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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.

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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.

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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

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