0
$\begingroup$

Let $\pi$ be the plane spanned up by vectors $v_1 = (1, 1, 1)$ and $v_2 = (2, 1, 3)$. Consider linear transformation that stretches the plane in the direction of $v_1$ by factor $2$ and in the direction of $v_2$ by factor $3$. Find all eigenvectors and eigenvalues for this transformation. In addition, find an explicit form of transformation matrix.

$\endgroup$
0
$\begingroup$

Hint:

The problem fix the eigenvalue $\lambda_1=2$ with eigenvector $v_1$ and the eigenvalue $\lambda_2=3$ with eigenvector $v_2$.

Without other conditions the other eigenvalue $\lambda_3$ can have any real value (note that it cannot have a complex value. Whay?).

$\endgroup$
  • $\begingroup$ based on the Av=λv we have Av1=2v1 and Av2= 3v2 am I right ? $\endgroup$ – user8523104 Sep 19 '17 at 9:12
  • $\begingroup$ Yes, but the third eigenvalue is free. $\endgroup$ – Emilio Novati Sep 19 '17 at 9:16
  • $\begingroup$ thank you when you say free, means Av3=0v3? or could you please tell me what do you mean exactly? Av1=2v1 and Av2= 3v2 then Av3=0v3 $\endgroup$ – user8523104 Sep 19 '17 at 9:18
  • $\begingroup$ Free means that $\lambda_3$ can be any real number. $\endgroup$ – Emilio Novati Sep 19 '17 at 19:42
  • $\begingroup$ so now let say our eigenvalues are 2,3,1 and these are vectors v1(1,0,2), v2(2,0,0), v3(0,1,0) then please how about the explicit form of transformation matrix? $\endgroup$ – user8523104 Sep 20 '17 at 6:41

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.