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
Best Answer
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).