Conditions on certain entries of a matrix to ensure one Jordan block per eigenvalue

Question

In preparation for a future exam, I found the following problem:

Let $$A = \begin{pmatrix} 1 & 0 & a & b \\ 0 & 1 & 0 & 0 \\ 0 & c & 3 & -2 \\ 0 & d & 2 & -1 \end{pmatrix}$$ Determine conditions on $a,b,c,d$ so that there is only one Jordan block for each eigenvalue of $A$ (in the Jordan canonical form of $A$).

I found, readily enough, that the only eigenvalue of the $A$ is $\lambda = 1$. So the problem is in fact to find conditions of $a,b,c,d$ such that $A$ is similar to a single-block Jordan matrix. Thus we want the dimension of the eigenspace $E_1$ to be 1, or equivalently, the rank of $A-I$ to be 3.

$$A-I = \begin{pmatrix} 0 & 0 & a & b \\ 0 & 0 & 0 & 0 \\ 0 & c & 2 & -2 \\ 0 & d & 2 & -2 \end{pmatrix}$$

So would the conditions $a,b$ not both zero, $a \neq -b$ and $c \neq d$ be sufficient? More to the point, what is the most economical way of stating the conditions on $a,b,c,d$ to ensure the desired outcome?

Answer 1

The question asks to determine conditions on $a$, $b$, $c$, and $d$ so that the Jordan form of $A$ has only one block. This is equivalent to determining conditions on $a$, $b$, $c$, and $d$ so that each eigenspace has dimension one. Since $\lambda=1$ is the only eigenvalue, we need only determine conditions on $a$, $b$, $c$, and $d$ so that $A-I$ has rank three (by the Rank-Nullity theorem).

One way to ensure that $A-I$ has rank three is to ensure that $A-I$ has a nonvanishing minor of size three. That is, we may demand that the determinant of the submatrix of $A-I$ obtained by deleting the first column and the second row is nonzero. Since $$ \det \begin{bmatrix} 0&a&b\\ c&2&-2\\ d&2&-2 \end{bmatrix}=2(a+b)(c-d) $$ our conditions are $a\neq -b$ and $c\neq d$.

Answer 2

You want that $A$ is a non-derogatory matrix, that is equivalent to $(A-I)³\not=0$.

Answer 3

Another approach: There is only one Jordan Block for each eigenvalue of a matrix when the minimum polynomial is the same as the characteristic polynomial. You have the characteristic equation: $(x-1)^4$, so you just need the minimum polynomial to be the same. This is equivalent to $(A-I)^3 \neq 0$. Now \begin{equation} (A-I)^3 = \begin{bmatrix} 0 & a (2 c - 2 d) + b (2 c - 2 d) &0 &0\\0&0&0&0\\0&0&0&0\\0&0&0&0\end{bmatrix}. \end{equation} So if we don't want this matrix to be the zero matrix, we must have $a (2 c - 2 d) + b (2 c - 2 d)\neq0$, which is the same conclusion arrived at by Brian Fitzpatrick...so this just shows a different way to get there...