Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Find an orthogonal basis for $\mathbb R^3$ consisting of the eigenvectors of the matrix $$\begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix}$$

Isn't this question basically just asking 'find the eigenvectors of this matrix'? And the part about finding 'an orthogonal basis' is irrelevant?

share|cite|improve this question
The insistence on orthgonality is not irrelevant. If some eigenspace has dimension greater than 1 then you need to be careful to pick a basis for that eigenspace consisting of mutually perpendicular eigenvectors. It definitely isn't always going to work if you don't make a good choice. To take an extreme case, try the 3x3 identity matrix: any nonzero vector is an eigenvector but it's far from true that all choices of eigenbases are mutually perpendicular. In your matrix, one eigenvalue has multiplicity 2, so the requirement that your basis be orthogonal has content. – KCd Apr 22 '12 at 10:46
The question tells me that this matrix contains a basis for $\mathbb R^3$ so I know it has 3 eigenvectors. It's not like I then have a choice of eigenvectors...I just work out the eigenvectors - as (1,1,1), (-1,1,0) and (-1,0,1) - and that's it. So I can't see what relevance there is to specifying orthogonality in this question? – Jim_CS Apr 22 '12 at 14:55
@KCd OK I see you are correct. I dotted those vectors are and they are not orthogonal. So how can I get an orthogonal basis? – Jim_CS Apr 22 '12 at 20:41

Sorry I got mixed up earlier with the statement of the Real Spectral Theorem. It tells you that there exists an orthogonal basis for $\Bbb{R}^3$ consisting of eigenvectors of your matrix $A$ with all eigenvalues real. So indeed finding the correct vectors in the eigenspace to be orthogonal is not immediate from the outset.

share|cite|improve this answer
Isnt it also the case that the question $tells$ me that this matrix will have 3 its not like I have a choice of choosing 'ones that are orthogonal'...I cant see why they bothered with the 'Find an orthogonal basis' part of the question. – Jim_CS Apr 22 '12 at 10:38
This is incorrect. The eigenvectors $[1,-1,0]$ and $[1,0,-1]$ are not orthogonal and are part of an eigenbasis. There really is work to do here to make sure the eigenbasis is an orthogonal basis. – KCd Apr 22 '12 at 10:49
@KCd I was confused about the statement of the real spectral theorem. I have corrected that now. – user38268 Apr 22 '12 at 10:51

The symmetric matrix (call is $A$) has two eigenvalues, one of multiplicity 2 at -1, and one of multiplicity 1 at 5. The eigenspaces $\ker(A+I)$ and $\ker (A-5I)$ are orthogonal complements, so the only issue is choosing a basis for $\ker(A+I)$ that is orthogonal.

Choose $\frac{1}{\sqrt{3}}(1,1,1)^T$ as a basis for $\ker(A-5I)$ (not a huge amount of choice there).

Since $\ker(A+I) = \ker (1,1,1)^T$, we can choose a element of the null space, say $\frac{1}{\sqrt{2}} (1,-1,0)^T$, and just find an orthogonal vector (in $\ker(A+I)$, of course), for example: $\frac{1}{\sqrt{6}} (1,1,-2)^T$.

It is easy to check that these vectors are orthogonal, in fact, orthonormal.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.