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Construct a linear transformation T : R4 → R4 such that Kernel(T) = Image(T). How about the same for a linear transformation S : R5 →R5.

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    $\begingroup$ Welcome to Mathematics Stack Exchange. How about $(a,b,c,d)\mapsto (c,d,0,0)$? Also, note dim(Ker)+dim(Im)=dim(space) $\endgroup$ Nov 3, 2019 at 13:20

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Take a base $v_i$ and a map $v_1 \mapsto 0$, $v_2\mapsto 0$, $v_3 \mapsto v_1$ and $v_4 \mapsto v_2$, the kernel and image seems isomorphic.

For odd dimensional spaces, the dimension will provide obstruction, because since isomorphic spaces have the same dimension:

$$ \dim \ker S + \dim \text{Im } S = 2 \dim \ker S = 5 $$

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Take $T:(a,b,c,d)\mapsto (c,d,0,0)$ for $\mathbb R^4$,

but for $\mathbb R^5$ there is none,

because dim(Ker)+dim(Im)=dim(space),

so if dim(Ker)=dim(Im) then dim(space) is even.

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