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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

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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).

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