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Find bases of kernel and image of $T$.

$T(M)=\begin{bmatrix} 1 & 1 \\ 2 & 2 \end{bmatrix} M$ from $\mathbb{R}^{2\times 2}$ to $\mathbb{R}^{2\times 2}$ with respect to the standard basis.

The transformation matrix is $T=\begin{bmatrix} 1 & 0 & 1 & 0 \\ 0 & 1 & 0 & 1 \\ 2 & 0 & 2 & 0 \\ 0 & 2 & 0 & 2 \end{bmatrix}$ and by inspection there's two linearly independent vector columns, so the rank is 2, and $ker(T)$ only occurs when $a=-c$ and $b=-d$ for a matrix $M=\begin{bmatrix} a & b \\ c & d \end{bmatrix}$, so ker(T) basis is $\begin{bmatrix} 1 & 0 \\ -1 & 0 \end{bmatrix}$, $\begin{bmatrix} 0 & 1 \\ 0 & -1 \end{bmatrix}$, but what's the image of $T$ then?

The image can only have two elements but how do you show all elements where $a\neq -c$ and $b\neq -d$ using only two matrices?

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The image of a linear transformation can be find as follows: you have found a matrix representation of your linear transformation. It follows that the image of this linear transformation is exactly the space spanned by the columns of your matrix, indeed: the image consists out of linear combinations of the columns of your matrix, so you find that the image in this case is spanned by the matrices $$\begin{pmatrix} 1 & 0\\2 & 0\end{pmatrix} \quad \text{ and } \quad \begin{pmatrix} 0 & 1\\0 & 2\end{pmatrix}.$$

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