Question

Every nonzero subspace of ${\mathbf{ℝ}}^{\mathbf{n}}$has an orthonormal basis.

Short answer

True.

Step-by-step solution

Step by step solution Step 1: Definition of an Orthonormal matrix

In linear algebra, an orthogonal matrix, or orthonormal matrix, is a real square matrix whose columns and rows are orthonormal vectors.

One way to express this is. Where QT is the transpose of Q and I is the identity matrix.

For example,

$Q^{T}Q=\left(\begin{array}{c}{{\overrightarrow{q}}_1}^T\\ {{\overrightarrow{q}}_2}^T\\ {{\overrightarrow{q}}_3}^T\\ .\\ .\\ {{\overrightarrow{q}}_n}^T\end{array}\right)({\overrightarrow{q}}_1,{\overrightarrow{q}}_2,{\overrightarrow{q}}_3,.......,{\overrightarrow{q}}_n)$

This subspace can be found using the Gram-Schmidt process.

Determine the Gram-Schmidt process

Consider a basis of a subspace Vof ${\mathit{R}}^{\mathbf{n}}$for$\mathit{j}\mathbf{=}\mathbf{2}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{,}\mathit{m}$we resolve the vector ${\overrightarrow{\mathbf{v}}}_{\mathbf{j}}$into its components parallel and perpendicular to the span of the preceding vectors ${\overrightarrow{\mathbf{v}}}_{\mathbf{1}}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{,}{\overrightarrow{\mathbf{v}}}_{\mathbf{j}\mathbf{-}\mathbf{1}}$,

Then,

${\overrightarrow{\mathbf{u}}}_{\mathbf{1}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{\left|}\mathbf{\right|}{\overrightarrow{\mathbf{v}}}_{\mathbf{1}}\mathbf{\left|}\mathbf{\right|}}{\overrightarrow{\mathbf{v}}}_{\mathbf{1}}\mathbf{,}{\overrightarrow{\mathbf{u}}}_{\mathbf{2}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{\left|}\mathbf{\right|}{{\overrightarrow{\mathbf{v}}}_{\mathbf{2}}}^{\mathbf{\perp }}\mathbf{\left|}\mathbf{\right|}}{{\overrightarrow{\mathbf{v}}}_{\mathbf{2}}}^{\mathbf{\perp }}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{,}{\overrightarrow{\mathbf{u}}}_{\mathbf{j}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{\left|}\mathbf{\right|}{{\overrightarrow{\mathbf{v}}}_{\mathbf{j}}}^{\mathbf{\perp }}\mathbf{\left|}\mathbf{\right|}}{{\overrightarrow{\mathbf{v}}}_{\mathbf{j}}}^{\mathbf{\perp }}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{,}{\overrightarrow{\mathbf{u}}}_{\mathbf{m}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{\left|}\mathbf{\right|}{{\overrightarrow{\mathbf{v}}}_{\mathbf{m}}}^{\mathbf{\perp }}\mathbf{\left|}\mathbf{\right|}}{{\overrightarrow{\mathbf{v}}}_{\mathbf{m}}}^{\mathbf{\perp }}$

Thus, by using Gram-Schmidt process the nonzero subspace of has an orthonormal basis.

Hence, the statement is true.