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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

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  • $\begingroup$ What do you know about the columns of a transformation matrix? Do you know how to construct a change-of-basis matrix? $\endgroup$ Commented Nov 13, 2019 at 19:09

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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.

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  • $\begingroup$ so can I say that the basis of $im(T)=span\left \{ {\begin{bmatrix} 2\\ 1\\ 0\\ 0 \end{bmatrix},\begin{bmatrix} 1\\ 2\\ 1\\ 1 \end{bmatrix},\begin{bmatrix} 2\\ 0\\ -1\\ -2 \end{bmatrix}}\right \}$ $\endgroup$ Commented Nov 13, 2019 at 21:33
  • $\begingroup$ Yes, provided you're convinced that these are indeed linearly independent, because of the rank-nullity theorem. $\endgroup$ Commented Nov 13, 2019 at 21:51

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