# Find a basis for KerT and ImT (T is a linear transformation)

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 $$KerT$$?

• 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 – 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 $$((0,0,0,0,1))$$
• The $\;\ker T\;$ cannot be trivial since then $\;T\;$ would be injective, which of course is nonsense. – DonAntonio Jun 15 '19 at 15:43