# Finding Jordan Canonical form for 3x3 matrix

I was looking at http://www.math.hkbu.edu.hk/~zeng/Teaching/math3407/Jordan_Form.pdf (section 2)

$A =\left(\begin{array}{ccc}4 & 0 & 1 \\2 & 3 & 2 \\1 & 0 & 4\end{array}\right)$

We find the Jordan Canonical Basis for $R^3 = \left \{\left(\begin{array}{c}1 \\2 \\1\end{array}\right), \left(\begin{array}{c}0 \\1 \\0\end{array}\right),\left(\begin{array}{c}-1 \\0 \\1\end{array}\right) \right\}$.

So far, so good.

But I don't see how they then arrive that the Jordan Canonical Form for A is:

$\left(\begin{array}{ccc}5 & 0 & 0 \\0 & 3 & 0 \\0 & 0 & 3\end{array}\right)$

• That last matrix will be the JCF iff the three vectors in the basis are respectively eigenvectors of $5, 3$ and $3$. So check to see if that is the case. – Simon S Apr 15 '15 at 17:38
• Are they just changing the basis of $A$ from the standard basis to a different basis? If so why can't you just do $A*$new basis for $R^3$. – larry Apr 15 '15 at 17:43
• See below for more discussion – Simon S Apr 15 '15 at 17:47

As a reminder of some theory:

Suppose $M$ is an arbitrary $2 \times 2$ matrix with two eigenvalues, $\lambda_1, \lambda_2$, and hence two corresponding eigenvectors, $v_1$ and $v_2$. Define a matrix $V$ whose columns are those two eigenvectors,

$$V =\left( \begin{matrix} v_1 & v_2 \\ | & | \end{matrix} \right)$$ where those vertical bars indicate the column.

Then the action of $M$ on $V$ is the same as the action of $M$ on the individual columns. As those columns are also eigenvectors we may write

$$MV =\left( \begin{matrix} Mv_1 & Mv_2 \\ | & | \end{matrix} \right) = \left( \begin{matrix} \lambda_1v_1 & \lambda_2v_2 \\ | & | \end{matrix} \right) = \underbrace{\left( \begin{matrix} v_1 & v_2 \\ | & | \end{matrix} \right)}_{V} \underbrace{\left( \begin{matrix} \lambda_1 & 0 \\ 0 & \lambda_2 \end{matrix} \right)}_{D}$$

where $D$ is the diagonal matrix of eigenvalues. That is, $MV = VD$. Or

$$M = VDV^{-1}$$

This is a change of basis, to/from a basis of eigenvectors.

In this notation, the Jordan normal form is the diagonal matrix $D$. The change of basis matrix $V$ comprises of eigenvectors.

All of this generalizes to $n \times n$ matrices. And there is the added complication that you can have Jordan blocks for eigenspaces where the geometric multiplicity is less than the algebraic multiplicity. However that last wrinkle doesn't apply to your example.

Back to your problem: the basis is an appropriate basis iff it corresponds to the eigenvalues on the diagonal, $\lambda = 5, 3, 3$. To check that, apply the matrix $A$ to each of them and confirm they are eigenvectors with those eigenvalues. (Hint: they are!)