Take the 2-minute tour ×
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.

Suppose matrix $A\in R^{3 \times 3} $ and $$A= \left[ \begin{array}{ccc} b_1 & b_2 & b_3 \end{array} \right] \left[ \begin{array}{ccc} c_1 & 0 & 0 \\\ 0 & c_2 & 0 \\\ 0 & 0 & c_3 \end{array} \right] \left[ \begin{array}{ccc} b_1^T \\\ b_2^T \\\ b_3^T \end{array} \right] $$ where $b_i\in R^3$ and $B=(b_1,b_2,b_3)$ is non-singular.

If $Ab_i=\lambda_i b_i, i=1,2,3$, that is $b_i$ is the eigenvectors of A, can we say $ b_1,b_2,b_3 $ are orthogonal eigenvectors of $A$ and $ c_1,c_2,c_3$ are eigenvalues of $A$?

Thank you very much. Shiyu

share|improve this question
    
Short answer yes (there might be some problem if some of the c's are equal). Long answer: what exactly is your question? Why they are orthogonal? or why b's are eigenvectors? or why c's are eigenvalues? –  Fabian Mar 5 '11 at 14:47
    
You have chosen the $b$'s yourself haven't you? So you should know if they're orthogonal? If the matrix $[b_1, b_2, b_3]$ is orthogonal, the vectors are orthonormal. –  Calle Mar 5 '11 at 14:57
    
@Fabian: if A is positive definite, then $A$ can be factorized as $A=U\Lambda U^T$ where $U$ is a orthogonal matrix with its column vectors as the eigenvectors of $A$, and $\Lambda$ has eigenvalues of $A$ on its diagonal entries. What I want to know actually is the converse problem. If A can be factorized as shown above. Does the $c_i$ is the eigenvalues? And are all the vectors $b_i$ orthogonal? –  Shiyu Mar 5 '11 at 15:05
    
@Calle: No. They are not chosen by myself. What I am given is $A=BCB^T, C=diag(c_1,c_2,c_3)$ and $b_i$ is eigenvectors of $A$. What I want to know is whether $b_i$ are orthogonal eigenvectors and $c_i$ are eigenvalues if $A$ can be factorized as above. –  Shiyu Mar 5 '11 at 15:13
    
If all the $\lambda$s are different, then the eigenvectors of $A$ are necessarily orthogonal and the $c$'s are eigenvalues. If some of the $\lambda$s are the same, this is not necessarily the case. –  Fabian Mar 5 '11 at 15:14

2 Answers 2

up vote 2 down vote accepted

We know that $A$ has a basis of eigenvectors, since $B$ is non-singular. Thus we can write:

$$A = BDB^{-1}$$

where $D$ is a diagonal matrix containing the eigenvalues. Since $A$ is positive definite, all eigenvalues are positive, so $D$ is invertible.

$$BDB^{-1} = BCB^T \Leftrightarrow DB^{-1} = CB^T \Leftrightarrow B^{-1} = D^{-1}CB^T$$

So $B^{-1}B^{-T} = D^{-1}C$. Since $B^{-1}B^{-T}$ is invertible, $D^{-1}C$ has to be invertible (all $c_i \neq 0$), so we can write $B^TB = C^{-1}D$. Since $C^{-1}D$ is a diagonal matrix, this shows that the vectors $b_1, b_2, b_3$ are orthogonal (but not necessarily orthonormal).

share|improve this answer
1  
Why $A$ is positive definite? –  Fabian Mar 5 '11 at 15:46
    
I thought that was assumed. Maybe it isn't? –  Calle Mar 5 '11 at 15:57
    
Thank you Calle! This is what I need. BTW, from $DB^{-1} =CB^T$ we can directly get $C^{-1} D=B^T B$. –  Shiyu Mar 5 '11 at 16:13
    
@Shiyu: You're absolutely right. This is just the way I worked things out (which isn't necessarily the straight way). –  Calle Mar 5 '11 at 16:31
1  
@Fabian and Calle: I think Calle's answer is right as long as $c_i$ is not zero. In fact, when some of $c_i$ are zeros, we can still get some useful results. Note $D=CB^T B$ always hold even if some $c_i$ are zeros. For example, $C=diag(c_1,c_2,0)$, then from $D=CB^T B$ we can still get $b_1,b_2,b_3$ are orthogonal to each other. And $D=diag(c_1 b_1^T b_1, c_2 b_2^T b_2, 0)$. –  Shiyu Mar 5 '11 at 16:44

If all the λs are different, then the eigenvectors of A are necessarily orthogonal and the cs are eigenvalues (assuming the $b$s are also normalized). If some of the λs are the same, this is not necessarily the case.

As an example what goes wrong when some $\lambda$ are equal, take the case $A= I_2$ (a $3\times 3$ example is also easy to construct). We have $$ A= \begin{pmatrix} b_1 & b_2 \end{pmatrix} \begin{pmatrix} c_1 & 0 \\\ 0 & c_2 \end{pmatrix} \begin{pmatrix} b_1^T \\\ b_2^T \end{pmatrix} $$ with $$ \begin{align} c_1 &=1 & c_2 &=2 \\ b_1 &= \frac{1}{\sqrt{2}}\begin{pmatrix} 1\\1\end{pmatrix} & b_2 & = \frac{1}{2} \begin{pmatrix} 1 \\ -1 \end{pmatrix} \end{align}, $$ so the $b$s are orthogonal but the $c$s are not the eigenvalues. (Calle post shows that in this case the $b$s have to be orthogonal)

To find an example where the $b$s are not orthogonal, take $A=0_2$. We can write the matrix in the form given above with $$ \begin{align} c_1 &=1 & c_2 &=-1 \\ b_1 &= \begin{pmatrix} 1\\1\end{pmatrix} & b_2 & = -b_1 \end{align}, $$ so neither are the $c$s the eigenvalues nor the $b$s orthogonal.

For the case when all the $\lambda$s are different, the $b$s are up to normalization uniquely determined. They are automatically orthogonal on each other. If they are not normalized the $c$s are not necessarily the eigenvalues. But if you assume them normalized then the $c$s correspond to the $\lambda$s.

share|improve this answer
    
Thank you for your answer and examples. Yes, you are right on the question whether $c_i$ are eigenvalues of $A$. If $b_i$ is normalized, $c_i$ is an eigenvalue of $A$. In fact, this can be seen from Calle's answer. From $B^T B=C^{-1}D$ we know $||b_i||^2 =\lambda_i / c_i$. And it seems this result holds as long as $c_i$ is not zero. When $c_i<0$, $\lambda_i<0$. Regarding to your second example, it goes a little far because $b_2=-b_1$ makes $B$ singular. –  Shiyu Mar 5 '11 at 16:33

Your Answer

 
discard

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.