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This question already has an answer here:

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?

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marked as duplicate by Marc van Leeuwen linear-algebra Feb 12 '15 at 9:32

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.

  • $\begingroup$ If you're missing just one eigenvalue, since the trace is the sum of the eigenvalues, ... $\endgroup$ – xavierm02 Jul 28 '14 at 15:11
  • $\begingroup$ @xavierm02 So the remaining eigenvalue is $<u,v>.$ But how do I get the minimal polynomial from this? $\endgroup$ – lovelesswang Jul 29 '14 at 7:03
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Hint: the vector $u$ is an eigenvector

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