Let $a,b,c$ be a positive real number such that $b^2+c^2<a<1$. Let
$A=\begin{bmatrix} 1&b&c\\ b&a & 0\\ c & 0 & 1\end{bmatrix}$. Then
Consider the above matrix. I want to comment about the nature of eigenvalues of the matrix in the sense that, they are all positive, all negative, mix of positive or negative, non zero, real or non real etc..
My efforts
We look at the matrix first and see if it looks like one of the familiar type introduced in standard Linear Algebra Text.
We can see, this matrix is symmetric.
As soon we hear the term "symmetric matrix" and there is already the term "eigenvalue" in the question. We go the next standard result which says that a real symmetric matrix is diagonalizable with all eigenvalues real.
Conclusion so far The given matrix has only real eigenvalues.
Another standard result is the sum of eigenvalues is equal to the trace of the matrix. Trace is positive here due to the conditions specified.
So not all eigenvalues can be negative.
So we are left with two choices
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all eigenvalues are positive
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Eigenvalues of A are either positive or negative.
I know this matrix is positive definite(I have already proved it, by showing that all sub determinant are positive) so all eigenvalues are positive.
My aim is to show that there are no negative eigenvalues without going into the theory of positive definite matrices.
Best Answer
The characteristic polynomial of your matrix is $$ p(x) = (a-x)(1-x)^2-c^2(a-x)-b^2(1-x). $$ Now, if $x < 0$, then \begin{align*} p(x) &> (a-x)(1-x)^2-c^2(1-x)-b^2(1-x) = (1-x)\cdot[(a-x)(1-x)-c^2-b^2]\\ &> (1-x)\cdot[x^2-(a+1)x] = x(1-x)(x-a-1) > 0. \end{align*} Therefore, $p$ cannot have zeros in $(-\infty,0)$. Also, $p(0) = \det A > 0$. Thus, the eigenvalues of $A$ are positive.