1
$\begingroup$

Let $B$ and $A$ be symmetric and positive definite matrices. Derive the matrix $A^{1/2}$ and show that for the eigenvalue problem $$A\vec{w} = \lambda B\vec{w}$$

all the eigenvalues are real and positive.

I know how to derive $A^{1/2}$ and don't need help with that.

I tried moving things around to get $(A - \lambda B)\vec{w} = 0$. Then $\vec{w}$ being an eigenvector, it cannot be zero, so we must have the determinant of $(A - \lambda B)$ be zero. Not sure what to do from there...

Also, since $A$ and $B$ are symmetric and positive definite, all of their eigenvalues are real and positive.

And I don't see where $A^{1/2}$ would come in use in the proof if the direction in which I'm going is the right one.

Thank you for your input!

$\endgroup$
  • $\begingroup$ I am not quite sure what this eigenvalue problem. Finding all $\lambda \in \mathbb{C}$ for which there is a solution to $Aw=\lambda Bw$? $\endgroup$ – Fimpellizieri Nov 21 '16 at 4:58
2
$\begingroup$

You showed that $\lambda$ is an eigenvalue for your problem is and only if $\det(A-\lambda B)=0$.

Note that $\lambda\ne0$, since $A$ is invertible.

You have \begin{align} 0&=\det(A-\lambda B)=\det[A^{1/2}(I-\lambda A^{-1/2}BA^{-1/2})A^{1/2}]\\ \ \\ &=\det(A)\,\det(I-\lambda A^{-1/2}BA^{-1/2}).\end{align} Thus $$ 0=\det(I-\lambda A^{-1/2}BA^{-1/2})=\det(\lambda^{-1}I- A^{-1/2}BA^{-1/2}).$$ So $\lambda^{-1}$ is an eigenvalue for the positive-definite matrix $A^{-1/2}BA^{-1/2}$, so $\lambda^{-1}>0$.

$\endgroup$
1
$\begingroup$

Martin's answer clearly shows how to use $A^{\tfrac12}$ in the problem, but if you are allowed to use other methods I think there's something simpler.

In the equality $Aw=\lambda Bw$, take the inner product with $w$ on both sides. On the LHS, we'll have $\langle w,Aw\rangle$, which must be positive because $A$ is positive definite. Hence, we'll have that the RHS too is positive, that is $\langle w,\lambda Bw\rangle = \lambda \langle w,Bw\rangle > 0$. Since $B$ is positive definite, it follows that $\langle w,Bw\rangle>0$ and so $\lambda$ too must be positive.

$\endgroup$
  • $\begingroup$ Nice! I totally forgot that taking inner products of equations is a tool in my arsenal, thank you for that! $\endgroup$ – MMtime Nov 21 '16 at 5:16
  • $\begingroup$ You're welcome! $\endgroup$ – Fimpellizieri Nov 21 '16 at 5:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.