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We know that the singular value decomposition of a matrix A is the factorization of A into the product of three matrices A = UDV T where the columns of U and V are orthonormal and the matrix D is diagonal with positive real entries.

Now, if we use SVD, does it tell if a matrix is singular or not?

If this is not the case, then how should I know if a matrix is singular or not?

Example: Find all values of 𝑎 for which the matrix is singular:

matrix:

MATRIX

(just a verification of terms question)

Should I use gaussian for this?

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    $\begingroup$ For this question I would choose to calculate the determinant of the matrix. $\endgroup$
    – Hirshy
    Aug 26, 2022 at 19:54
  • $\begingroup$ The SVD will tell you if a matrix is singular or not (it's singular iff $0$ is a singular value) but this isn't very helpful in practice. For this problem you can compute the determinant; sometimes doing row or column operations will be easier. $\endgroup$ Aug 26, 2022 at 20:13

1 Answer 1

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Doing the singular value decomposition of a matrix with unknowns would be particularly painful (it is already painful enough for a fully numerical matrix), as it is almost never an easy process.

In this case, you are expected to notice that if you do the determinant by the second column, you immediately get $$ \det A=a(a^2-9). $$ So $A$ will be singular precisely when $a(a^2-9)=0$, which is the case when $a=0$, $a=3$, and $a=-3$.

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