0
$\begingroup$

Find matrix P that orthogonally diagonalizes C. $$ C= \begin{pmatrix} 1&2&0\\2&1&0\\ 0&0&5\\ \end{pmatrix} $$

I have worked through this problem by taking the eigenvectors and then normalizing the answer. The answer I get is

$$ P= \begin{pmatrix} -1/\sqrt2&1/\sqrt2&0\\1/\sqrt2&1/\sqrt2&0\\ 0&0&1\\ \end{pmatrix} $$

My book though gives the following answer of

$$ P= \begin{pmatrix} 1/\sqrt2&1/\sqrt2&0\\1/\sqrt2&-1/\sqrt2&0\\ 0&0&1\\ \end{pmatrix} $$

The eigenvalues I got were $5,3,$and $-1$. The eigenvectors and normalization I got were $(0,0,1)=(0,0,1)$ $(1,1,0)=(1/\sqrt2,1/\sqrt2,0)$ and $(-1,1,0)=(-1/\sqrt2,1/\sqrt2,0)$.

Which one is correct?, if any.

$\endgroup$
6
  • 3
    $\begingroup$ Why not compute $P^{-1}CP$ for both choices of $P$ to see what you get? $\endgroup$
    – Jason DeVito
    Aug 5 at 20:45
  • $\begingroup$ I get the same answer for both, the original matrix. $\endgroup$
    – user242559
    Aug 5 at 20:51
  • 1
    $\begingroup$ Why have you chosen to order your basis as $\{(0,0,1),(1,1,0),(-1,1,0)\}$ (well, the ON basis of this) and not any of the other ways? Would that change your matrix $P$? How? And does the ordering of the basis affect if $P$ can orthogonally diagonalise $C$? $\endgroup$
    – asking questions
    Aug 5 at 20:52
  • 1
    $\begingroup$ Diagonalizing matrices are not unique, since you can modify the basis you want to get another basis (e.g., multiplying an eigenvector by $1$ or $-1$ will not change orthonormality). Here, it's a matter of the order in which you put the eigenvectors, and possibly multiplying an eigenvector by $-1$. $\endgroup$
    – Arturo Magidin
    Aug 5 at 20:53
  • $\begingroup$ I’m not quite sure how the textbook got the y value to be negative and not the x-value.Unless they did multiply the eigenvector by -1, which seems unnecessary. $\endgroup$
    – user242559
    Aug 5 at 20:56

0

Reset to default

Your Answer

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

Browse other questions tagged or ask your own question.