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From the matrix $$A=\frac{1}{3}\begin{bmatrix}1&-2&2\\-2&-2&-1\\2&-1&-2\end{bmatrix}$$

I managed to find the eigenvalues $-1,-1,$ and $\,1$. For the eigenvalue $1$, I found the eigenvector $$E_1=\begin{bmatrix}2\\-1\\1\end{bmatrix}$$

And whilst working to find the eigenvectors for the eigenvalue $-1$, I was confused by how I could pick those two vectors:

$$(R+I)X=0 \iff \begin{bmatrix}2&-1&1\\0&0&0\\0&0&0\end{bmatrix}X=0$$

  • How do I pick two non parallel, in this case orthagonal because $A$ is symmetric, eigenvectors?

  • An interesting phenomenon occured. Whilst trying to find eigenvectors for eigenvalue $-1$ the top row of the matrix is the same as the eigenvector for the eigenvalue $1$, see $(2,-1,1)$. Is there any geometrical or algebraic reason behind this?

For the second question, I should mention that $A$ is the rotation matrix of $\pi$ radians about an axis with directional vector $E_1$.


Added the matrix $S$ formed by the eigenvectors of $R$.

$\left( \begin{array}{ccc} \frac{2}{\sqrt{6}} & \frac{1}{\sqrt{5}} & \frac{1}{\sqrt{5}} \\ -\frac{1}{\sqrt{6}} & \frac{2}{\sqrt{5}} & 0 \\ \frac{1}{\sqrt{6}} & 0 & -\frac{2}{\sqrt{5}} \\ \end{array} \right)$

But the equation $S^TRS = D$ is not satisfied.

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  • $\begingroup$ I made an error at first with my $(R+I)X=0$ matrix, fixed it now so that it shows the right one. $\endgroup$ – B. Lee Mar 7 '15 at 16:08
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You did something wrong in row-reducing: note that $$ A + I=\frac{1}{3} \pmatrix{ 4&-2&2\\ -2&1&-1\\ 2&-1&1 } $$ and each row is simply a multiple of the first. We therefore end up with $$ \pmatrix{ 2&-1&1\\ 0&0&0\\ 0&0&0 }x = 0 $$ We now want an orthonormal basis for the nullspace of this matrix. We begin with any basis, such as $$ \pmatrix{1\\2\\0},\pmatrix{1\\0\\-2} $$ and apply the Gram-Schmidt process.

The geometric reason for your observation is as follows: the null space for $A + I$ must be orthogonal to the null space of $A - I$. So, $A+I$ is a rank-one matrix whose nullspace is everything perpendicular to the vector forming the basis of the other eigenspace. All such matrices must reduce to the transpose of that vector.


Applying the Gram-Schmidt process to get an orthonormal basis for the eigenspace: take $$ v_1 = \frac{1}{\sqrt 5}\pmatrix{1\\2\\0}\\ u_2 = \pmatrix{1\\0\\-2} - \left(v_1 \cdot \pmatrix{1\\0\\-2}\right) v_1 = \pmatrix{4/5\\-2/5\\-2}\\ v_2 = u_2/\|u_2\| = \frac{1}{\sqrt{30}} \pmatrix{2\\-1\\-5} $$ Using this $v_1,v_2$ will give you the $S$ you want.

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    $\begingroup$ Yes, sorry. I corrected it. Actually where you ended is where I am confused. How do I go about choosing two eigenvectors (that aren't the same) from the last matrix you wrote? $\endgroup$ – B. Lee Mar 7 '15 at 16:12
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    $\begingroup$ See my latest edit $\endgroup$ – Omnomnomnom Mar 7 '15 at 16:19
  • $\begingroup$ You should look back in your notes to see how you went about finding the null space (AKA kernel) of a linear transformation. $\endgroup$ – Omnomnomnom Mar 7 '15 at 16:21
  • $\begingroup$ Actually, I would love to learn about kernel and such, but it has not been in our curriculum. Also, was your choices of vectors correct? Their dot product is 1, which means they aren't orthogonal, right? (See edit) $\endgroup$ – B. Lee Mar 7 '15 at 16:39
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    $\begingroup$ You have to apply the Gram Schmidt process $\endgroup$ – Omnomnomnom Mar 7 '15 at 17:29

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