Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
1  
also you could take $f \equiv 0$ –  clark Aug 25 '12 at 1:38

1 Answer 1

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.