Generalized eigenvalue problem with symmetrical real matrices

Question

I am investigating the generalized eigenvalue problem

$$(\lambda\,\boldsymbol{A}+\boldsymbol{B})\,\boldsymbol{x}=\boldsymbol{0}$$

where $\boldsymbol{A}$ and $\boldsymbol{B}$ are real-valued symmetrical matrices, $\lambda$ are the eigenvalues and $\boldsymbol{x}$ are the eigenvectors.

According to Wikipedia, the eigenvalues $\lambda$ are all real-valued if $\boldsymbol{A}$ is positiv definite. Is there a simple proof for that?

What happens if $\boldsymbol{A}$ is only positiv semidefinite? Are the eigenvalues are still all real-valued?

I assume, from a mechanical point of view, that for positiv definite matrices $\boldsymbol{A}$ and $\boldsymbol{B}$, the eigenvalues are all negative real-valued. Is there a simple proof for that statement? And again, what happens if the matrices (or one of them) is only positiv semidefinite?

Answer 1

Regarding the specific question about positive definite. Suppose $A$ is positive definite symmetrical matrix. So $x^TAx$ is positive for all non zero vector $x$.

For symmetrical matrices eigenvectors constitute orthogonal basis. Say here the eigen vectors are $x_1,x_2,,,x_n$, we can represent $x$ as $x = a_1x_1+a_2x_2+...+a_nx_n$.
So $Ax = a_1\lambda_1x_1+a_2\lambda_2x_2+...+a_n\lambda_nx_n$. Now as the vectors are orthogonal and $x^TAx$ is the dot product of $x$ and $Ax$. Assume all $\lambda_i$ are positive, so here we get

$x^TAx= {a_1}^2\lambda_1{(x_1.x_1)}+{a_2}^2\lambda_2{(x_2.x_2)}+...+{a_n}^2\lambda_n{(x_n.x_n)}$
Note that this is always positive as not all $a_i$ are zero and all$\lambda_i$ are positive. So A is positive definite.

Now say $A$ is positive definite so for any non zero vector $x$, $x^TAx$ is positive. We can put eigenvactors in place of $x$ and get $x^TAx = x^T\lambda x = \lambda x^Tx$
Now the sign of this completely depends on $\lambda$ and as $x^TAx$ is positive, $\lambda$ has to be positive (true for all eigenvalues).

edit:
So 1) $x^TAx$ is positive for all non zero vector $x$
2) All eigenvalues of $A$ are positive
Both are equivalent statement to define a positive definite matrix.

Answer 2

Assuming that $A$ is positive definite, then we have that it is also nonsingular. Thus, we can express the eigenvalue problem as follows:

$$ \lambda x = -A^{-1}Bx. $$

But notice that since $A$ is PD, then so is $A^{-1}$, and so for all nonzero $x$: $$ \lambda^2x^\top x =x^\top BA^{-2}Bx > 0, $$ and thus $\lambda$ must be real, see https://math.stackexchange.com/a/354119/443030

EDIT: Regarding your second question, let $A=B=\begin{pmatrix}1&0 \\ 0&0\end{pmatrix}$ and $\lambda\in\mathbb{C}$. Clearly, $A$ is PSD, then

$$ \det(\lambda A+B) = \det\begin{pmatrix}\lambda+1&0\\0&0\end{pmatrix} = 0. $$ It appears that if $A$ is PSD with $\ker A \cap\ker B \neq \{0\}$, then $\lambda$ is a free variable, i.e. we can choose any $\lambda\in\mathbb{C}$, so that $$ (\lambda A + B)x = 0, $$ for any $x = \begin{pmatrix}0\\ t\end{pmatrix},~ t\in\mathbb{C}$.

So yes, in this case, the generalized eigenvalue can be complex.