A question on orthogonal eigenvectors of positive definite matrices 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
 A: 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).
A: 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.
