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Given a $3\times3$ matrix "A" which is written below, find the geometric multiplicity (GM) of the largest eigenvalue of "A". $$\begin {bmatrix} a&2f&0\\ 2f&b&3f\\ 0&3f&c\\ \end{bmatrix}$$ $a,b,c$ and $f$ are real numbers and $f$ is non zero.

If I choose $a,b,c$ distinct and $f$ very small , I get $3$ non intersecting Gershgorin discs which imply that all three eigenvalues are distinct. In this case, I get the geometric multiplicity to be one. But for other cases is it possible to find the GM from the given data? The matrix is real symmetric and hence orthogonally diagonalisable. How do I exploit this property to conclude the problem?

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    $\begingroup$ the characteristic polynomial has a repeated root precisely when it has a common factor with its derivative. With your matrix of symbols, finding that polynomial gcd will not be nice, but there will be no need to come up with closed forms for the eigenvalues. $\endgroup$
    – Will Jagy
    Commented Apr 4, 2021 at 16:40

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Suppose there is some $\lambda$ that has geometric multiplicity $\geq 2$.
$B:=\big(A-\lambda I \big)$ then
$2\leq\dim \ker B$

but the 1st column of $B$ is linearly independent of the 2nd one, and neither column is zero $\implies \text{rank}\big(B\big)\geq 2\implies \dim \ker B\leq 1$ by rank nullity. We conclude
$2\leq\dim \ker B\leq 1$ which is a contradiction.

Thus all eigenvalues must be simple.

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    $\begingroup$ Cool! What a simple and elegant answer!! Thank you. $\endgroup$
    – Lawrence Mano
    Commented Apr 4, 2021 at 23:47

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