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If $\varphi ,\psi$ are linear mapping $\mathbb{R}^2\rightarrow \mathbb{R}^2$, who are through $\varphi(x_1,x_2)=(x_1+x_2,0)^T$ and $\psi(x_1,x_2)=(-x_2,x_1)^T$ defined.

How do I find the matrices of the following mapping: $ \psi,\varphi, \varphi \circ \psi ,\psi\circ\varphi$?

So I think that $\varphi = \begin{pmatrix} 1 &1 \\ 0 &0 \end{pmatrix}$ and $\psi= \begin{pmatrix} 0 &-1 \\ 1 &0 \end{pmatrix}$, is this good?

If yes can someone help me how to get composition of this two?

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Your matrices look good, but $\varphi$ and $\psi$ shouldn't be assigned to them. Maybe call them $A$ and $B$. Anyway, $(\varphi \circ \psi)(x_1,x_2)$ just means 'apply $\psi$ to $(x_1,x_2)^T$, and then apply $\varphi$ to your result'.

So, we already have that $\psi(x_1,x_2) = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = B \mathbf{x} = \begin{pmatrix} -x_2 \\ x_1 \end{pmatrix}$.

Now we want to apply $\varphi$ to this vector to obtain $(\varphi \circ \psi)(x_1,x_2) = \varphi(\psi(x_1,x_2))$.

Again, we already know that we can represent $\varphi(x_1,x_2)$ by the matrix-vector product, $A \mathbf{x}$, for any vector $\mathbf{x} \in \mathbb{R}^2$. Our vector $B\mathbf{x}$ above is definitely in $\mathbb{R}^2$ by the definition of $\psi$, so to apply $\varphi$ to it, we can simply multiply by $A$. Hence,

$(\varphi \circ \psi)(x_1,x_2) = A(B\mathbf{x}) = (AB)\mathbf{x}$

So the matrix to represent $(\varphi \circ \psi)(x_1,x_2)$ is simply $AB = \begin{pmatrix} 1 &1 \\ 0 &0 \end{pmatrix} \begin{pmatrix} 0 &-1 \\ 1 &0 \end{pmatrix} = \begin{pmatrix} 1 & -1 \\ 0 &0 \end{pmatrix}$.

The case for $\psi \circ \varphi$ is pretty much identical, you can just multiply the matrices to get the composition.

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