= Solution
Under an orthogonal change of coordinates represented by $Q$, the component matrix changes by
$$
T'=QTQ^T=QTQ^{-1}.
$$
Thus $T'$ is <similar matrices>[similar] to $T$, so its <characteristic polynomial>, <eigenvalues>, and their <algebraic multiplicities> are unchanged.
If $T$ is <symmetric matrix>[symmetric], the <spectral theorem for real symmetric matrices> makes it <diagonalizable matrix>[diagonalizable], so algebraic and geometric multiplicities agree. Consequently the dimensions of its eigenspaces are also independent of the coordinate frame.
For the stated tensor, choose a unit eigenvector $n$ belonging to the simple eigenvalue $\mu$. Its perpendicular plane is the eigenspace of $\lambda$, so
$$
T=\lambda(I-nn^T)+\mu nn^T
=\lambda I+(\mu-\lambda)nn^T.
$$
Hence
$$
\boxed{T_{ij}=\alpha\delta_{ij}+\beta n_in_j},
\qquad
\alpha=\lambda,\quad\beta=\mu-\lambda.
$$
Back to article page
