91Ó°ÊÓ

$$\begin{gathered} \mathrm{Let}\mathrm{A}\mathrm{be}\mathrm{an}\mathrm{n}×\mathrm{n}\mathrm{matrix}.\mathrm{Also},\mathrm{let}{\mathrm{σ}}_1,{\mathrm{σ}}_2,…,{\mathrm{σ}}_{\mathrm{n}}\mathrm{be}\mathrm{the}\mathrm{singular}\mathrm{values}\mathrm{of}\mathrm{A}. \\ \mathrm{Then}{\mathrm{σ}}_1^2,{\mathrm{σ}}_2^2,......,{\mathrm{σ}}_{\mathrm{n}}^2\mathrm{are}\mathrm{the}\mathrm{eigen}\mathrm{values}\mathrm{of}{\mathrm{A}}^{\mathrm{t}}\mathrm{A}. \end{gathered}$$

Now we know that, the the product of the eigen values of a matrix is equal to the determinant of the matrix.

Hence we have

$$\begin{gathered} {\mathrm{σ}}_1^2{\mathrm{σ}}_2^2,.......,{\mathrm{σ}}_{\mathrm{n}}^2=\det \left({\mathrm{A}}^{\mathrm{t}}\mathrm{A}\right)=\det \left({\mathrm{A}}^{\mathrm{t}}\right)\det \left(\mathrm{A}\right) \\ =\det \left(\mathrm{A}\right)\det \left(\mathrm{A}\right)[\mathrm{Since} \det (\mathrm{A})=\det ({\mathrm{A}}^{\mathrm{t}}\left)\right] \\ =\left[\det \right(\mathrm{A}){]}^2 \end{gathered}$$

$$\begin{gathered} \mathrm{Thus}, ({\mathrm{σ}}_1{\mathrm{σ}}_2…{\mathrm{σ}}_{\mathrm{n}}{)}^2=[\det \left(\mathrm{A}\right){]}^2\mathrm{which}\mathrm{means} \\ {\mathrm{σ}}_1{\mathrm{σ}}_2…{\mathrm{σ}}_{\mathrm{n}}=\left|\det \right(\mathrm{A}\left)\right| \end{gathered}$$

Now let A be a $2×2$matrix and let ${\mathrm{σ}}_1,{\mathrm{σ}}_2$be the singular values of A. Then by using Theorem 8.3 .2 we get that, the image of the unit circle under A is an ellipse whose semi-major and semi-minor axes have lengths ${\mathrm{σ}}_1$and ${\mathrm{σ}}_1$respectively. Also, we know that $\left|\det \right(\mathrm{A}\left)\right|$gives the area or volume magnification factor for the linear transformation corresponding to the matrix A, which is $x→Ax$Now the area of the unit circle is 1 and the area of the ellipse, which is the image of the unit circle under $\mathrm{A}\mathrm{is}{\mathrm{σ}}_1{\mathrm{σ}}_2$since the semi-major and semi-minor axes this ellipse have lengths ${\mathrm{σ}}_1\mathrm{and}{\mathrm{σ}}_2$respectively. Therefore, the area magnification factor for the linear transformation corresponding to the matrix $\mathrm{A}\mathrm{is}{\mathrm{σ}}_1{\mathrm{σ}}_2$Thus we have${\mathrm{σ}}_1{\mathrm{σ}}_2…{\mathrm{σ}}_{\mathrm{n}}=\left|\det \right(\mathrm{A}\left)\right|$