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I want to show the equivalence of two definition of positivity. Let $A \in \mathcal{L}(H)$, where $\mathcal{H}$ is the $n-$dimensional Hilbert Space $\mathbb{C}^n$. $A$ is positive if $\langle x,Ax \rangle \geq 0 $ for all $x \in \mathcal{H}$.

Show that $A$ is positive if and only if it is Hermitian ($A=A^*$) and all its eigenvalues are positive.

It is a classical exercise to show that an Hermitian matrix is positive definite iff its eigenvalues are positive. The difference in this question is that one only assumes the operator is positive and has to deduce that it is Hermitian and its eigenvalues are positive, which cannot be solved using the same approach.

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Arindam
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  • For all real vectors $x$ or for all complex vectors $x$? – Erick Wong Aug 17 '15 at 06:30
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    This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level. – Carl Mummert Aug 18 '15 at 11:19
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    I did not vote to close this as a duplicate, but instead because there is so much room to improve the question by adding background, motivation, and effort. – Carl Mummert Aug 18 '15 at 14:16

1 Answers1

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To be clear, the heart of the question is the following assertion: $$\text{positive definite} + \text{complex field} \implies \text{Hermitian}.$$

The key idea here is to do a complex variation on the classic symmetric + skew-symmetric splitting. That is, $$A = \frac{A + A^*}{2} + i \frac{A - A^*}{2i}.$$ Since $\frac{1}{i} = -i$, we have, $$\frac{A - A^*}{2i}~ \text{is Hermitian!}$$ In a sense this splitting is a matrix analog of splitting a complex number into it's real and imaginary parts.

Now since $A$ is positive definite we have, $$0 \le \langle z, A z \rangle = \underbrace{\langle z, \frac{A + A^*}{2} z \rangle}_{\text{real}} + i ~ \underbrace{\langle z, \frac{A - A^*}{2i} z \rangle}_{\text{real}}.$$

The only way this whole quantity can be real for all $z$ is if $$\frac{A - A^*}{2i} = 0.$$ In other words, $A = A^*$, so $A$ is Hermitian.

It is interesting to note that this result does NOT hold when restricted to the real field, as can be seen with the following counterexample:

$$M = \begin{bmatrix}1 & 2 \\ & 1\end{bmatrix}.$$

For real $x,y$, we have, $$\begin{bmatrix}x & y\end{bmatrix}\begin{bmatrix}1 & 2 \\ & 1\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix} = (x+y)^2 \ge 0,$$ so $M$ is real-positive-definite.

On the other hand, $M$ is not complex-positive-definite since we can take $x=i$, $y=1$ to get, $$\begin{bmatrix}-i & 1\end{bmatrix}\begin{bmatrix}1 & 2 \\ & 1\end{bmatrix}\begin{bmatrix}i \\ 1\end{bmatrix} = 2 - 2i \notin \mathbb{R}.$$

Nick Alger
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