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 $KerT$?

  • $\begingroup$ 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

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 $((0,0,0,0,1))$

  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.