# Dimension of eigenspace

Given that a $3 \times 3$ matrix has only one eigenvalue, what is the dimension of its corresponding eigenspace? It says that the answer is 3. But couldn't we have some matrix

$A = \begin{pmatrix} \lambda & 0 & 0 \\ 0 & \lambda & 1 \\ 0 & 0 &\lambda \end{pmatrix}$

in Jordan canonical form. Then $\lambda$ is its only eigenvalue, but there are two Jordan blocks, hence the geometric multiplicity should be 2?

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"It says"... Who or what is "it"? –  Arturo Magidin Oct 1 '11 at 3:31
You can only say that the dimension is 3 (full dimension) iff the matrix is diagonalizable (eg , if it's normal). –  leonbloy Oct 1 '11 at 12:25

A $3 \times 3$ matrix with a single eigenvalue $\lambda$ can have 3 possible canonical Jordan structures: i) $J_1(\lambda) \bigoplus J_1(\lambda) \bigoplus J_1(\lambda)$ (dimension of eigenspace $3$), ii) $J_1(\lambda) \bigoplus J_2(\lambda)$ (dimension of eigenspace $2$) and iii) $J_3(\lambda)$ (dimension of eigenspace $1$). By $J_k(\lambda)$ i mean a $k \times k$ Jordan block corresponding to eigenvalue $\lambda$.
If a 3x3 matrix $A$ has only one real eigenvalue, then we can deduce that the eigenvalue's algebraic multiplicity is either 1 ($A$'s other eigenvalues are complex) or 3 ($A$'s characteristic polynomial is of the form $(\lambda-a)^3$).
p.s. The eigenspace is 3-dimensional if and only if $A=kI$ (in which case $k=\lambda$).