I'm in a machine learning course, but one of the questions requires additional knowledge in linear algebra that wasn't covered in the course. The question is:
According to the spectral decomposition theorem, every real symmetric matrix has a spectrum (it can be diagnolised by an orthonormal matrix). Moreover, it can be shown that the diagnolising matrix is its' eigen matrix. Use this to diagnolise the matrix C and elaborate the calculations.
Now I really don't have a clue what is the spectral decomposition theorem, and I'm having a hard time in figuring out how to approach this question. C is not given so I guess this should be a general answer.
After searching about this, I found that the spectral decomposition theorem says that for every real symmetric matrix $C$ there exists an orthogonal matrix $Q$ that it’s columns are the eigenvectors of $C$, and a diagonal matrix $A$ that the values of its diagonal are the eigenvalues, and thus we get that $C=(Q^T AQ)$. Here they are asking us about a symmetric matrix and if I'm not mistaken, it is always diagonalisable, and has orthogonal eigenvectors.
How do I continue from here to diagnolise the matrix C as they asked? It seems they want us to really diagonalise C and not just prove it's diagonalisable (according to what I found if it's symmetric, it's diagonalisable, please correct me if I'm wrong).
The spectral theorem for a real, symmetric matrix $A$ states that $A$ has an orthonormal basis of eigenvectors $\{ v_1,v_2,\cdots,v_n \}$ with corresponding real eigenvalues $\lambda_1,\lambda_2,\cdots,\lambda_n$, which may or may not be distinct. That is, $$ v_j^{\perp} v_k = 0,\;\; j \ne k \\ v_j^{\perp} v_j = 1 \\ Av_j = \lambda_j v_j $$ Using $\{ v_1,v_2,\cdots,v_n\}$ as a basis for the underlying vector space results in a diagonal matrix $D$ representing $A$ with $\lambda_1,\lambda_2,\cdots,\lambda_n$ along the main diagonal of the representing matrix $D$.