Minimal polynomial of a matrix whose elements have a certain form

Question

Find the minimal polynomial of the $n$-dimensional matrix $(a_{ij})$ when the matrix elements $a_{ij}$ have the form $a_{ij} = u_i v_j.$

Let $A=uv^T$ where $u,v$ are column vectors.

Then rank$(A)\leq$rank$(u)\leq1.$ So kernal$(A)\geq n-1.$ That is, the geometric multiplicity $\geq n-1.$ According to Jordan decomposition theorem, the number of Jordan blocks w.r.t. $0\ \geq n-1.$ Therefore, the algebraic multiplicity of $0$ $\geq n-1.$

Suppose rank$(A)=1,$ how do I find the other eigenvalue?

If you're missing just one eigenvalue, since the trace is the sum of the eigenvalues, ... — xavierm02, Jul 28 '14 at 15:11
@xavierm02 So the remaining eigenvalue is $<u,v>.$ But how do I get the minimal polynomial from this? — lovelesswang, Jul 29 '14 at 07:03

score 0 · Answer 1 · answered Jul 28 '14 at 17:06

0

Hint: the vector $u$ is an eigenvector

answered Jul 28 '14 at 17:06

Ben Grossmann

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Minimal polynomial of a matrix whose elements have a certain form

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