Find the value of $s$ for which the matrix is positive definite I am given the following Matrix 
$$
A=        \begin{pmatrix}
        s & -4 & -4 \\
        -4 & s & -4 \\
        -4 & -4 & s \\
        \end{pmatrix}
$$
and asked to find $s$ such that $A$ is positive definite. In finding it, I simply set the determinant to $0$ and solved for $s$, and in doing so, I got $0 = s^2-4s-32$ which solves to $s=-4,8$. I'm not sure whether this is the correct process though, and if I use either value for $s$ I have at least one $0$ for each eigenvalue of $A$. Am I on the right track? Thanks in advance. 
 A: You could also approach it like this: A symmetric matrix is postive definite iff all eigenvalues are positive.
Take $s=-4$ and we have a rank $1$-matrix $A = \begin{pmatrix}-4&-4&-4\\-4&-4&-4\\-4&-4&-4\end{pmatrix}$. Such a matrix has two different eigenvalues: $0$ and the trace, which is $-12$.
For arbitrary $s$ our matrix is of the form $(s+4)\operatorname{Id}+A$, hence the eigenvalues are $s+4$ and $s+4-12=s-8$. This shows that the matrix is positive definite iff $s > 8$.
A: The matrix $A$ will be positive definite if and only if all its eigenvalues are positive. You have
$$ A = \left( \begin{matrix} s & -4 & -4 \\ -4 & s & -4 \\ -4 & -4 & s \end{matrix} \right) = \left( \begin{matrix} -4 & -4 & -4 \\ -4 & -4 & -4 \\ -4 & -4 & -4 \end{matrix} \right) + \left( \begin{matrix} s + 4 & 0 & 0 \\ 0 & s + 4 & 0 \\ 0 & 0 & s + 4 \end{matrix} \right) = B + (s+4)I. $$
The eigenvalues of $B$ are $0$ (with multiplicity $2$) and $-12$ (with multiplicity $1$) and so the eigenvalues of $A$ are $s + 4$ (with multiplicity $2$) and $s - 8$ (with multiplicity $1$). Thus, $A$ will be positive definite if and only if $s + 4 > 0$ and $s - 8 > 0$ - that is, if and only if $s > 8$.

In the answer, we used the observation that if $B$ is a diagonalizable matrix with eigenvalues $\lambda_1, \ldots, \lambda_n$ and $p(X) = a_nX^n + \ldots + a_0$ is a polynomial, then $p(B)$ (which is also diagonalizable) has eigenvalues $p(\lambda_1), \ldots, p(\lambda_n)$. This is easily proved by checking that if $v_i$ is an eigenvector of $B$ with $Bv_i = \lambda_i v_i$ then it is also an eigenvector of $p(B)$ and $p(B)v_i = p(\lambda_i) v_i$. In our case, $p(X) = X + (s + 4)$.
