-3
$\begingroup$

Is Lambda=4 is eigenvalue of

$$ \left( \begin{matrix} 3 & 0 & -1 \\ 2 & 3 & 1 \\ -3 & 4 & 5 \\ \end{matrix} \right) $$ If so find one corresponding eigenvector.

I know how to determine if the given lambda value is eigenvalue. I just solve $ (A-\lambda I)x=0$. But I'm not sure what approach to use in finding eigenvector of that matrix.

Thanks

$\endgroup$
  • $\begingroup$ I don't know a best way than an other to find eigenvector... $\endgroup$ – Hexacoordinate-C Mar 14 '15 at 19:20
  • $\begingroup$ Set the determinant of the relevant matrix equal to zero? $\endgroup$ – Mark Bennet Mar 14 '15 at 23:00
1
$\begingroup$

when i row reduce the matrix $A - 4I$ i get $$\pmatrix{1&0&1\\0&1&1\\0&0&0} $$ therefore an eigenvector corresponding to the eigenvalue $4$ is $$\pmatrix{1\\1\\-1}. $$

$\endgroup$
  • $\begingroup$ Thanks! But why is it -x3 as a common?Why not just x3? $\endgroup$ – user3273345 Mar 14 '15 at 19:32
  • $\begingroup$ @user3273345, any nonzero multiple of an eigenvector is also an eigenvector. i just picked $x_3 = -1$ so that we can have $x_1 = x_2 = 1.$ had i picked $x_3 = 1,$ then $x_1 = x_2 = -1$ $\endgroup$ – abel Mar 14 '15 at 19:35

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.