I am working on this problem:
For which numbers $c$ and $d$ does $A$ have real eigenvalues and orthonormal eigenvectors?
$$A= \begin{bmatrix}1 & 2 & 0\\2 & d & c\\0 & 5&3\end{bmatrix}$$
Because the matrix we are looking for is orthonormal, I figured A must be a symmetric matrix, therefore $c=5$. I tried solving for the eigenvalues of the following updated matrix:
$$A= \begin{bmatrix}1 & 2 & 0\\2 & d & 5\\0 & 5&3\end{bmatrix}$$
I have checked the algebra multiple times and I continue to get the following matrix which I can't factor and get eigenvalues for:
$$0=d\lambda^2-4d\lambda+3d-\lambda^3+4\lambda^2+26\lambda-37$$
I tried solving for $d$ and got a massive ratio that didn't make any sense. The goal of this is to eventually state examples of values for $c$ and $d$ that we can use to find orthonormal vectors that are combinations of the columns.
I thought perhaps the trace or determinant may have some relevance, but I can't figure out how.
Best Answer
Suppose that $A$ indeed has real eigenvalues $\lambda_j$ and orthogonal eigenvectors $\boldsymbol{v}_j$. Put the eigenvectors into a matrix $O = (\boldsymbol{v}_1, \dots, \boldsymbol{v}_n )$ and then we find \begin{align*} O^T A O = \operatorname{diag}(\lambda_1,\dots,\lambda_n) =:D \end{align*} and thus $A^T = (ODO^T)^T = OD^TO^T=ODO^T= A$, which shows that $A$ is symmetric and thus $c=5$.
Continuing: No matter the choice of $d$, the matrix will now be symmetric, so by the spectral theorem for real symmetric matrices it has real eigenvalues and an orthonormal basis.