Showing that two tensors are coaxial Given two second-order tensors
$\mathbf{C} = C_{ij}\mathbf{e}_i\otimes\mathbf{e}_j$ and $\mathbf{U} = U_{ij}\mathbf{e}_i\otimes\mathbf{e}_j$ with the following relation between $\mathbf{C}$ and $\mathbf{U}$: $\mathbf{C} = \mathbf{U}^T\mathbf{U}$
Additionally, we know that both $\mathbf{C}$ and $\mathbf{U}$ are symmetric.
I know that this can be done by showing that the two tensors are commutative ($\mathbf{UC} = \mathbf{CU}$), but I'd like to show it by using the eigenvalue problem. This is what I have tried so far:
$(\mathbf{U}-\lambda_i^U\mathbf{I})\mathbf{n}_i^U = \mathbf{0} \ \ \mathrm{and} \ \ (\mathbf{C}-\lambda_i^C\mathbf{I})\mathbf{n}_i^C = \mathbf{0}$
So that $\mathbf{U} = \mathbf{N}^U\bar{\mathbf{U}}(\mathbf{N}^U)^T$ and $\mathbf{C} = \mathbf{N}^C\bar{\mathbf{C}}(\mathbf{N}^C)^T$ where $\bar{\mathbf{U}} \ \mathrm{and} \ \bar{\mathbf{C}}$ are the eigenmatrices and $\mathbf{N}$ contains the eigenvectors $\mathbf{N} = [\mathbf{n}_1 \ \mathbf{n}_2 \ \mathbf{n}_3]$
If we insert $\mathbf{C} = \mathbf{U}^T\mathbf{U}=\mathbf{U}\mathbf{U}$ in the eigenvalue problem, we get:
$\left(\mathbf{N}^U\bar{\mathbf{U}}\underbrace{(\mathbf{N}^U)^T\mathbf{N}^U}_{=\mathbf{I}}\bar{\mathbf{U}}(\mathbf{N}^U)^T-\lambda_i^C\mathbf{I}\right)\mathbf{n}_i^C = \mathbf{0}$
$\left(\mathbf{N}^U\bar{\mathbf{C}}(\mathbf{N}^U)^T-\lambda_i^C\mathbf{I}\right)\mathbf{n}_i^C = \mathbf{0}$
For this to hold, $\mathbf{N}^U\bar{\mathbf{C}}(\mathbf{N}^U)^T = \mathbf{C}$ must hold, which implies $\mathbf{N}^U = \mathbf{N}^C$. It is then easy to show that the eigenvalues of $\mathbf{U}$ equals the square root of the eigenvalues of $\mathbf{C}$
Is this sufficient?
Any assistance, or alternative routes appreciated.

Since this question didn't get much response (4 days later), I'll post my own answer.
$$\bf{U}\bf{V} = \bf{V}\bf{D}$$
where $\bf{V}$ is the tensor with the eigenvectors, and $\bf{D}$ is the diagonal tensor with the eigenvalues on the diagonal.
One has $$\bf{U} = \bf{V}\bf{D}\bf{V}^{-1}$$
and since $$\bf{U} = \bf{U}^{T}$$ it follows that $$\bf{V}\bf{D}\bf{V}^{-1} = (\bf{V}\bf{D}\bf{V}^{-1})^{T} = \bf{V}^{-T}\bf{D}\bf{V}^{T}$$ which yields $$\bf{U}\bf{U}^{T} = \bf{V}\bf{D}\bf{V}^{-1}\bf{V}\bf{D}\bf{V}^{-1} = \bf{V}\bf{D}^{2}\bf{V}^{-1} = \bf{C}$$As such, the two tensors have the same eigenvectors.
