How to find the geometric multiplicity of the largest eigenvalue of real symmetric matrix which is non singular?
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If all the eigenvalues are distinct then GM of each eigenvalue must be 1. But what about if one of the eigenvalue repeats or there may be all theree eigenvalues are same? I am not getting how to tackle the problem?
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If a 3 by 3 matrix has 2 eigenvalues the same and one different then there are two possibilities. There may be two independent eigenvectors corresponding to that "double" eigenvalue or there may be just one eigenvector (and multiples). In the first case, the "geometric multiplicity" of that eigenvalue is 2 and in the second it is 1. In the first case the matrix is similar to the diagonal matrix $\displaystyle \begin{bmatrix}a & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & b \end{bmatrix}$ and in the second case the matrix is similar to the Jordan Normal Form $\displaystyle \begin{bmatrix} a & 1 & 0 \\ 0 & a & 0 \\ 0 & 0 & b \end{bmatrix}$.
If a 3 by 3 matrix has a single eigenvalue, then there are 3 possibilities. There might be three independent eigenvectors. In that case the geometric multiplicity is 3 and the matrix is similar to the diagonal matrix $\displaystyle \begin{bmatrix}a & 0 & 0 \\0 & a & 0\\ 0 & 0 & 0 \end{bmatrix}$. There might be two independent eigenvectors. In that case the geometric multiplicity is 2 and the matrix is similar to the Jordan Normal Form $\displaystyle \begin{bmatrix}a & 0 & 0 \\ 0 & a & 1 \\ 0 & 0 & a\end{bmatrix}$. Finally, there might be a single eigenvector (and its multiples). In that case the geometric multiplicity is 1 and the matrix is similar to the Jordan Normal Form $\displaystyle \begin{bmatrix}a & 1 & 0 \\0 & a & 1 \\ 0 & 0 & a \end{bmatrix}$.
Let $\lambda $ be an eigenvalue of $A$
$\displaystyle A-\lambda I = \begin{bmatrix}a-\lambda & 2f & 0 \\ 2f & b-\lambda & 3f \\ 0 & 3f & c-\lambda \end{bmatrix}$
since $f\neq 0$, rank $(A-\lambda I) = 2$
therefore nullity $(A-\lambda I) =1$