3
$\begingroup$

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?

$\endgroup$

3 Answers 3

4
$\begingroup$

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...

$\endgroup$
1
  • $\begingroup$ Thanks -- good point about the minimal polynomial being the same as the characteristic polynomial. $\endgroup$
    – bosmacs
    Apr 29, 2014 at 13:53
3
$\begingroup$

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 $c\neq d$.

$\endgroup$
1
$\begingroup$

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

$\endgroup$

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.