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I recently solved a problem in which i found a matrix to have four eigenvalues, three eigenvalues of 2 and one eigenvalue of 3. What is the significance of having repeating answers of eigenvalues. Don't you need just one eigenvector to span the entire eigenspace of an eigenvalue? That was my initial thought but when i tested these three eigenvectors for linear independence, i determined they were in fact linearly independent.

This is the matrix i solved:

{{2, 0, 0, 0}, {0, 2, 0, 0}, {0, 0, 2, 2}, {0, 0, 0, 3}}

Eigenvalues: {2,2,2,3}

Eigenvectors: {{0, 0, -2, -1}, {0, 0, 1, 0}, {0, 1, 0, 0}, {1, 0, 0, 0}}

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  • $\begingroup$ Don't you need just one eigenvector to span the entire eigenspace of an eigenvalue? Not necessarily. $\endgroup$ – Omnomnomnom Aug 6 '15 at 9:19
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A matrix is diagonalizable if and only if there is a basis of eigenvectors, so if an eigenvalue has algebraic multiplicity $m$ (i.e. its multiplicity as a root of the characteristic polynomial) it also has geometric multiplicity $m$ (i.e. it has $m$ linearly independent eigenvectors). On the other hand, there are matrices that are not diagonalizable, and they have eigenvalues with geometric multiplicity less than the algebraic multiplicity. For example,

$$ \pmatrix{0 & 1\cr 0 & 0\cr}$$

has algebraic multiplicity $2$ but geometric multiplicity $1$. The geometric multiplicity is always at least $1$, but need not be any greater than that.

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