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Let $A\in M_n(\mathbb{R})$, a symmetric positive-definite matrix. Prove that for every $k\in\mathbb{N}$, $A^k$ is also positive definite.

So since $A\in M_n(\mathbb{R})$ is symmetric and positive definite, we know there's an invertible $P$ such that $P^tAP = \text{Diag}(c_1,\ldots,c_n)$ such that $c_1,\ldots ,c_n > 0$.

I think it's sufficent to prove it for $k=2$ and the rest is given by induction.

I also know that Matrix congruence is an equivalence relation so I tried to "play" with the definitions to infer something about $A^2$ but havn't got something useful.

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    $\begingroup$ For such matrix $A$ all eigenvalues are real (and this property is equivalent to $A$ being positive definite). So...? $\endgroup$ Commented Jul 28, 2015 at 11:13
  • $\begingroup$ You could also prove it for $k = 1$ as your base case. It's easier, since that case is trivial, and it's more correct, because proving $k = 2$ but not $k = 1$, then proving an induction step, technically only proves it for $k \ge 2$. $\endgroup$ Commented Jul 28, 2015 at 11:16
  • $\begingroup$ possible duplicate of If $A$ is positive definite then so is $A^k$ $\endgroup$ Commented Jul 28, 2015 at 11:21

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To prove that $A^k$ is positive definite, it's enough to prove that its eigenvalues $\lambda_1,\dots,\lambda_n$ are strictly positive.

On the other hand, if $\mu_1,\dots,\mu_n$ are the eigenvalues of $A$ then $\mu_i>0$ and $\lambda_i=\mu_i^k$, for all $i$. Therefore $\lambda_i>0$ for all $i$.

To show that $\lambda_i=\mu_i^k$, you can proceed by induction. The idea is that if $v_i$ is the eigenvector associated with $\mu_i$, then

\begin{align} A^kv_i=&A^{k-1}(Av_i)=A^{k-1}(\mu_i v_i)=\mu_i A^{k-2}(A v_i)\\&=\mu_i^2 A^{k-3}v_i=\dots =\mu_i^{k-1}Av_i=\mu_i^{k}v_i. \end{align}

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Alternate approach:

  • For any $M$, $MM^T$ is positive definite
  • Every symmetric positive definite matrix has a unique symmetric positive definite square root

From there, we have $$ A^k = \sqrt{A}^k[\sqrt{A}^k]^T $$ The conclusion follows.

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Here is a more elementary approach. It applies when $A$ is Hermitian positive definite (and in particular, when $A$ is symmetric positive definite).

  1. $A^0=I$ is clearly positive definite. By assumption, so is $A^1=A$.
  2. Positive powers of $A$ are positive definite, because when $r=0$ or $1$, $A^r$ is positive definite and hence $x^\ast A^{2k+r}x=(A^kx)^\ast A^r(A^kx)>0$ for any vector real $x$.
  3. $A^{-1}$ is positive definite because $x^\ast A^{-1}x=(A^{-1}x)^\ast A(A^{-1}x)>0$.
  4. By a similar argument to (2), negative powers of $A$ (i.e. positive powers of $A^{-1}$) are positive definite.
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