Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I got stuck in this problem from Spring 99, Berkeley Problems in Mathematics:

Let $A$ be a $n\times n$ matrix such that $a_{ij}\not=0$ if $i=j+1$ but $a_{ij}=0$ if $i\ge j+2$. Prove that $A$ cannot have more than one Jordan block for any eigenvalue.

I thought the matrix would satisfy some obvious relationship like $A^{2}=0$, but I realized the entries not listed are not even specified; thus such a gross simplification cannot hold. Working on toy examples does not tell me much, so I decided to ask in here.

share|cite|improve this question
up vote 3 down vote accepted

So, let $\lambda$ be an eigenvalue. The number of Jordan Blocks for $\lambda$ is well known to be the geometric multiplicity of $\lambda$. As the geometric multiplicity of an eigenvalue is the dimension of the nullspace of $A - \lambda I$, therefore, we need to show that this dimension is $1$. Hence, we only need to show that the nullspace of $B = A - \lambda I$ has only one vector in it (i.e. all the vectors in the null space are multiples of that vector).

To show this, let $v_1, v_2, \cdots, v_n$ be the column vectors of $B$. Note that if $v = (a_1, a_2, \cdots, a_n)$ is a vector in the nullspace, then $a_1 v_1 + a_2 v_2 + \cdots + a_nv_n = 0$. Let's fix $a_1 =1$. Comparing the first component of the columns, gives a non trivial relationship between $a_1,a_2$ as the rest of the components are $0$. Hence, $a_2$ is fixed. Similarly, comparing second component fixes $a_3$. Continuing this procedure, we get the assertion.

share|cite|improve this answer
Excuse me for my low level, but why we only need to show the nullspace of $B$ has only one vector in it? – Bombyx mori Jul 31 '12 at 16:01
I've edited that part. Hope it's clearer now. – Rijul Saini Jul 31 '12 at 16:06
Very slick proof. Sorry for the late endorsement. – Bombyx mori Aug 31 '12 at 9:30

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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