0
$\begingroup$

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.

  • 1
    also you could take $f \equiv 0$2012-08-25

1 Answers 1

3

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.