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I have a problem in my studies and am trying to prove it. I have worked so far, but need some advice as to whether I'm handling it correctly:

Let $v$ be a column vector of size $n$ and let $A = vv^t$ be an $n x n $ matrix. How many non-zero eigenvalues does this matrix A have? Also what are the eigenvalues of A and what are the corresponding eigenvectors?

so far I have the following $$ A = vv^t $$ $$ Av=\lambda v $$ $$vv^t (v) = \lambda v$$ the non-zero eigenvalues is the n power of its characteristic polynomial given by equation $$P_A(t) = \prod_{i=1}^{n} (t - \lambda_i)$$ n is the amount of eigenvalues.

Because this matrix is diagonalizable, the eigenvalues can be found with matrix $P$ as the eigenvectors $$P^{-1} vv^tP = D$$ the eigenvalues are the diagonal of $D$

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  • $\begingroup$ I suppose your matrix is $A=v^tv$ and, if so, notice that any row is proportional to $v$ so that his rank is at most $1$.. $\endgroup$ – Emilio Novati Feb 16 '15 at 14:58
  • $\begingroup$ a similar question: math.stackexchange.com/questions/1143472/… $\endgroup$ – Emilio Novati Feb 16 '15 at 15:04
  • $\begingroup$ I know that the rank is the dimension of its image and is related to its nullity. Does that mean that if the rank is 1 it has 1 eigenvalue and 1 eigenvector? I am sorry but can you please explain further? $\endgroup$ – liujm Feb 16 '15 at 15:19
  • $\begingroup$ Yes. If the rank is $1$ then $0$ is an eigenvalue of multilicity $n-1$. $\endgroup$ – Emilio Novati Feb 16 '15 at 15:39

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