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"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?

3 Answers 3

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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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    You are right, how do I prove it then? – 2017-02-16
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    **Hint:** it's very easy to put a diagonal matrix in row echelon form. What could the row echelon form of a diagonal matrix look like? Or if you prefer, what does the kernel of a diagonal matrix look like? Once you have either: given a diagonal matrix, how can you quickly find its rank? Why does squaring the diagonal matrix give you something of the same rank, then? – 2017-02-16
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    Once you have the right fact about invertible matrices, the last steps are obvious. – 2017-02-16
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    @Omnomnomnom Sure, but I don't know what facts the OP knows. Of course some of the steps that I mentioned can be replaced by the / turned into appropriate lemmas. – 2017-02-16
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    @MichaelBurr that was meant for the OP, if that wasn't clear. – 2017-02-16
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    @Omnomnomnom Thanks for the clarification, I thought you meant it towards me! – 2017-02-16
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Why don't you use the fact that multiplying a matrix by on the left side by the inverse of an invertible and on the right side by an invertible matrix is a rank preserving operation? So, we'd have

$rank(P^{-1}DP) = rank(P^{-1}D^2P^{-1})\quad \Longrightarrow \quad rank(D) = rank(D^2)$.

It is clear that the rank of $D$ and $D^2$ are equivalent, since the all the rows are clearly linearly independent. Or in other words, the entry in each pivot of the matrix is a non-zero number, which implies that the nullity of the linear operator that induces the diagonal representation is zero. Hence, the ranks are the same.

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Alternatively, if $A$ is diagonalisable, $A= P^{-1}DP$ implies $A^2= P^{-1}D^2P$ so $A^2$ is diagonalisable and if $c_i$ is an eigenvalue of $A$ with multiplicity $a_i,$ then $c_i^{2}$ is an eigenvalue of $A^2$ with multiplicity $a_i.$

So if $p(x)= \displaystyle \prod_{i=1}^n(x-c_i)^{a_i}$ is the characteristic polynomial of $A,$ then $q(x)= \displaystyle \prod_{i=1}^n(x-c_i^{2})^{a_i}$ is the characteristic polynomial of $A^2.$ The nullity of $A$ is the geometric multiplicity of $0$ as an eigenvalue and since $A$ is diagonalisable, this coincides with the algebraic multiplicity and hence the nullity of $A$ is $a_i$ if $c_i =0$ and $0$ otherwise. Similarly, the nullity of $A^2$ is $a_i$ if $c_i^{2}=0$ and $0$ otherwise. Hence nullity of $A$ = nullity of $A^2$ and thus $\text{rank}(A)= \text{rank}(A^2).$