# Product of two symmetric matrices with eigenvalues all 0 or 1 is idempotent

Assume $A$, $V$ are symmetric and $V$ is positive definite.

If $AV$ has eigenvalues that are all zero or one, show $AV$ is idempotent.

My proof so far (haven't used the symmetric property or the fact that $V$ is positive definite - where do those come in?):

If the eigenvalue $\lambda=0$, and assume there exists a nonzero $u$ such that

$$AVu=\lambda u=0u=0$$ $$(AV)^2u=AVAVu=AV0=0$$

Then $(AV)^2u=AVu=0$ implies $(AV)^2=AV$ since $u\ne 0$.

If $\lambda=1$, and assume there exists nonzero $u$ such that

$$AVu=\lambda u=u$$

$$(AV)^2u=AVAVu=AVu$$

$$\implies (AV)^2=AV$$

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You've only proven that $(AV)^2 u = (AV) u$ so long as $u$ can be written as the sum of eigenvectors of $AV$, which in a more general case may or may not be any vector. The real question is this: how do we know that $AV$ must be diagonalizable? – Omnomnomnom Jan 14 '14 at 13:44
Note that if $\sigma(A) = \{0,1\}$, then $A$ is idempotent if and only if it is diagonalizable. – Omnomnomnom Jan 14 '14 at 13:56

As I stated in my comment, it is sufficient to prove that $AV$ is diagonalizable. Note that if $M$ is a matrix whose eigenvalues are $0$ and $1$, then $M$ is idempotent if and only if $A$ is diagonalizable.

### Proof that $A$ is idempotent $\implies$ $A$ is diagonalizable:

Suppose $A$ is idempotent. Then $A^2 - A = 0$. It follows that the minimal polynomial of $A$ divides $x^2 - x = x(x-1)$. It follows that $A$ is diagonalizable (why?)

I will leave it to you to prove the converse, which is necessary for the proof of your statement (Hint: if $A = SDS^{-1}$, what does $A^2$ look like? If $\lambda = 0,1$ are the only eigenvalues, what does $D$ look like?).

With that in mind...

### Proof that $AV$ is diagonalizable

Because $A$ is positive definite, it has a unique positive definite square root given by $R = A^{1/2}$. We note that $$R^{-1}AVR= A^{-1/2}AVA^{1/2} = A^{1/2}VA^{1/2} = RVR$$ We note that $(RVR)^*=R^*V^*R^* = RVR$ (${}^*$ here denotes the conjugate transpose). It follows that $RVR$ is diagonalizable, i.e., that $RVR = S D S^{-1}$ for some invertible matrix $S$.

Thus, we have $R^{-1}(AV)R = S^{-1}DS$, which we rearrange to find $$AV = (RS^{-1})D(RS^{-1})^{-1}$$ Thus, we conclude that $AV$ is diagonalizable, as desired. Since $AV$ has only $0$ and $1$ as its eigenvalues, we may further conclude that it is idempotent.

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thank you for such a clear argument. do i still need to use what i wrote above or just state since $AV$ has only 0 and 1 as eigenvalues, it is then idempotent? – lightfish Jan 15 '14 at 5:46
What you wrote above is a fine proof that if a matrix is diagonalizable and its eigenvalues are either $0$ or $1$, then that matrix is idempotent. The key is that in such a case (and in fact only in such a case), every $u$ can be written as the linear combination of eigenvectors. – Omnomnomnom Jan 15 '14 at 12:36
@omnomnomnomm , is $RVR$ diagonalizable because there exists an invertible $R$ such that $RVR$ can be written in the form $R^{-1}WR$? Also, why do we need to show it is conjugate transposable? thanks! – lightfish Jan 25 '14 at 21:22
What I showed is that $RVR$ is equal to its own conjugate transpose. That is, $RVR$ is Hermitian. Hermitian matrices are normal, and all normal matrices are unitarily diagonalizable. – Omnomnomnom Jan 25 '14 at 21:31
Also, it is a useful and well known fact that a matrix is Hermitian if and only if it is unitarily diagonalizable and all of its eigenvalues are real. – Omnomnomnom Jan 25 '14 at 21:32