Let $T$ be a linear operator $\mathbb{R}^4\rightarrow\mathbb{R}^4$ such that
$$
T\begin{bmatrix} 2\\1\\0\\0 \end{bmatrix}=\begin{bmatrix} 2\\1\\0\\0 \end{bmatrix},\;
T\begin{bmatrix} 3\\1\\0\\0 \end{bmatrix}=\mathbf{0}\:,\:
T\begin{bmatrix} 1\\2\\1\\1 \end{bmatrix}=\begin{bmatrix} 1\\2\\1\\1 \end{bmatrix},\:
T\begin{bmatrix} 2\\0\\-1\\-2 \end{bmatrix}=\begin{bmatrix} -2\\0\\1\\2 \end{bmatrix} $$
Find the standard matrix of the linear operator T. Find the basis of the image of T, and find the basis of the kernel of T.
I'm not sure where to even start in terms of finding the standard matrix
Best Answer
Hint: By linearity, calculate $Te_i$ with the standard basis vectors $e_i$, these give the $i$th columns of the standard matrix.
You already have given $3$ linearly independent vectors in the image of $T$, and a vector in the kernel.