# Example of eigenvectors in different bases (follow-up question)

This is a follow-up question on this one: Connection between eigenvalues and eigenvectors of a matrix in different bases

Assume I have matrix $$B=\left( \begin{array}{cccc} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \end{array} \right).$$

It's eigenvalues are $-1,-1,1,1$ with corresponding eigenvectors $(0,-1,0,1)^T,(-1,0,1,0)^T,(0,1,0,1)^T,(1,0,1,0)^T$.

Now if I chose a different basis, e.g. $d_1=(2,1,0,0)^T,d_2=(0,0,2,1)^T,d_3=(2,-1,0,0)^T,d_4=(0,0,2,-1)^T$ it becomes: $$[B]_{\mathcal{D}}=\left( \begin{array}{cccc} 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \end{array} \right).$$

Now, the eigenvalues stay the same (as they should) but I have difficulties with the transformation of the eigenvectors. When I calculate them with $\mathcal{D}^{-1} v$ I e.g. get for the eigenvalue -1 $(-0.5,0.5,0.5,-0.5)^T$ and $(-0.25,0.25,-0.25,0.25)^T$ instead of the correct $(0,0,-1,1)^T$ and $(-1,1,0,0)^T$. Both are obviously not multiples of each other.

Can anybody help - what is wrong here (it must be some careless mistake, I guess)? Thank you!

EDIT
There was a nasty typo in the first matrix - the last $1$ should have been a $0$ - now everything seems correct. I am really sorry!

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Calculation error presumably. Eyeballing, it looks like $(0,-1,0,1)^T$ is not an eigenvector for eignvalue $-1$. Maybe that's one source of error. – André Nicolas Jun 24 '11 at 21:46
@user6312: I think this was due to my unfortunate typo (see my edit) - I am so sorry! – vonjd Jun 25 '11 at 8:15

The (-1) eigenvectors you got span the same space as the vectors you were expecting to get, so your answer is correct. Note that if $v_1$ and $v_2$ are eigenvectors with the same eigenvalue, then so is any linear combination $av_1 + bv_2$. In your case, if $v_1=(0,0,-1,1)^T$ and $v_2=(-1,1,0,0)^T$, then the vectors you computed are $\frac{1}{2}(v_2 - v_1)$ and $\frac{1}{4}(v_2 + v_1)$.

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I am so sorry - I just found a typo in matrix A - does this change your answer? Sorry for the inconvenience! – vonjd Jun 25 '11 at 8:13
...I read through your answer again and I think it shouldn't be affected by the typo - Thank you in any case! – vonjd Jun 25 '11 at 8:39

First: your computations are incorrect. The eigenvectors of $B$ corresponding to $-1$ are the nullspace of $$\left(\begin{array}{rrrr} -1 & 0 & 1 & 0\\ 0 & -1 & 0 & 1\\ 1 & 0 & -1 & 0\\ 0 & 1 & 0 & 0 \end{array}\right),$$ so they are all vectors of the form $(a,b,c,d)$ with $a+c=0$, $b=d=0$. That is, they are all multiples of $(1,0,-1,0)$ alone.

Second: note that I did not say that $(1,0,-1,0)$ is the eigenvector corresponding to $-1$; that's because there is more than one eigenvector corresponding to $-1$. Every nonzero linear combination of eigenvectors corresponding to $-1$ is an eigenvector corresponding to $-1$.

If the eigenspace has dimension $1$, then of course any two eigenvectors you find will necessarily be multiples of each other; but if the eigenspace has dimension greater than $1$, then two bases need not be related to one another as multiples, just like the vectors in the bases $\beta_1 = [(1,0), (0,1)]$ and $\beta_2 = [(2,1), (1,2)]$ of $\mathbb{R}^2$ are not multiples of each other, even though they span the exact same space. As Alex B. points out, the two vectors you got and the two vectors you were expecting span the same subspace, so the both sets of vectors give you the same set of eigenvectors.

Remember: when you found eigenvectors, you didn't finding the complete list of eigenvectors, you only found a basis for the eigenspace.

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I am really, really sorry - there was a typo in the first matrix - does this change your answer? Sorry for the inconvenience! – vonjd Jun 25 '11 at 8:12
...I read through your answer again and I think the first part will be affected by the typo but the more general and important second part shouldn't be - thank you again! – vonjd Jun 25 '11 at 8:40
@vonjd: Indeed: the part that starts "Second" is not affected. The point is that there aren't just two (or one) eigenvectors, but an entire subspace of eigenvectors (excluding the zero vector). When you took the basis of eigenvectors and used the translation matrix, you got one possible basis for that space; when you computed the basis directly from the transformed matrix, you got another basis for that space. But as Alex B. pointed out, the two vectors you got by translating span the exact same space as the two vectors you got "directly", so the two answers are equivalent. – Arturo Magidin Jun 25 '11 at 17:57