0
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ The nullity being 3 tells you more than just that $0$ is an eigenvalue.... $\endgroup$ – Gerry Myerson Aug 24 '15 at 13:05
4
$\begingroup$

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

$\endgroup$
  • $\begingroup$ $A\cdot v^t=v^t\cdot v\cdot v^t=0\cdot v^t$? 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.