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Let $T:\mathbb{R}^5\rightarrow \mathbb{R}^4$ Be a linear transformation such that $T\left(\begin{bmatrix} x\\ y\\ z\\ t\\ w\end{bmatrix}\right)$ = $\begin{bmatrix} x-2y\\ y-2z\\ z-2t\\ t-2x\end{bmatrix}$

I managed to find the basis and the dimension for $ImT$ pretty easily, however how do I formally prove the dimension and the basis for $KerT$?

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  • $\begingroup$ You can, for instance, use the "dimensions. thm", and you have that $5=\dim(Ker) + \dim(Im)$ , so you know which is the dimension of the kernel. In order to find a basis for the kernel, you just need to solve the linear system $Ax=0$, where $A$ is the representative matrix w.r.t the canonical basis $\endgroup$ – VoB Jun 15 '19 at 14:52
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For the kernel, notice that you must have $x=2y=4z=8t=16x$ so that $x=0$ and from the other equations, $x=y=z=t=0$. It follows that the kernel is $ \{(0,0,0,0,w), w\in\mathbb{R}\}$ so that the kernel is of dimension 1 and a basis is $((0,0,0,0,1))$

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    $\begingroup$ The $\;\ker T\;$ cannot be trivial since then $\;T\;$ would be injective, which of course is nonsense. $\endgroup$ – DonAntonio Jun 15 '19 at 15:43
  • $\begingroup$ Indeed, didn't pay attention to the last variable. Corrected my post, thanks ! $\endgroup$ – aleph0 Jun 15 '19 at 15:45

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