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I was kinda crushed to discover that two different matrices with different properties can actually share the same characteristic polynomial ($-\lambda^3-3\lambda^2+4$):

$A=\begin{pmatrix} 1 & 2& 2\\ -3 &-5 &-3 \\ 3& 3 & 1 \end{pmatrix} , B=\begin{pmatrix} 2 & 4& 3\\ -4 &-6 &-3 \\ 3& 3 & 1 \end{pmatrix}$

$A$ has an eigenline and an eigenplane (and thus an eigenbasis), whereas $B$ has two eigenlines (so no eigenbasis). The repeated eigenvalue -2 of B corresponds to an eigenspace with basis {(-1,1,0)}.

When is the geometric multiplicity of an eigenvalue smaller than its algebraic multiplicity (as in case B)? Are there general conditions to look for?

Thanks!

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    @JM Ah, thanks for the concrete example and link:)2012-08-07

2 Answers 2

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The general condition is the presence of nontrivial Jordan blocks.

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As far as I know, there are no necessary and sufficient algebraic conditions of the kind you might be looking for.

However, there are some sufficiency conditions; the spectral theorem (finite dimensional version) states that if the matrix is real symmetric it's eigenvectors form an orthonormal basis for the space the matrix maps from/to, which implies the algebraic and geometric multiplicity of the eigenvalues are the same.

The result can also be extend normal matrices, a matrix A is said to be normal iff A*A=AA* (A* denotes the conjugate transpose of A). If a matrix is normal then it is diagonalisable, which by the diagonalisation theorem implies that the matrix's eigenvectors span space the matrix maps to and from. This said its probably still quicker to work out eigenvectors yourself than to be working out A*A and AA* in the hope it A*A=AA* holds true...