# Does SVD(SINGULAR VALUE DECOMPOSITION) tell if a matrix is singular or not?

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 š¯‘ˇ for which the matrix is singular:

matrix:

(just a verification of terms question)

Should I use gaussian for this?

• For this question I would choose to calculate the determinant of the matrix. Aug 26, 2022 at 19:54
• 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. Aug 26, 2022 at 20:13

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 $$a=-3$$.