# Jordan Canonical Form matrices

I have just been introduced to the notion of the Jordan Canonical Form, in the context of matrices.Let A $\in \mathbb{C^{n\times n}}$, $X$ non singular, then the Jordan Canonical form is; $X^{-1}AX = J$, where $J$ is the matrix whose diagonal entries are the Jordan blocks.

It is stated that the number of Jordan blocks is the number of linearly independent eigenvectors of A? I can't see why this is true?

• I would reformulate "..., $X$ non singular, then the JCF is; ..." to "then there is a non-singular $X$ so that $X^{-1}AX=J$ is in JCF, ...". As it is now it sounds like this is true for all $X$. – M. Winter Nov 2 '17 at 13:23

Maybe an example will make this clearer. Suppose the Jordan canonical form of $A$ consists of a single $3 \times 3$ Jordan block $$J = \begin{pmatrix} \lambda & 1 & 0\\ 0 & \lambda & 1\\ 0 & 0 & \lambda \end{pmatrix}$$ and let $\beta = \{v_1, v_2, v_3\}$ be the basis that puts $A$ in this form (i.e., the $v_i$ are the columns of $X$). Then $$A v_1 = \lambda v_1, \qquad Av_2 = v_1 + \lambda v_2, \qquad Av_3 = v_2 + \lambda v_3 \, .$$ So it looks like $v_1$, the vector corresponding to first column of $J$, is the only eigenvector of $A$. And indeed, $J - \lambda I$ has two pivots, hence $\ker(J - \lambda I)$ and thus $\ker(A - \lambda I)$ has dimension $1$. So $\ker(A - \lambda I) = \DeclareMathOperator{\spn}{span} \spn\{v_1\}$. Similarly for any Jordan block, the vector corresponding to the first column is an eigenvector, and the rest are generalized eigenvectors, but not eigenvectors. The column $v_i$ of $X$ corresponding to the first column of the block satisfies $A v_i = \lambda v_i$, and if $J$ is $m \times m$, then $J - \lambda I$ has $m - 1$ pivots, so its kernel is $1$-dimensional.