Prove that the eigenvalues for $A\vec{w} = \lambda B\vec{w}$ are real and positive

Question

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!

Answer 1

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 $\lambda^{-1}>0$.

Answer 2

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.