Finding eigenvectors of symmetric matrices when the general solutions of eigenvalues inserted return several vectors

Given a symmetric matrix

$A = \left(\begin{array}{rrr} 2 & 1 & 1\\ 1 & 2 & 1\\ 1 & 1 & 2 \end{array}\right).$

By solving the characteristic equation I found out that the eigenvalues are as follows: $\lambda{_1} = 4$, $\lambda{_2} = 1$, $\lambda{_3} = 1$.

Solving $(A-\lambda{_1}I) = 0$ and the general solution to it's RREF returns:

$\vec w_2 = x_3 \left(\begin{array}{rrr} 1 \\ 1 \\ 1 \end{array}\right).$

Solving $(A-\lambda{_2}I) = 0$ and the general solution to it's RREF returns:

$\vec w_3 = x_2 \left(\begin{array}{rrr} 1 \\ -1 \\ 0 \end{array}\right).$

$\vec w_4 = x_3 \left(\begin{array}{rrr} 1 \\ 0 \\ -1 \end{array}\right).$

So since this is a symmetric matrix as we can see easily, the eigenvectors are supposed to be orthogonal to each other. I can immediately see that $\vec w_4$ is orthogonal to $\vec w_2$. The question now is in order to find the 3rd eigenvector, do you simply take the cross-product of $\vec w_4$ and $\vec w_2$ finding a vector orthogonal to both of them? Doing this I found this vector $\left(\begin{array}{rrr} 1 \\ -2 \\ 1 \end{array}\right).$

The last vector happen to be the right answer, but my question is if this really is the correct way to do it, or if that was simply luck (will it work in every case)? If the last is the case what's the correct method?

Correct answer is supposed to be:

• You should not get two independent vectors for $\lambda_1$, since it only has algebraic multiplicity $1$. There must be some mistake there. – KittyL May 7 '15 at 18:16
• $A - \lambda_1 E$ has rank 2, so general solution has to have dimension of 1. How did you get $w_1$? – uranix May 7 '15 at 18:19
• aah yea used the "wrong" notes, you're right, removed it from post. – jibo May 7 '15 at 18:27
• The correct statement is that the eigenspaces corresponding to distinct eigenvalues will be orthogonal. So two eigenvectors corresponding to different eigenvalues are necessarily orthogonal, but if they have the same eigenvalue, they may not be. – Joppy Sep 22 '18 at 7:52

Note that the matrix can be written as $I+ee^T$, where $e$ is a vector of ones.

The $I$ just shifts the eigenvalues by one, just concentrate on $A=e e^T$.

Note that $A e = n e$, so this gives one eigenvalue. Let $v$ be any vector such that $v \bot e$, then $Av=0$. Hence all other eigenvalues are zero.

Hence the eigenvalues of the original matrix are $1,....,1,n+1$.

In the problem above, you can pick any two vectors that are orthogonal to $e$, they are not unique.

• k so we could say this is a special case, elegant. But in general, when this is not the case, what would be the general procedure for finding the last eigenvector? – jibo May 7 '15 at 21:00
• @Mollart: Your question is a bit vague. – copper.hat May 7 '15 at 21:05
• Ill see if I can come up with specific example if I find one. – jibo May 8 '15 at 18:02
• But lets say generalize the above example, need 3 eigenvectors cause we have 2 eigenvalues, one of which has multiplicity 2. So since we have a symmetric matrix, its eigenvectors are supposed to be orthogonal to each other. In the case we end up in the same situation as here with 2 vectors from the general solutions that are orthogonal, but a 3rd one that isn't. What would be the procedure for finding the 3rd eigenvector, assuming we are not dealing with a matrix that can be written as $I+ee^T$. Do you understand what I mean now? – jibo May 8 '15 at 18:10
• Find the eigenvector corresponding to the multiplicity 1 eigenvalue, and the the orthogonal complement must be the eigenspace corresponding to the multiplicity 2 eigenvalue. – copper.hat May 8 '15 at 18:12

In given matrix sum of the elements in all the rows is 2 + 1 + 1 is equal to 4. Since this sum is same in all the rows so one of the eigen values is 4 Now, the trace of the matrix is equal to the sum of the eigenvalues so, sum of all the eigenvalues is equal to 6. Also, the determinant of the matrix is equal to the product to the eigenvalues which is equal to 4. Now, λ₁+λ₂+λ₃=6 & λ₁*λ₂*λ₃=4 so, λ₁=4, λ₂=1, λ₃=1