Can a tensor of rank-2 be diagonalized by two different set of basis?

Question

I have two inertia tensors for a body, both calculated in two different points of it, but both gave me a diagonalized matrix, is this possible? the tensor is a symmetric one.

Answer 1

Hint: Consider the identity matrix. Seems to be diagonalizable with respect to any basis.

Additional Just in case the other example is not enlightening. Consider the matrix \begin{align} A = \begin{pmatrix} 2& 1\\ 1 & 2 \end{pmatrix} \end{align} then by the spectral theorem we see that $A$ can be diagonalize by an orthogonal matrix, i.e. \begin{align} A = Q^T DQ. \end{align} Here by direct computation, we see that \begin{align} Q = \frac{1}{\sqrt{2}} \begin{pmatrix} -1 & 1\\ 1 & 1 \end{pmatrix}. \end{align} However, we could consider \begin{align} A = \lambda Q^T D(1/\lambda)Q = P^{-1}DP \end{align} where we used the fact that \begin{align} P = \frac{1}{\lambda} Q. \end{align} Thus, we have infinitely many ways of diagonalizing $A$.