Solve eigenvalue problem with known constraint on one of the Eigenvalues

I have the following problem and would appreciate any help.

I have a real, symmetric matrix M given by $$M=\begin{pmatrix} m_{11} & m_{12} & m_{13} & m_{14} \\ m_{12} & m_{22} & m_{23} & m_{24} \\ m_{13} & m_{23} & m_{33} & m_{34} \\ m_{14} & m_{24} & m_{34} & m_{44} \\ \end{pmatrix}$$ I know all entries of $M$ apart from $m_{11}$, which I would like to find numerically.

I would like to diagonalise $M$ such that it is equal to the diagonal matrix $$M_{D}=\begin{pmatrix} m_1 & 0 & 0 & 0 \\ 0 & m_2 & 0 & 0 \\ 0 & 0 & m_3 & 0 \\ 0 & 0 & 0 & m_4 \\ \end{pmatrix}$$

$m_{2}$, $m_{3}$ and $m_{4}$ can take any value but $m_{1}=x$, where $x$ is a fixed value I know. In general one cannot analytically derive the eigenvector/eigenvalues for a 4x4 matrix but I would like to know is there an efficient way of finding $m_{2}$, $m_{3}$ and $m_{4}$ and the eigenvectors i.e. the diagonalising matrix of M?

One thing I thought to do would be to numerically minimise the distance between the known value for $m_{1}$ and the resulting eigenvalue for a given m_{11}.

Does anyone know a better algorithm than this?

• Would Computational Science be a better home for this question? – Qmechanic Jan 9 '18 at 6:44
• I think it's a Math SE question (solved in one stroke by @Guy Gur-Ari in his answer). – Frobenius Jan 9 '18 at 7:42

The characteristic polynomial of $M$ has a root $x$, namely $det(M-xI)=0$. This expression is linear in $m_{11}$, so you can solve for it analytically.