Question

If Ais a $\mathbf{2}\mathbf{\times }\mathbf{2}$matrix with singular values 3 and 5, then there must exist a unit vector $\overrightarrow{\mathbf{u}}$in ${\mathbf{R}}^{\mathbf{2}}$such that $||A\overrightarrow{u}||\mathbf{=}\mathbf{4}$.

Short answer

The given statement is TRUE.

Step-by-step solution

Definition of a singular value decomposition

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

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

Check whether the given statement is TRUE or FALSE

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

Any unit vector $\dot{\mathrm{v}}$in ${\mathrm{ℝ}}^2$can be written as role="math" localid="1664193104216" $\dot{\mathrm{v}}=\mathrm{cos\theta }\dot{{\mathrm{v}}_1}+\mathrm{sin\theta }\dot{{\mathrm{v}}_1}$. Hence,

role="math" localid="1664193350764" $\mathrm{A}\dot{\mathrm{v}}=\mathrm{cos\theta A}\dot{{\mathrm{v}}_1}+\mathrm{sin\theta A}\dot{{\mathrm{v}}_1}$

$=5\mathrm{cos\theta }\dot{{\mathrm{u}}_1}+3\mathrm{sin\theta }\dot{{\mathrm{u}}_2}$

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

$||\mathrm{A}\dot{\mathrm{v}}||=\sqrt{25{\cos }^2\mathrm{\theta }+{\sin }^2\mathrm{\theta }}$

When $\mathrm{\theta }=0,||\mathrm{A}\dot{\mathrm{v}}||=5$and when $\mathrm{\theta }=\frac{\mathrm{\pi }}{2},||\mathrm{A}\dot{\mathrm{v}}||=3$. Thus, varies continuously from 5 to 3 as 0 varies from 0 to $\frac{\mathrm{\pi }}{2}$.

Final Answer

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

Thus, the given statement is TRUE.