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Matrix A describes linear mapping $\varphi : \mathbb{R}^n\rightarrow \mathbb{R}^m$. How to find base of the kernel and of the image of mapping?

$A=\begin{pmatrix} 1 &3 &1 &0 \\ 0& 0 & 1 &3 \end{pmatrix}$

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Your matrix $A:\mathbb{R}^4 \to \mathbb{R}^2$ has full rank (columns $2$ and $4$ are linearly independent) so the image is $\mathbb{R}^2$ and any two linearly independet vector of this space are a basis.

Fo the kernel solve the equation $A\vec x=0$ that is: $$ \begin{cases} x+3y+z=0\\ z+3t=0 \end{cases} $$ that gives the solution $\vec x=(x,-x/3+t,-3t,t)^T$ now chose two different values for $x$ and $t$ and you have a basis.

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