# Example of a linear map $T\epsilon \mathcal{L}(\mathbb{R}^4 \to \mathbb{R}^4)$ such that $\text{dim}[\text{null}(T)] = \text{dim}[\text{range}(T)]$

Give an example of a linear map $T\epsilon \mathcal{L}(\mathbb{R}^4 \to \mathbb{R}^4)$ such that $\text{dim}[\text{null}(T)] = \text{dim}[\text{range}(T)]$.

I came up with $$\text{Let}\ (e_1, e_2, e_3, e_4)\ \text{be a basis of}\ \mathbb{R}^4\ \text{and let}\ T \epsilon \mathcal{L}(\mathbb{R}^4 \to \mathbb{R}^4)\ \text{such that}$$ $$T(e_1)=T(e_2)=0$$$$T(e_3)=e_3$$$$T(e_4)=e_4$$.

Is this example so trivial as to be invalid, and if so, what would have been a better example?

• Seems wrong. $null(T) = span(e_1,e_2)$, $range(T) = span(e_3,e_4)$. Oct 21, 2016 at 20:37
• This is wrong. You should put $T(e_3)=e_2$ and similarly for $e_4$ (for example) to make it valid. Besides that, no, it's not that trivial, as in normal form they very much look like the one you wanted to produce Oct 21, 2016 at 20:38
• Many apologies: I forgot to mention that I'm interested in dimension of the kernel and the image. Oct 21, 2016 at 20:42
• By the rank-nullity theorem, any linear map with $\dim(\operatorname{null}(T)) = 2$ will suffice. Oct 21, 2016 at 20:48

$\textbf{Answer to originally phrased question.}$

You are very close. You can fix it with:

$$T(e_3) = T(e_4) = 0$$ $$T(e_1) = e_3$$ $$T(e_2) = e_4$$

Now you have the $\text{null}(T) = \text{span}(e_3,e_4) = \text{range}(T).$

$\textbf{Answer to editted question.}$

Any $4 \times 4$ matrix with two columns equal to zero and the other two linearly independent will give you a transformation with this property.