3
$\begingroup$

Suppose you have the matrix $A = \begin{bmatrix} 1 & 0 & 1 & -1 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix}$

Put it into Jordan Canonical Form.

I get that the eigenvalue is 1 with multiplicity 4, the corresponding eigenvectors are $x = \begin{bmatrix}1\\0\\0\\0\end{bmatrix}$ and $y = \begin{bmatrix}0\\1\\0\\0\end{bmatrix}$.

When I try to construct the Jordan Block, however, I cannot figure out if the Jordan matrix has to consist of two 2x2 blocks or a 3x3 and 1x1 block. Because $(A-I)w = x$ and $(A-I)w=y$ are both inconsistent (for w$\in \mathbb{R^4}$). So by the theory we had so far about the Jordan Canonical Form, there can be no 2x2 blocks, let alone 3x3 blocks, for $(A - I)w$ is inconsistent for both x and y. How would you solve this problem?

Thanks!

$\endgroup$
  • $\begingroup$ The size of the largest block is the smallest natural number $k$ such that $\text{rank}\left(A-I_4\right)^k=\text{rank}\left(A-I_4\right)^{k+1}$, which is $3$. You must have made some mistake when finding 4w$. $\endgroup$ – Git Gud Apr 9 '14 at 23:23
1
$\begingroup$

Note that a good understanding of the theory behind Jordan forms will often provide many short cuts. You already know that $A$ has eigenvalue $1$ with algebraic multiplicity $4$ and that $A-I$ has nullity $2$. Therefore $(A-I)^2$ has nullity $3$ or $4$. Doing a bit of calculation, we have $$(A-I)^2=\pmatrix{0&0&0&1\cr\cdot&\cdot&\cdot&\cdot\cr\cdot&\cdot&\cdot&\cdot\cr\cdot&\cdot&\cdot&\cdot\cr}$$ and without computing any further entries we can see that this does not have nullity $4$. Therefore $(A-I)^2$ has nullity $3$. So, without any computation, $(A-I)^3$ will have nullity $4$ and the Jordan form will be $$J=J_3(1)\oplus J_1(1)\ .$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.