I am supposed to find the Jordan canonical form of a couple of matrices, but I was absent for a few lectures.

\begin{bmatrix} 1 & 1 & 0 \\ 0 & 1 & 2 \\ 0 & 0 & 3 \end{bmatrix}

Since this is an upper triangular matrix, its eigenvalues are the diagonal entries. Hence $\lambda_{1,2}=1$ and $\lambda_3 = 3$, with corresponding eigenvectors $(1,2,2)$ and $(1,0,0)$. Now what? I do not know how to proceed, nor what it means that my matrix is built up by Jordan blocks.

  • $\begingroup$ Good question, OP. $\endgroup$ – Don Larynx Dec 6 '13 at 3:50

Since this is a very small matrix there's actually no need to explicitly find the Jordan form by going through the routine tedious procedure. You know the characteristic polynomial is $$p(x) = (x-1)^2(x-3)$$ and you can check that the minimal polynomial is the same. This means a few things.

  1. You have a single Jordan block corresponding to $3$. This is just $(3)$.
  2. You cannot have two Jordan blocks corresponding to $1$ since that would make the matrix diagonalizable (which it is not since the minimal polynomial does not split into distinct factors). Therefore you must have a single block of the form $\begin{pmatrix}1 & 1 \\ 0 & 1\end{pmatrix}$

These combined gives you the form (up to order of the blocks) $$J=\begin{pmatrix}1 & 1 & 0 \\ 0 & 1 & 0 \\ 0& 0 & 3\end{pmatrix}$$

In general, for small matrices like these (anything up to $6\times 6$) you can find the Jordan form through similar types of analysis using the facts

  1. The geometric multiplicity of an eigenvalue is the number of blocks corresponding to it.
  2. The algebraic multiplicity of an eigenvalue is the sum of the total sizes of the blocks.
  3. The exponent of the term corresponding to an eigenvalue in the minimal polynomial is the size of the largest block.
  • 1
    $\begingroup$ I take it what goes wrong at $7\times7$ is you need more information to distinguish $3-3-1$ from $3-2-2$? $\endgroup$ – Gerry Myerson Nov 6 '12 at 6:15
  • $\begingroup$ @GerryMyerson Yes, exactly. $\endgroup$ – EuYu Nov 6 '12 at 6:23

The Jordan Canonical form of this matrix is $$ \begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 3 \end{pmatrix} .$$

You can read about Jordan Canonical forms here. Especially in section Complex matrices the properties listed there can help you find the Jordan Canonical form of most matrices.


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