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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

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  • $\begingroup$ 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? $\endgroup$
    – Fabian
    Mar 5, 2011 at 14:47
  • $\begingroup$ 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. $\endgroup$
    – Calle
    Mar 5, 2011 at 14:57
  • $\begingroup$ @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? $\endgroup$
    – Shiyu
    Mar 5, 2011 at 15:05
  • $\begingroup$ @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. $\endgroup$
    – Shiyu
    Mar 5, 2011 at 15:13
  • $\begingroup$ 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. $\endgroup$
    – Fabian
    Mar 5, 2011 at 15:14

2 Answers 2

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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).

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    $\begingroup$ Why $A$ is positive definite? $\endgroup$
    – Fabian
    Mar 5, 2011 at 15:46
  • $\begingroup$ I thought that was assumed. Maybe it isn't? $\endgroup$
    – Calle
    Mar 5, 2011 at 15:57
  • $\begingroup$ Thank you Calle! This is what I need. BTW, from $DB^{-1} =CB^T$ we can directly get $C^{-1} D=B^T B$. $\endgroup$
    – Shiyu
    Mar 5, 2011 at 16:13
  • $\begingroup$ @Shiyu: You're absolutely right. This is just the way I worked things out (which isn't necessarily the straight way). $\endgroup$
    – Calle
    Mar 5, 2011 at 16:31
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    $\begingroup$ @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 $D=diag(c_1 b_1^T b_1, c_2 b_2^T b_2, 0)$. $\endgroup$
    – Shiyu
    Mar 5, 2011 at 16:44
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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.

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  • $\begingroup$ 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. $\endgroup$
    – Shiyu
    Mar 5, 2011 at 16:33

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