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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).

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  • $\begingroup$ It's "diagonalise" or "diagonalize", not "diagnolise". It comes from the word "diagonal". $\endgroup$
    – Giuseppe Negro
    Commented Feb 13, 2023 at 17:32
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    $\begingroup$ @GiuseppeNegro Thanks for the correction. English is not my mother tongue so I still get some stuff wrong. $\endgroup$
    – Kev
    Commented Feb 13, 2023 at 17:36

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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 $D$.

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  • $\begingroup$ I'll edit a little bit what I said earlier, according to what I found: Let’s assign $C$ as the symmetric matrix. If there is an invertible $n×n$ matrix Q and a diagonal matrix $D$ such that $C = QDQ^-1$, then an $n×n$ matrix $C$ is diagonalizable. I did those steps: 1) detect the eigenvalues by using $det(C-\lambda I) = 0$, this will tell us what are the two values of \lambda.. and those values are the in the $D$ matrix. 2) find the eigenvectors with $(C-\lambda I)x = 0$. I got stuck at this because I'm getting that $Q$ is an all zero matrix, which is non invertible, not good. $\endgroup$
    – Kev
    Commented Feb 15, 2023 at 9:02
  • $\begingroup$ Please correct me if I'm doing something wrong, or if I have the false approach to this question. $\endgroup$
    – Kev
    Commented Feb 15, 2023 at 9:05
  • $\begingroup$ @Kev : $Q$ is the change of basis matrix that diagonalizes $C$. That is, $Q^{-1}CQ=D$. $\endgroup$
    – Disintegrating By Parts
    Commented Mar 4, 2023 at 17:35

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