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

linear algebramatricesmatrix decompositionmatrix equationssoft-question

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?

Best Answer

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 $a=-3$.