Question from a past exam: Prove $dim(Im(T))=Tr(T)$

linear algebralinear-transformationsprojection

Found this question from a past exam.

Let $E:V \rightarrow V$ be the projection operator on a finite-dimensional vector-space $V$ over $F$.

1) Prove $V=Im(E) \oplus Ker(E)$

(I believe I proved this one but I'd like to see a more certain proof.)

2) Prove $dim(Im(E))=Tr(E)$.

Specifically: find a base $B$ of $V$ such that $dim(Im(E))=Tr([E]_B)$

I am clueless about number 2.

Best Answer

  1. a. For $v \in V$, $v = Ev + v - Ev$ and use the fact that $E^2 = E$.

    b. $Im(E) \cap Ker(E) = \{0\}$ is easy to see.

  2. (Based on the assumption that V is finite dim.) Use the fact that the eigenvalues of $E$ are all $1$ or $0$ because it's a projection.

If you aren't able to finish the proof based on this info, do comment and I'll add more hints.

Cheers!

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