Let's say I have two matrices $A$ and $B$. Both as square symmetric and of same dimension $n \times n$. I can have a primary decomposition in terms of Eigen-spaces and Eigenvalues. Lets say $u_1,\cdots,u_n$ are the eigen-vector decomposition for matrix $A$.
Now there is a simultaneous diagonalizability theorem which states that if $A$ and $B$ commute then there exists a common orthonormal basis in which both these matrices are diagonalizable. However does this mean that if $u_1$ is an eigenvector correponding to $\lambda_1(A)$, (the highest eigenvalue corresponding to $A$) then $u_1$ would also be the eigenvector corresponding to $\lambda_1(B)$, the highest eigenvalue of $B$, where $u_1,\cdots,u_n$ are the common orthonormal basis which diagonalizes both matrices?
If not, does there exist any result which asserts that not only there exists a common orthonormal basis, but also the order of the eigen-values corresponding to this basis remains the same for both matrices.
Thanks
Suppose that$$A=\begin{bmatrix}1&0\\0&-1\end{bmatrix}\quad\text{and that}\quad B=\begin{bmatrix}-1&0\\0&1\end{bmatrix}.$$Then the only orthonormal basis of eigenvectors of each of them are those in which the first vector is $\pm(1,0)$ and the second vector is $\pm(0,1)$, and those with the same vectors by the inverse order. But then the eigenvalue corresponding to the first vector with respect to the matrix $A$ is the eigenvalue corresponding to the second vector with respect to the matrix $B$. So, the answer is negative.