Question

If both singular values of a $\mathbf{2}\mathbf{\times }\mathbf{2}$matrix$\mathbf{A}$are less than 5, then all the entries ofmust be less than 5.

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${\mathbf{Av}}_{\mathbf{i}}\mathbf{=}{\mathbf{\sigma }}_{\mathbf{i}}{\mathbf{u}}_{\mathbf{i}}$ role="math" localid="1664177451234" ${\mathbf{\sigma }}_{\mathbf{i}}$= singular value.

To Find TRUE or FALSE

Consider$\mathrm{A}\dot{{\mathrm{v}}_1}=3\dot{{\mathrm{u}}_1}$and$\mathrm{A}\dot{{\mathrm{v}}_2}=2\dot{{\mathrm{u}}_2}$where$\left\{\dot{{\mathrm{v}}_1},{\dot{\mathrm{v}}}_2\right\}$and$\left\{{\dot{\mathrm{u}}}_1,\dot{{\mathrm{u}}_2}\right\}$are pairs of orthogonal unit vectors and ${\sigma }_1,{\sigma }_2<5$.

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

$\mathrm{A}\dot{\mathrm{v}}=\mathrm{cos\theta A}\dot{{\mathrm{v}}_1}+\mathrm{sin\theta A}\dot{{\mathrm{v}}_1}$

Since,

$\mathrm{A}\dot{{\mathrm{v}}_1}$$={\mathrm{\sigma }}_1\dot{{\mathrm{u}}_1}$

$\mathrm{A}\dot{{\mathrm{v}}_2}$$={\mathrm{\sigma }}_2\dot{{\mathrm{u}}_2}$

$\mathrm{A}\dot{\mathrm{v}}$$=\mathrm{cos\theta }{\sigma }_1\dot{{\mathrm{u}}_1}$$+{\mathrm{sin\theta \sigma }}_2\dot{{\mathrm{u}}_2}$

$\mathrm{A}\dot{\mathrm{v}}$$={\sigma }_1\mathrm{cos\theta }\dot{{\mathrm{u}}_1}$$+{\mathrm{\sigma }}_2\mathrm{sin\theta }\dot{{\mathrm{u}}_2}$

Find $||\mathrm{A}\dot{\mathrm{v}}||$as follows:

$=5$

This implies that $||{\mathrm{Ae}}_1||<5,||{\mathrm{Ae}}_2||<5$, where ${\mathrm{e}}_1,{\mathrm{e}}_2$are standard basis vectors.

Let $\mathrm{A}=\left[\begin{array}{cc}\mathrm{a}& \mathrm{b}\\ \mathrm{c}& \mathrm{d}\end{array}\right],$then ${\mathrm{Ae}}_1={\left(\mathrm{a},\mathrm{c}\right)}^{\mathrm{T}}.$

Then, $||{\mathrm{Ae}}_1||<5$implies that $\sqrt{{\mathrm{a}}^2+{\mathrm{c}}^2}<5$, which means $\mathrm{a},\mathrm{c}<5.$

Similarly, $\mathrm{b},\mathrm{d}<5.$

Hence, all the entries of $\mathrm{A}$are less than 5.

Final Answer

The given statement is TRUE.