2 = \left(\begin{array}{c}1\\ 2\end{array}\right)$. And that is the matrix equation you have.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .