Question

The product of the nsingular values of an$\mathbf{n}\mathbf{\times }\mathbf{n}$matrix A must be det$\left|\mathbf{A}\right|$.

Short answer

The given statement is TRUE.

Step-by-step solution

Step 1: Definition of the singular value decomposition

The diagonal entries of the matrix are singular values, which are arranged in ascending order.

Singular values are non-negative, which means it’s all positive real numbers.

Step 2: To Find TRUE or FALSE

Let${\mathrm{\sigma }}_1,{\mathrm{\sigma }}_2,\cdots ,{\mathrm{\sigma }}_{\mathrm{n}}$be singular values of$\mathrm{A}$. Then,role="math" localid="1664186808738" ${\mathrm{\sigma }}_1^2,{\mathrm{\sigma }}_2^2,\cdots ,{\mathrm{\sigma }}_{\mathrm{n}}^2$are eigenvalues of${\mathrm{A}}^{\mathrm{T}}\mathrm{A}$.

Thus, ${\mathrm{\sigma }}_1^2{\mathrm{\sigma }}_2^2\cdots {\mathrm{\sigma }}_{\mathrm{n}}^2=\det \left({\mathrm{A}}^{\mathrm{T}}\mathrm{A}\right)=\left(\det {\left(\mathrm{A}\right)}^2\right)$, as${\mathrm{A}}^{\mathrm{T}}=\mathrm{A}$.

Take square roots on both sides.

${\mathrm{\sigma }}_1{\mathrm{\sigma }}_2\cdots {\mathrm{\sigma }}_{\mathrm{n}}=\left|\det \left(\mathrm{A}\right)\right|$

That is, the determinant of a matrix is equal to the product of the eigen values thus $\det \left|\mathrm{A}\right|$will be the absolute value of the product of the eigen values which is in fact equal to the product of the singular values.

Final Answer

Thus, the given statement is TRUE.