[Math] If $A$ is a diagonalizable matrix then rank($A$)=rank($A^2$)

diagonalizationlinear algebramatricesmatrix-rank

"Prove that if $A$ is a diagonalizable matrix then rank($A$)=rank($A^2$)".

This is what I had in mind:

$ D = $$ P^{-1} $ $A$ $P$ is diagonal.

$ D^2 = P^{-1} A P P^{-1} A P= P^{-1} A^2 P$ is diagonal.

Therefore, rank($D$)=rank($D^2$).

rank($P^{-1}AP$)=rank($P^{-1}A^2P$)

$P$ is invertible, therefore rank($P$)=rank($P^{-1}$)

and rank($A$)=rank($A^2$).

Is this proof legitimate? Or is something missing?

Best Answer

You're missing a lot of details:

  1. Why is $\operatorname{rank}(D)=\operatorname{rank}(D^2)$?

  2. How does $\operatorname{rank}(P)=\operatorname{rank}(P^{-1})$ help you? Perhaps you need a different fact about the rank of invertible matrices.

  3. There seems to be a big jump to the last line of $\operatorname{rank}(A)=\operatorname{rank}(A^2)$. How did the facts that you list combine to give this equality?

Hints :

  1. No where in your "proof" do you do any work. You either need to cite theorems and show how they apply or actually do a computation.

  2. Show that $D$ and $D^2$ have nonzero entries in exactly the same positions. Use this to argue that their column spans are identical.

  3. Explain why a invertible matrix will take a subspace of dimension $k$ to another subspace of dimension $k$.

  4. Use 2 and 3 together to argue that the column span of $A$ and $A^2$ are identical.

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