I'm trying to complete a question on symmetric rank 2 tensors in 3D. So far I have shown that the eigenvalues (and their multiplicities) of $T_{ij}$ (more strictly of the 3x3 matrix with entries given by $T_{ij}$ i suppose but i will abuse the terminology here) is independent of coordinate frame. I'm okay up to here.
The question then asks us to show that, given $T_{ij}$ has eigenvalues $\lambda,\lambda,\mu$ that we can write $$T_{ij}=\alpha\delta_{ij}+\beta n_in_j$$ where $n$ is a unit vector.
My attempt:
$T_{ij}$ is real and symmetric and hence diagonalisable by an orthogonal matrix, so $\begin{pmatrix} \lambda&0&0 \\ 0&\lambda&0 \\ 0&0&\mu \\ \end{pmatrix} = P^TTP$ where P is the matrix formed by placing the orthonormal vectors a,b,c in the columns of P. Rearranging for $T_{ij}$ and expanding out the summation we find $$T_{ij}=\lambda a_i a_j+\lambda b_i b_j + \mu c_i c_j$$ But I fail to see how this agrees with the given form? I suppose i could let $\alpha = 0$, $\beta = 2\lambda+\mu$ and $n_i n_j = \frac{\lambda a_i a_j+\lambda b_i b_j + \mu c_i c_j}{2\lambda + \mu}$ but i doubt this is what the question was looking for seeing as i'm completely ignoring the delta part. So where have i gone wrong?
Your computation of $T_{ij}$ is incorrect. You should find $T_{11} = \lambda a_1^2 + \lambda a_2^2 + \mu a_3^2$ for instance. I have no idea how you arrived to your formula.
This being said, rather than doing a brute-force computation, you should rather observe that the formula you want to prove amounts to $T=\alpha I_3 + \beta N$ where $N$ is the orthogonal projection on the unit vector $n$. Then observe that $$ \begin{pmatrix} \lambda & 0 & 0 \\ 0 & \lambda & 0 \\ 0 & 0 & \mu \end{pmatrix} = \lambda I_3 + \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & \mu-\lambda \end{pmatrix}.$$