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I saw this property on Wikipedia. But they didn't give a proof, and I failed to work out one. Is it easy to prove?

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Use eigen-decomposition –  chaohuang Oct 3 '12 at 4:44

3 Answers 3

up vote 2 down vote accepted

As $A$ is positive definite, so $A$ is Hermitian($A^{H}=A$) and for arbitrary vector $X$, we have $X^{H}AX\geq0$ and $X^{H}AX=0$ only when $X=0$.

Assume that $X_{0}$ is a solution for $AX=0$, i.e. $AX_{0}=0$, then $X_{0}^{H}AX_{0}=0$, and we can get that $X_{0}=0$ by above conclusion. This shows that $AX=0$ only have the trivial solution, so $A$ is invertible.

For $A^{-1}$, we can compute as following:


so $A^{-1}$ is also Hermitian.

For arbitrary vector $X$,


and $X^{H}A^{-1}X=(A^{-1}X)^{H}A(A^{-1}X)=0$ only when $A^{-1}X=0$, equivalent to saying $X=0$.

So $A^{-1}$ is also positive definite.

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The proof for $A$ operating on a finite-dimensional space is essentially that $A$ has a basis of positive eigenvalues. This is because it's symmetric (Hermitian,) and so orthogonally (unitarily) diagonalizable over the reals (complexes,) $A=P^{-1}\hat{A}P$. Then if $\hat{A}$ has an entry $\lambda_i,$ $$(P^{-1}e_i)^TA(P^{-1}e_i)=e_i^T(P^{-1})^TP^{-1}\hat{A}e_i=e_i^TPP^{-1}\lambda_ie_i= \lambda_ie_i^Te_i=\lambda_i$$ So if $\lambda_i$ were nonpositive, $A$ couldn't be positive definite.

Now we naturally find $A^{-1}=P\hat{A}^{-1}P^{-1}$, where we can invert $\hat{A}$ just by inverting the $\lambda_i$, leaving all eigenvalues strictly positive. Since $P^{-1}$ is orthogonal or Hermitian $\{P^{-1}e_i\}$ forms a basis, and the same computation from above shows $(P^{-1}e_i)^Te_i^TA^{-1}P^{-1}e_i$ is positive. Now $x^TA^{-1}x$ can be written as a sum of those positive terms, and we see $A^{-1}$ is positive definite.

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A matrix is positive definite if it is Hermitian and every eigenvalue is positive. And since the eigenvalues of $A^{-1}$ are the reciprocals of the eigenvalues of $A$, they are also positive. So you just need to check that $A$ is indeed invertible to conclude $A^{-1}$ is positive definite. This follows since $A$ is injective (and hence bijective, by Rank-Nullity): If $v\neq 0$ then $Av\neq 0$ since if it was, $v^T A v = v^T 0 = 0$ which contradicts $A$ being positive definite.

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