Let $v=(3 ,1 ,3 ,-4)$ and $A=v^t\cdot v$. Find the eigenvalues and eigenvectors of $A$ without calculating the whole matrix $A$

$Rank(AB) \leq min((Rank(A),Rank(B))$ so $Rank(A)\leq 1$ but because $A$ isn't the zero matrix $Rank(A)\neq 0$ and therefore $Rank(A)=1$.

$A$ is $4 \times 4$ size matrix so by rank-nullity theorem we get $4 = 1 + Null(A)$ $\rightarrow$ $Null(A)=3$ so $A$ is not invertible and $0$ is one of its eigenvalues.

How can I find the other eigenvalues and eigenvectors?

  • $\begingroup$ The nullity being 3 tells you more than just that $0$ is an eigenvalue.... $\endgroup$ – Gerry Myerson Aug 24 '15 at 13:05

HINT: What happens if you compute $A \cdot v^t$?

  • $\begingroup$ $A\cdot v^t=v^t\cdot v\cdot v^t=0\cdot v^t$? So I need to look for $v^t\cdot v=0$? $\endgroup$ – gbox Aug 24 '15 at 11:42
  • $\begingroup$ No, $v \cdot v^t = \left\Vert v \right\Vert^2 = 35$, i.e. $A \cdot v^t = v^t \cdot 35 = 35 \cdot v^t$. $\endgroup$ – GenericNickname Aug 24 '15 at 11:43

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