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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?


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When you matrix is not diagonalisable – M Turgeon Aug 6 '12 at 19:16
@M I know this result/theorem, but I'm hoping for a more fundamental answer. Like some algebraic property or somesuch that will maybe unfurl during row operations. Or something. – Ryan Aug 6 '12 at 19:22
It's simple to build examples of matrices with repeated eigenvalues whose behavior with respect to diagonalization differs, as Qiaochu alludes to. To use $3\times 3$ matrices as an example, $$\begin{pmatrix}\lambda_1&1&\\&\lambda_1&\\&&\lambda_2\end{pmatrix}$$ will only have two eigenvectors (i.e. it is a defective matrix), while $$\begin{pmatrix}\lambda_1&&\\&\lambda_1&\\&&\lambda_2\end{pmatrix}$$ is certainly diagonalizable. – J. M. Aug 7 '12 at 0:23
@JM Ah, thanks for the concrete example and link:) – Ryan Aug 7 '12 at 4:05
up vote 4 down vote accepted

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...

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