If the principle invariants of two second order tensors $A$ and $B$ are equal, then the eigenvalues of $A$ and $B$ are equal. How to prove? Problem is in the title. I found this problem in an a continuum mechanics textbook intended for applied mathematicians, so naturally everything in this problem is defined over $\mathcal{R}^3$.
The principle invariants of a second order tensor $M$ are defined as
$$
\def\tr{\operatorname{tr}}
\begin{align*} 
I_1 &= \tr(M) \\
I_2 &= \frac{1}{2}\bigl[ (\tr(M))^2 - \tr(M^2) \bigr] \\
I_3 &= \det(M)
\end{align*}
$$
Defining the principle invariants in terms of eigenvalues and assuming the matrix is symmetric (which is reasonable given the context of continuum mechanics), we can write
$$
\begin{align*}
a_1 + a_2 + a_3 &= b_1 + b_2 + b_3 \\
a_1a_2 + a_1a_3 + a_2a_3 &= b_1b_2 + b_1b_3 + b_2b_3 \\
a_1a_2a_3 &= b_1b_2b_3,
\end{align*}
$$
where $a_i$ represents the eigenvalues of $A$ and the same for $B$. Now, my plan to show the equality of these eigenvalues was to expand the equation
$$
(a_1 - b_1)(a_2 - b_2)(a_3 - b_3) = 0
$$
and represent the product in terms of the prior equations. However, the algebra has taken up many pages and hours so far and I'm no longer so certain that I'm doing this right. Is my strategy flawed? Perhaps there's a theorem in linear algebra that I've forgotten that might help out?
Thank you for reading! :)
 A: Note that $a_{1,2,3}$ are the three roots of the polynomial $$f_a(x)=(x-a_1)(x-a_2)(x-a_3),$$ and similarly for the $b_i$.  Can you show that the two polynomials are equal, and therefore have the same roots?
A: Choose a basis so that the tensor is represented by a matrix. The characteristic polynomial doesn't depend on the choice of basis, and can be represented in two ways. Expanded, its coefficients are the principal invariants. Factored, its roots are the eigenvalues.
$$
p_A(t) = t^3 - I_1 t^2 + I_2 t - I_3 = (t - \lambda_1)(t - \lambda_2)(t - \lambda_3).
$$
It's clear that if two tensors have same invariants, then they have same eigenvalues.
By the way, you can interpret all of the principal invariants in terms of determinants of principal minors (square submatrices formed by keeping a subset of the rows and the same subset of the columns):

*

*$I_1$ is sum of determinants of $1 \times 1$ principal minors, i.e. the trace

*$I_2$ is sum of determinants of $2 \times 2$ principal minors, which is equivalent to given expression

*$I_3$ is sum of determinants of $3 \times 3$ principal minor, i.e. the determinant of the entire matrix

