Question

Consider a subspace V of${\mathbf{ℝ}}^{\mathbf{n}}{\mathbf{ℝ}}^{\mathbf{n}}$. Let${\overrightarrow{\mathbf{v}}}_{\mathbf{1}}\mathbf{,}{\overrightarrow{\mathbf{v}}}_{\mathbf{2}}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{,}{\overrightarrow{\mathbf{v}}}_{\mathbf{p}}$ be a basis of V and${\overrightarrow{\mathbf{w}}}_{\mathbf{1}}\mathbf{,}{\overrightarrow{\mathbf{w}}}_{\mathbf{2}}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{,}{\overrightarrow{\mathbf{w}}}_{\mathbf{q}}$ a basis of. Is$\left\{{\overrightarrow{v}}_1,{\overrightarrow{v}}_2,....,{\overrightarrow{v}}_p,{\overrightarrow{w}}_1,{\overrightarrow{w}}_2,...,{\overrightarrow{w}}_q\right\}$ a basis of${\mathbf{ℝ}}^{\mathbf{n}}$? Explain.

Short answer

The set $\{{\overrightarrow{\mathrm{v}}}_1,{\overrightarrow{\mathrm{v}}}_2,....,{\overrightarrow{\mathrm{v}}}_{\mathrm{p}},{\overrightarrow{\mathrm{w}}}_1,{\overrightarrow{\mathrm{w}}}_2,...,{\overrightarrow{\mathrm{w}}}_{\mathrm{q}}\}$spanned${\mathrm{ℝ}}^n$ .

Step-by-step solution

Determine the spanning property.

Consider a basis of $\{{\overrightarrow{\mathrm{v}}}_1,{\overrightarrow{\mathrm{v}}}_2,....,{\overrightarrow{\mathrm{v}}}_{\mathrm{p}}\}$and role="math" localid="1660116504622" $\{{\overrightarrow{\mathrm{w}}}_1,{\overrightarrow{\mathrm{w}}}_2,...,{\overrightarrow{\mathrm{w}}}_{\mathrm{q}}\}$ for$V^{\perp }$ in${\mathrm{ℝ}}^n{\mathrm{ℝ}}^n$ .

Assumerole="math" localid="1660116499579" $\{{\overrightarrow{\mathrm{v}}}_1,{\overrightarrow{\mathrm{v}}}_2,....,{\overrightarrow{\mathrm{v}}}_{\mathrm{p}},{\overrightarrow{w}}_1,{\overrightarrow{w}}_2,....,{\overrightarrow{w}}_q\}$ and$\overrightarrow{v}\in {\mathrm{ℝ}}^n$ .

As ${\mathrm{ℝ}}^n=v+v^{\perp }{\mathrm{ℝ}}^n=v+v^{\perp }$and$\left\{{\overrightarrow{\mathrm{v}}}_1,{\overrightarrow{\mathrm{v}}}_2,....,{\overrightarrow{\mathrm{v}}}_{\mathrm{p}}\right\},\{,{\overrightarrow{w}}_1,{\overrightarrow{w}}_2,....,{\overrightarrow{w}}_q\}$ are basis of V and$V^{\perp }$ respectively.

There exists constant$a_i,b_i\in \mathrm{ℝ}$such that$\sum _{i=1}^p{a}_i{\overrightarrow{v}}_i+\sum _{i=1}^q{b}_i{\overrightarrow{w}}_i=v$ .

Therefore, the point$\overrightarrow{v}\in B$ .

Determine the property of linear independence.

Assume $\sum _{i=1}^p{a}_i{\overrightarrow{v}}_i+\sum _{i=1}^q{b}_i{\overrightarrow{w}}_i=0$where$\sum _{i=1}^p{a}_i{\overrightarrow{v}}_i\in V$ and$\sum _{i=1}^q{b}_i{\overrightarrow{w}}_i=0$ .

As$V\cap V^{\perp }=\left\{0\right\}$ implies$\sum _{i=1}^p{a}_i{\overrightarrow{v}}_i=0$ and$\sum _{i=1}^q{b}_i{\overrightarrow{w}}_i=0$ .

By the definition of linear independency, the value of the coefficients$a_i=0$ and$b_i=0$ for all$i$ .

If$\sum _{i=1}^p{a}_i{\overrightarrow{v}}_i+\sum _{i=1}^q{b}_i{\overrightarrow{w}}_i=0$ implies$a_i=0$ and$b_i=0$ , by the definition of linear independence the setrole="math" localid="1660117205931" $B=\{{\overrightarrow{\mathrm{v}}}_1,{\overrightarrow{\mathrm{v}}}_2,....,{\overrightarrow{\mathrm{v}}}_{\mathrm{p}},{\overrightarrow{w}}_1,{\overrightarrow{w}}_2,....,{\overrightarrow{w}}_q\}$ is linear independence.

Hence, the set $\{{\overrightarrow{\mathrm{v}}}_1,{\overrightarrow{\mathrm{v}}}_2,....,{\overrightarrow{\mathrm{v}}}_{\mathrm{p}},{\overrightarrow{w}}_1,{\overrightarrow{w}}_2,....,{\overrightarrow{w}}_q\}$spanned ${\mathrm{ℝ}}^n{\mathrm{ℝ}}^n$if $\{{\overrightarrow{\mathrm{v}}}_1,{\overrightarrow{\mathrm{v}}}_2,....,{\overrightarrow{\mathrm{v}}}_{\mathrm{p}}\}$is basis of V and$\{{\overrightarrow{w}}_1,{\overrightarrow{w}}_2,....,{\overrightarrow{w}}_q\}$ is basis for$V^{\perp }$ .