0
$\begingroup$

In a linear algebra book one problem is as follows: "given a linear map from $\Bbb R^2$ to $\Bbb R^2$ defined in terms of standard basis the problem is to find the image of $(x,y)$. "

The solution I know is to express $(x,y)$ as a linear combination of standard bases and then use linearity of the map and then put the values of the images of stanard bases.

In the book I see the following solution : represent the map in matrix form with respect to the standard bases and multiply it with the vector (x,y) represented as column vector. I could not understand how these two are equivalent. How is matrix multiplication coming into picture. I am aware of the fact that if $T$ and $S$ are two linear maps then wrt to a basis the matrix representation is same as the multiplication of the two matrices for each of the linear maps wrt to the same basis

$\endgroup$
1
$\begingroup$

I will write elements of $\mathbb{R}^2$ as pairs, to distinguish them from column vectors. The matrix $A$ of a linear map $T$ (w.r.t. the standard basis) is defined so that if $T(1,0)=(a_{11},a_{21})$ and $T(0,1)=(a_{12},a_{22})$, then:

$$A\begin{pmatrix}1\\0\end{pmatrix}=\begin{pmatrix}a_{11}\\a_{21}\end{pmatrix}$$ $$A\begin{pmatrix}0\\1\end{pmatrix}=\begin{pmatrix}a_{12}\\a_{22}\end{pmatrix}$$

Now given $(x,y)=x(1,0)+y(0,1)$, we also have that the column $\binom{x}{y}$ can be written as $x\left(\begin{smallmatrix}1\\0\end{smallmatrix}\right)+y\left(\begin{smallmatrix}0\\1\end{smallmatrix}\right)$.

Then $T(x,y)=xT(1,0)+yT(0,1)=(xa_{11}+ya_{12},xa_{21}+ya_{22})$ as you know, and:

$$A\begin{pmatrix}x\\y\end{pmatrix}=xA\begin{pmatrix}1\\0\end{pmatrix}+yA\begin{pmatrix}0\\1\end{pmatrix}=\begin{pmatrix}xa_{11}+ya_{12}\\xa_{21}+ya_{22}\end{pmatrix}$$

so the two calculations agree (up to swapping pairs for columns).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.