Question

Consider two subspaces$\mathit{V}$and $\mathit{W}$of ${\mathbf{ℝ}}^{\mathbf{n}}$whose intersection consists only of the vector$\overrightarrow{\mathbf{0}}$.

a.Consider linearly independent vectors role="math" localid="1664198538004" ${\overrightarrow{\mathbf{v}}}_1,{\overrightarrow{\mathbf{v}}}_2,\dots ,{\overrightarrow{\mathbf{v}}}_p$ in$\mathit{V}$and ${\overrightarrow{\mathbf{w}}}_{\mathbf{1}}\mathbf{,}{\overrightarrow{\mathbf{w}}}_{\mathbf{2}}\mathbf{,}\mathbf{\dots }\mathbf{,}{\overrightarrow{\mathbf{w}}}_{\mathbf{q}}$ in W. Explain why the vectors ${\overrightarrow{\mathbf{v}}}_1,{\overrightarrow{\mathbf{v}}}_2,\dots ,{\overrightarrow{\mathbf{v}}}_p$ , ${\overrightarrow{\mathbf{w}}}_{\mathbf{1}}\mathbf{,}{\overrightarrow{\mathbf{w}}}_{\mathbf{2}}\mathbf{,}\mathbf{\dots }\mathbf{,}{\overrightarrow{\mathbf{w}}}_{\mathbf{p}}$ are linearly independent.

b.Consider a basis ${\overrightarrow{\mathbf{v}}}_1,{\overrightarrow{\mathbf{v}}}_2,\dots ,{\overrightarrow{\mathbf{v}}}_p$ of $\mathit{V}$and a basis ${\overrightarrow{\mathbf{w}}}_{\mathbf{1}}\mathbf{,}{\overrightarrow{\mathbf{w}}}_{\mathbf{2}}\mathbf{,}\mathbf{\dots }\mathbf{,}{\overrightarrow{\mathbf{w}}}_{\mathbf{q}}$of $\mathit{W}$ . Explain why ${\overrightarrow{\mathbf{v}}}_{\mathbf{1}}\mathbf{,}{\overrightarrow{\mathbf{v}}}_{\mathbf{2}}\mathbf{,}\mathbf{\dots }\mathbf{,}{\overrightarrow{\mathbf{v}}}_{\mathbf{p}}\mathbf{,}{\overrightarrow{\mathbf{w}}}_{\mathbf{1}}\mathbf{,}{\overrightarrow{\mathbf{w}}}_{\mathbf{2}}\mathbf{,}\mathbf{\dots }\mathbf{,}{\overrightarrow{\mathbf{w}}}_{\mathbf{q}}$ , is a basis of $\mathit{V}\mathbf{+}\mathit{W}$ . See Exercise 50.

Short answer

1. As both the vectors, ${\overrightarrow{\mathrm{v}}}_1,{\overrightarrow{\mathrm{v}}}_2,\dots ,{\overrightarrow{\mathrm{v}}}_{\mathrm{p}}$ and ${\overrightarrow{w}}_1,{\overrightarrow{w}}_2,\dots ,{\overrightarrow{w}}_q$ are in $V$ and $W$thus, the vectors are linearly independent.
2. The $\overrightarrow{v}$is a linear combination of the ${\overrightarrow{\mathrm{v}}}_i$ , and$\overrightarrow{w}$ is a linear combination of the ${\overrightarrow{w}}_j$. This shows that the vectors${\overrightarrow{\mathrm{v}}}_1,{\overrightarrow{\mathrm{v}}}_2,\dots ,{\overrightarrow{\mathrm{v}}}_{\mathrm{p}},{\overrightarrow{w}}_1,{\overrightarrow{w}}_2,\dots ,{\overrightarrow{w}}_q$ span $V+W$.

Step-by-step solution

Consider the condition and proof part (a)

A subset $W$of the vector space ${\mathrm{ℝ}}^n$ is called a (linear) subspace of ${\mathrm{ℝ}}^n$if it has the following three properties:

1. $W$contains the zero vector in ${\mathrm{ℝ}}^n$.
2. $W$is closed under addition: If ${\overrightarrow{w}}_1$and ${\overrightarrow{w}}_2$are both in $W$, then so is ${\overrightarrow{w}}_1+{\overrightarrow{w}}_2$.
3. $W$is closed under scalar multiplication: If $\overrightarrow{w}$is in $W$and $k$is an arbitrary scalar, then $k\overrightarrow{w}$is in $W$.

Consider a relation $c_1{\overrightarrow{\mathrm{v}}}_1+\dots +c_p{\overrightarrow{\mathrm{v}}}_{\mathrm{p}}+d{\overrightarrow{w}}_1+\dots +d_q{\overrightarrow{w}}_q=\overrightarrow{0}$.

Then $c_1{\overrightarrow{\mathrm{v}}}_1+\dots +c_p{\overrightarrow{\mathrm{v}}}_{\mathrm{p}}=-d{\overrightarrow{w}}_1-\dots -d_q{\overrightarrow{w}}_q$is $\overrightarrow{0}$ , because this vector is both in $V$and in $W$.

As both the vectors, ${\overrightarrow{\mathrm{v}}}_1,{\overrightarrow{\mathrm{v}}}_2,\dots ,{\overrightarrow{\mathrm{v}}}_{\mathrm{p}}$ and ${\overrightarrow{w}}_1,{\overrightarrow{w}}_2,\dots ,{\overrightarrow{\mathrm{w}}}_q$ are in data-custom-editor="chemistry" $\mathrm{V}$and $W$ thus, the vectors are linearly independent.

Proof for part (b)

The vectors ${\overrightarrow{\mathrm{v}}}_1,{\overrightarrow{\mathrm{v}}}_2,\dots ,{\overrightarrow{\mathrm{v}}}_{\mathrm{p}}$, ${\overrightarrow{w}}_1,{\overrightarrow{w}}_2,\dots ,{\overrightarrow{w}}_q$are linearly independent.

Consider a vector $\overrightarrow{x}$in $V+W$.

By the definition of $V+W$, we can write $\overrightarrow{x}=\overrightarrow{\mathrm{v}}+\overrightarrow{w}$for a $\overrightarrow{\mathrm{v}}$ in Vand a $\overrightarrow{w}$in W.

The $\overrightarrow{\mathrm{v}}$ is a linear combination of the ${\overrightarrow{\mathrm{v}}}_i$, and $\overrightarrow{w}$ is a linear combination of the ${\overrightarrow{w}}_j$.

This shows that the vectors ${\overrightarrow{\mathrm{v}}}_1,{\overrightarrow{\mathrm{v}}}_2,\dots ,{\overrightarrow{\mathrm{v}}}_{\mathrm{p}},{\overrightarrow{w}}_1,{\overrightarrow{w}}_2,\dots ,{\overrightarrow{w}}_q$span $V+W$.