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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?

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    $\begingroup$ Seems wrong. $null(T) = span(e_1,e_2)$, $range(T) = span(e_3,e_4)$. $\endgroup$
    – Simon
    Commented Oct 21, 2016 at 20:37
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    $\begingroup$ 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 $\endgroup$
    – b00n heT
    Commented Oct 21, 2016 at 20:38
  • $\begingroup$ Many apologies: I forgot to mention that I'm interested in dimension of the kernel and the image. $\endgroup$ Commented Oct 21, 2016 at 20:42
  • $\begingroup$ By the rank-nullity theorem, any linear map with $\dim(\operatorname{null}(T)) = 2$ will suffice. $\endgroup$
    – arkeet
    Commented Oct 21, 2016 at 20:48

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

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

$\textbf{Answer to editted question.}$

Your example is good :)

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.

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  • $\begingroup$ Apologies: I forgot to mention that I was comparing the dimension of the kernel to the dimension of the image $\endgroup$ Commented Oct 21, 2016 at 20:43
  • $\begingroup$ @HandsomeGorilla No worries, I just didn't want someone to come in and berate me for saying your example was wrong. $\endgroup$
    – Ken Duna
    Commented Oct 21, 2016 at 20:44
  • $\begingroup$ Explicitly what one of these transformations would look like? $\endgroup$
    – Hopmaths
    Commented Feb 8, 2022 at 16:33

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