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The problem: Being given the matrix: $$ \begin{bmatrix} 0 & -1 & -1 \\ 1 & 2 & 1 \\ -1 & -1 & 0 \end{bmatrix}$$ and two of its eigenvalues $0,1$. Obtain matrix $P$ (whose columns are associated to the eigenvalues)

My attempts: I know the formula first I use 0 and subtract it from the entries across the diagonal and get:

the same matrix. then I add row 2 to row 3 and get:

$$ \begin{bmatrix} 0 & -1 & -1 \\ 1 & 2 & 1 \\ 0 & 1 & 1 \end{bmatrix} $$

Then I added row one to row 3, row flipped one and two and got: $$ \begin{bmatrix} 1 & 2 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{bmatrix} $$ and the solution that I got was:

x = $$ \begin{bmatrix} 1 \\ -1\\ 1 \end{bmatrix} $$

but the book says no, it's:

$$ \begin{bmatrix} -1 \\ 1 \\ -1 \end{bmatrix} $$

For the corresponding lambda =1, after subtracting the diagonals I get: $$ \begin{bmatrix} -1 & -1 & -1 \\ 1 & 1 & 1 \\ -1 & -1 & 0 \end{bmatrix} $$

Then I added row one and row two, multiplied row one by -1: $$ \begin{bmatrix} 1 & 1 & 1 \\ 0 & 0 & 0 \\ -1 & -1 & 0 \end{bmatrix} $$

All I can do here is add row one and three and get: $$ \begin{bmatrix} 1 & 1 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix} $$ leaving one free variable and well quite honestly it doesn't matter what my answer is from here because I only have two columns for P. Something is seriously wrong I don't know if I am making a logical error in my row operations and happen to be committing a logical no no or what? I have checked over it and I just can't for the life of me figure out the what problem is. Please give a concise explanation. Thank you.

As for eigenvalues being defined only up to a constant I don't understand what you mean by my answer is okay if its negative of the book moreover I lack the proper rows.

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    $\begingroup$ Your inability to do this does not mean you are stupid... $\endgroup$ – John Martin Apr 27 '16 at 15:59
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    $\begingroup$ eigenvectors are only well-defined up to a constant, so if your answer is the negative of the book's answer you're fine $\endgroup$ – hunter Apr 27 '16 at 16:01
  • $\begingroup$ Okay I will adjust accordingly looks like the community would prefer it but it is how I feel. $\endgroup$ – K. Gibson Apr 27 '16 at 16:01
  • $\begingroup$ For matrix $A$ with an eigenvalue $\lambda$, if $\mathbf v$ is an eigenvector that satisfies $A\mathbf v = \lambda \mathbf v$, then for non-zero scalar $k$, it is also true that $$A(k\mathbf v) = \lambda (k\mathbf v)$$ So $k\mathbf v$ is also an eigenvector associated with $\lambda$, but obviously that is linearly dependent to $\mathbf v$. $\endgroup$ – peterwhy Apr 27 '16 at 16:12
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The eigenvector you obtained and the one in the book differ only in sign, that is by multiplication by $-1$. Both vectors are eigenvectors for eigenvalue $0$ and equally correct answers.

For the other eigenvalue, you just forgot to subtract in the last line. If you correct this you will get two free variables.

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  • $\begingroup$ Okay now I see it. Thank you. I don't know why errors like that in the last line always evade me. $\endgroup$ – K. Gibson Apr 27 '16 at 16:14
  • $\begingroup$ You are welcome. In this case the problem may have been that you thought you were making a more serious error. But, in fact it was just a minor glitch and you seem on a good way to understand the material. :-) $\endgroup$ – quid Apr 27 '16 at 16:16

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