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I am trying to prove using Eigen values that if a symmetric matrix $A$ is positive definite then so is $A^2$.

Now let $\lambda$ be an Eigen value of $A^2$. I want to show that it is positive. Now if it is a square of an Eigen value of $A$ then I am done.

Clearly if $\mu$ is an Eigen value of $A$ then $\mu^2$ is an Eigen value of $A^2$.

But it is not clear to me why any Eigen value of $A^2$ is the square of an Eigen value of $A$.

Any help would be appreciated.

R_D
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    If $v$ is an eigenvector of $A$, what does $A^2v$ equal? – amd May 30 '18 at 07:25
  • Thank you Guido and egreg for your answers. This is a question from Strang's book and it specifically asks to use Eigen values. That is why I asked for it in this way. If you could still answer as per my requirement I would greatly appreciate it. – R_D May 30 '18 at 07:38
  • @R_D I added a final paragraph before seeing your previous request. – egreg May 30 '18 at 07:39
  • @amd The eigenvalues of $A$ are only strictly positive if $A$ is symmetric –  May 30 '18 at 07:39
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    @James False and irrelevant. – amd May 30 '18 at 07:41

5 Answers5

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This is a case where going with the definitions is better.

If $A$ is any matrix, then $A^TA$ is positive semidefinite, because $$ x^T(A^TA)x=(Ax)^T(Ax)\ge0 $$ This also shows that if $A$ is square and invertible, then $A^TA$ is positive definite, because $x^T(A^TA)x=0$ implies $Ax=0$ and so $x=0$.

Since your $A$ is symmetric and invertible…

If you prefer to do it with the eigenvalues, recall that a symmetric matrix is diagonalizable (with an orthogonal matrix, but this is not needed here). Then $A=SDS^{-1}$ and $A^2=SD^2S^{-1}$.

egreg
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There exists $P$ invertible and $J$ in the canonical Jordan form such that

$$ A = PJP ^{-1} $$

and therefore

$$ A^2 = PJP^{-1}PJP^{-1} = PJ^2P^{-1} $$

you can square $J$ by blocks, and $J^2$ will be upper triangular with $(J^2)_{ii} = (J_{ii})^2$. Hence, the eigenvalues of $J~^2$ will be those of $J$, squared. It suffices then to note that the eigenvalues of $A^2$ are those of $J^2$.

To prove this last claim, we can more generally assert the following: if $C$ and $D$ are conjugate matrices, they share the same eigenvalues. Since being conjugates is a symmetric relation, it will suffice to prove that if $\lambda$ is an eigenvalue of $C$, it is an eigenvalue of $D$. Let $v$ be an eigenvector of eigenvalue $\lambda$ for $C$, that is

$$ Cv = \lambda v $$

Since $C$ and $D$ are conjugates, there exists $Q$ invertible with $C = QDQ^{-1}$, and

$$ Cv = \lambda v \iff QDQ^{-1}v = \lambda v \iff DQ^{-1}v = \lambda Q^{-1}v $$

so $Q^{-1}v$ is an eigenvector of eigenvalue $\lambda$ for D.

qualcuno
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If you go with the classical definition.

Since $A$ is symmetric, so $A^T=A$. In other words $A^2=AA^T$. So it's enough to prove that $AA^T$ is positive definite, i.e $\langle x,A^TAx \rangle > 0$. But $\langle x,A^TAx \rangle=||Ax||^2 >0(Why?).$Therefore $A^2$ is positive semidefinite.

Topology
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A direct proof: let $( \cdot| \cdot)$ be the usual inner product on $ \mathbb R^n$ and $|| \cdot||$ the induced norm.

If $ \mu$ is an eigenvalue of $A^2$, then there is $v \in \mathbb R^n$ sucht that $||v||=1$ and $A^2v= \mu v$. Since $A$ is invertible $A^2$ is invertible, hence $ \mu \ne 0$.

Furthermore

$$ \mu= \mu ||v||^2=\mu (v|v)=( \mu v|v)=(A^2v|v)=(Av|Av)=||Av||^2 >0.$$

Fred
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If $A$ is symmetric, then so is $A^2$.

Let $\lambda \in \mathbb{\mathbb{R}}$ be an eigenvalue of $A^2$, and let $\sqrt{\lambda} \in \mathbb{C}$ be a square root of $\lambda$. We have$$(A-\sqrt{\lambda} I)(A-\sqrt{\lambda} I)= A^2 - \mu I$$ $A^2 - \lambda I$ is not invertible so $A-\sqrt{\lambda} I$ is also not invertible, which implies that $\sqrt{\lambda}$ is an eigenvalue of $A$. Since $A$ is positive definite, we have $\sqrt{\lambda} > 0$ so $\lambda > 0$.

Therefore, $A^2$ is a symmetric matrix with positive eigenvalues, and hence positive definite.

mechanodroid
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