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Suppose we have a linear transformation $f: V \to V$. How is it possible that $\dim(\operatorname{im}(f \circ f))$ is larger than $\dim(\operatorname{im}(f))+\dim(\operatorname{im}(f)) - \dim(V)$?

It's a question on a past exam, so there should be an example that proves this.

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also you could take $f \equiv 0$ – clark Aug 25 '12 at 1:38
up vote 3 down vote accepted

For example, take $f$ on ${\mathbb R}^2$ given by the matrix $\pmatrix{1 & 0\cr 0 & 0\cr}$. Then $im(f \circ f) = im(f) = {\mathbb R} \times \{0\}$, and $1 > 1 + 1 - 2$. I don't know what you mean by "increase" in the title of this question, there's nothing increasing here.

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