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Find the characteristic polynomial and minimal polynomial of the rank-$1$ matrix $uv^*\in\mathbb{C}^{n\times n}$

I wasn't able to do any and couldn't understand the parts of the answer which are highlighted:

$(uv^*)u = (v^*u)u$ so $\lambda = v^*u$ is an eigenvalue.

Let $v_2,...,v_n$ be $n-1$ linearly independent vectors such that $v^*v_i=0, i=2,...,n$

(I thought $v$ is a column vector with $n$ entries. But this line is making each entry a column vector itself which confuses me. Especially since I proved the first line of the answer going through the definition of the inner and outer product for $n\times1$ column vectors.) Then $(uv^*)v_i=0$ so $\lambda=0$ is an eigenvalue of multiplicity $n-1$. We deduce that the char poly is $p(\lambda )=\lambda^{n-1}(\lambda-v^*u)$ and that the

min poly is $q(\lambda )=\lambda (\lambda-v^*u)$.

(I know the min poly is the same as the char poly except the powers are different. They are determined by the largest size of the Jordan block rather than the algebraic multiplicity as in the char poly. I'm just not sure how it was done here).

Any help would be appreciated.


marked as duplicate by Marc van Leeuwen linear-algebra Oct 10 '16 at 7:49

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.


For the first point: $v^*$ is a nonzero linear form $\Bbb C^n\to\Bbb C$, so its kernel has dimension $n-1$ (by rank-nullity). They are just choosing a basis $[v_2,\ldots,v_n]$ for that kernel here. (Later they should in fact either say "geometric multiplicity $n-1$", or (algebraic) "multiplicity at least $n-1$", since multiplicity $n$ is possible).

In the second point, they don't actually give any argument why exactly that power of (the indeterminate) $\lambda$ is needed for the minimal polynomial. But you can fairly easily check that the given polynomial actually annihilates your matrix $uv^*$; see this answer. In addition one should check in the special case $v^*u=0$ that the minimal polynomial is indeed $\lambda^2$ and not $\lambda$. But a minimal polynomial of $\lambda$ would imply that the matrix is zero, which is incompatible with the assumption that it has rank $1$.


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