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?


  • $\begingroup$ 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

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 $$J=J_3(1)\oplus J_1(1)\ .$$


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