Question

(a) Let $\mathit{V}$be a subset of role="math" localid="1660109056998" ${\mathbf{ℝ}}^{\mathbf{n}}$. Let $\mathbf{m}$be the largest number of linearly independent vectors we can find in $\mathbf{V}$. (Note $\mathbf{m}\mathbf{\le }\mathbf{n}$, by Theorem 3.2.8.) Choose linearly independent vectors ${\overrightarrow{\mathbf{\upsilon }}}_{{}_{\mathbf{1}}}\mathbf{,}\mathbf{\hspace{0.17em}}\mathbf{\hspace{0.17em}}{\overrightarrow{\mathbf{\upsilon }}}_{{}_{\mathbf{2}}}\mathbf{,}\mathit{\hspace{0.17em}}\mathbf{\dots }\mathbf{,}\mathbf{\hspace{0.17em}}\mathbf{\hspace{0.17em}}{\overrightarrow{\mathbf{\upsilon }}}_{{}_{\mathbf{m}}}$ in$\mathit{V}$. Show that the vectors ${\overrightarrow{\mathit{\upsilon }}}_{{}_{\mathbf{1}}}\mathbf{,}\mathbf{\hspace{0.17em}}\mathbf{\hspace{0.17em}}{\overrightarrow{\mathit{\upsilon }}}_{{}_{\mathbf{2}}}\mathbf{,}\mathit{\hspace{0.17em}}\mathbf{\dots }\mathbf{,}\mathbf{\hspace{0.17em}}\mathbf{\hspace{0.17em}}{\overrightarrow{\mathit{\upsilon }}}_{{}_{\mathit{m}}}$span$\mathit{V}$ and are therefore a basis of $\mathit{V}$. This exercise shows that any subspace of${\mathbf{ℝ}}^{\mathbf{n}}$ has a basis.

If you are puzzled, think first about the special case when role="math" localid="1660109086728" $\mathit{V}$is a plane in ${\mathbf{ℝ}}^3$. What is$\mathbf{m}$ in this case?

(b) Show that any subspace$\mathit{V}$ of ${\mathbf{ℝ}}^{\mathbf{n}}$can be represented as the image of a matrix.

Short answer

(a)We proved that the vectors span $\mathrm{V}$i.e., can written as linear combination of the vectors from $B^{\text{'}}$and are a basis of$V$ .

(b) We proved that any subspace of$V$ can be represented as the image of a matrix.

Step-by-step solution

(a) To show that the vectors span V and forms a basis

Let$V$ be a subset of ${\mathrm{ℝ}}^n$.

Let $m$be the largest number of linearly independent vectors.

The set $B$of vectors is defined as $\mathrm{B}=\{{\overrightarrow{\mathrm{\upsilon }}}_1,{\overrightarrow{\mathrm{\upsilon }}}_2,\dots ,{\overrightarrow{\mathrm{\upsilon }}}_{\mathrm{m}}\}$.

We need to show that the vectors that spans$V$are basis of $V$.

We will use here the Theorem 3.2.7 given as follows:

“The vectors${\overrightarrow{\upsilon }}_1,{\overrightarrow{\upsilon }}_2,\dots ,{\overrightarrow{\upsilon }}_m$ in ${ℝ}^n$are linearly independent if (if and only if) there are nontrivial relations among them.”

Now, the set$B^{\text{'}}$containing the vector$\overrightarrow{\omega }$is given by,

$B^{\text{'}}=\{{\overrightarrow{\upsilon }}_1,{\overrightarrow{\upsilon }}_2,\dots ,{\overrightarrow{\upsilon }}_m,\overrightarrow{\omega }\}$, where$\overrightarrow{\omega }\in V$is an arbitrary vector.

Thus, we have$\mathrm{m}+1$vectors in the above set.

Any vector, say,${\overrightarrow{\mathrm{\upsilon }}}_{\mathrm{k}}$, in the set $B^{\text{'}}$can be written as linear combination of other vectors of the same set.

Therefore,

$\begin{array}{l}{\mathrm{x}}_1{\overrightarrow{\mathrm{\upsilon }}}_1+{\mathrm{x}}_2{\overrightarrow{\mathrm{\upsilon }}}_2+\dots +{\mathrm{x}}_{\mathrm{m}}{\overrightarrow{\mathrm{\upsilon }}}_{\mathrm{m}}+\mathrm{a}\overrightarrow{\mathrm{\omega }}={\overrightarrow{\mathrm{\upsilon }}}_{\mathrm{k}}\\ \Rightarrow {\mathrm{x}}_1{\overrightarrow{\mathrm{\upsilon }}}_1+{\mathrm{x}}_2{\overrightarrow{\mathrm{\upsilon }}}_2+\dots +{\mathrm{x}}_{\mathrm{m}}{\overrightarrow{\mathrm{\upsilon }}}_{\mathrm{m}}+-{\overrightarrow{\mathrm{\upsilon }}}_{\mathrm{k}}=-\mathrm{a}\overrightarrow{\mathrm{\omega }}\\ \Rightarrow \overrightarrow{\mathrm{\omega }}=-\frac{{\mathrm{x}}_1{\overrightarrow{\mathrm{\upsilon }}}_1}{\mathrm{a}}-\frac{{\mathrm{x}}_2{\overrightarrow{\mathrm{\upsilon }}}_2}{\mathrm{a}}-\dots -\frac{{\mathrm{x}}_{\mathrm{m}}{\overrightarrow{\mathrm{\upsilon }}}_{\mathrm{m}}}{\mathrm{a}}+\frac{{\overrightarrow{\mathrm{\upsilon }}}_{\mathrm{k}}}{\mathrm{a}}\end{array}$

Hence, an arbitrary vector$\overrightarrow{\omega }$ from$V$can be written as the linear combination of the vectors from $B\text{'}$.

We know that they are independent, they can form a basis.

(b) To show that any subspace V can be represented as the image of matrix

From part (a), we can choose maximum number of linearly independent vectors from$V$ which are given by,

${\overrightarrow{\upsilon }}_1,{\overrightarrow{\upsilon }}_2,\dots ,{\overrightarrow{\upsilon }}_m$, which spans$V$.

The image of a matrix is spanned by its column vectors.

Therefore, the matrix $A$is given by,

$A=\left[\begin{array}{cccc}|& |& & |\\ {\overrightarrow{\upsilon }}_1& {\overrightarrow{\upsilon }}_2& \cdots & {\overrightarrow{\upsilon }}_m\\ |& |& & |\end{array}\right]$