91影视

The single value decomposition produces orthonormal bases of v鈥檚 and u鈥檚 for the four fundamental sub-spaces.

Using those bases, A becomes a diagonal matrix$鈭�$and${\mathbf{Av}}_{\mathbf{i}}\mathbf{=}{\mathbf{蟽}}_{\mathbf{i}}{\mathbf{u}}_{\mathbf{i}}\mathbf{,}\mathbf{}{\mathbf{蟽}}_{\mathbf{ii}}\mathbf{=}$singular value.

From the given statement, write $\mathrm{A}\stackrel{藱}{{\mathrm{v}}_1}=5\stackrel{藱}{{\mathrm{u}}_1}$and $\mathrm{A}\stackrel{藱}{{\mathrm{v}}_2}=3\stackrel{藱}{{\mathrm{u}}_2}$, where role="math" localid="1664192723959" $\left\{\stackrel{藱}{{\mathrm{v}}_1,\stackrel{藱}{{\mathrm{v}}_2}}\right\}$and $\left\{\stackrel{藱}{{\mathrm{u}}_1,\stackrel{藱}{{\mathrm{u}}_2}}\right\}$are pairs of orthogonal unit vectors.

Any unit vector $\stackrel{藱}{\mathrm{v}}$in ${\mathrm{鈩�}}^2$can be written as role="math" localid="1664193104216" $\stackrel{藱}{\mathrm{v}}=\mathrm{肠辞蝉胃}\stackrel{藱}{{\mathrm{v}}_1}+\mathrm{蝉颈苍胃}\stackrel{藱}{{\mathrm{v}}_1}$. Hence,

role="math" localid="1664193350764" $\mathrm{A}\stackrel{藱}{\mathrm{v}}=\mathrm{肠辞蝉胃A}\stackrel{藱}{{\mathrm{v}}_1}+\mathrm{蝉颈苍胃A}\stackrel{藱}{{\mathrm{v}}_1}$

$=5\mathrm{肠辞蝉胃}\stackrel{藱}{{\mathrm{u}}_1}+3\mathrm{蝉颈苍胃}\stackrel{藱}{{\mathrm{u}}_2}$

Now, find $||\mathrm{A}\stackrel{藱}{\mathrm{v}}||$as follows:

$||\mathrm{A}\stackrel{藱}{\mathrm{v}}||=\sqrt{25{\cos }^2\mathrm{胃}+{\sin }^2\mathrm{胃}}$

When $\mathrm{胃}=0,||\mathrm{A}\stackrel{藱}{\mathrm{v}}||=5$and when $\mathrm{胃}=\frac{\mathrm{蟺}}{2},||\mathrm{A}\stackrel{藱}{\mathrm{v}}||=3$. Thus, varies continuously from 5 to 3 as 0 varies from 0 to $\frac{\mathrm{蟺}}{2}$.

Hence, by Intermediate Value Property of continuous functions, there exists $0<\mathrm{蝇}<\frac{\mathrm{蟺}}{2}$such that $\stackrel{藱}{\mathrm{u}}=\cos 蝇\stackrel{藱}{v_1}+\sin 蝇\stackrel{藱}{v_2}$and $||\mathrm{A}\stackrel{藱}{\mathrm{v}}||=4$.

Thus, the given statement is TRUE.