91影视

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

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

The set $B$of vectors is defined as $\mathrm{B}=\{{\stackrel{鈫�}{\mathrm{蠀}}}_1,\text{鈥�}{\stackrel{鈫�}{\mathrm{蠀}}}_2,\text{鈥�}鈥�,\text{鈥�}{\stackrel{鈫�}{\mathrm{蠀}}}_{\mathrm{m}}\}$.

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:

鈥淭he vectors${\stackrel{鈫�}{蠀}}_1,\text{鈥�}{\stackrel{鈫�}{蠀}}_2,\text{鈥�}鈥�,\text{鈥�}{\stackrel{鈫�}{蠀}}_m$ in ${\mathrm{鈩�}}^n$are linearly independent if (if and only if) there are nontrivial relations among them.鈥�

Now, the set$B^{\text{'}}$containing the vector$\stackrel{鈫�}{蝇}$is given by,

$B^{\text{'}}=\{{\stackrel{鈫�}{蠀}}_1,\text{鈥�}{\stackrel{鈫�}{蠀}}_2,\text{鈥�}鈥�,\text{鈥�}{\stackrel{鈫�}{蠀}}_m,\text{鈥�}\stackrel{鈫�}{蝇}\}$, where$\stackrel{鈫�}{蝇}鈭�V$is an arbitrary vector.

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

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

Therefore,

$\begin{array}{l}{\mathrm{x}}_1{\stackrel{鈫�}{\mathrm{蠀}}}_1+{\mathrm{x}}_2\text{鈥�}{\stackrel{鈫�}{\mathrm{蠀}}}_2+\text{鈥�}鈥�+{\mathrm{x}}_{\mathrm{m}}{\stackrel{鈫�}{\mathrm{蠀}}}_{\mathrm{m}}+\mathrm{a}\text{鈥�}\stackrel{鈫�}{\mathrm{蝇}}={\stackrel{鈫�}{\mathrm{蠀}}}_{\mathrm{k}}\\ 鈬�{\mathrm{x}}_1{\stackrel{鈫�}{\mathrm{蠀}}}_1+{\mathrm{x}}_2\text{鈥�}{\stackrel{鈫�}{\mathrm{蠀}}}_2+\text{鈥�}鈥�+{\mathrm{x}}_{\mathrm{m}}{\stackrel{鈫�}{\mathrm{蠀}}}_{\mathrm{m}}+鈭�{\stackrel{鈫�}{\mathrm{蠀}}}_{\mathrm{k}}=鈭�\mathrm{a}\text{鈥�}\stackrel{鈫�}{\mathrm{蝇}}\\ 鈬�\stackrel{鈫�}{\mathrm{蝇}}=鈭�\frac{{\mathrm{x}}_1{\stackrel{鈫�}{\mathrm{蠀}}}_1}{\mathrm{a}}鈭�\frac{{\mathrm{x}}_2\text{鈥�}{\stackrel{鈫�}{\mathrm{蠀}}}_2}{\mathrm{a}}鈭�鈥�鈭�\frac{{\mathrm{x}}_{\mathrm{m}}{\stackrel{鈫�}{\mathrm{蠀}}}_{\mathrm{m}}}{\mathrm{a}}+\frac{{\stackrel{鈫�}{\mathrm{蠀}}}_{\mathrm{k}}}{\mathrm{a}}\end{array}$

Hence, an arbitrary vector$\stackrel{鈫�}{蝇}$ 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.

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

${\stackrel{鈫�}{蠀}}_1,\text{鈥�}{\stackrel{鈫�}{蠀}}_2,\text{鈥�}鈥�,\text{鈥�}{\stackrel{鈫�}{蠀}}_m$, which spans$V$.

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

Therefore, the matrix $A$is given by,

$A=\left[\begin{array}{cccc}|& |& & |\\ {\stackrel{鈫�}{蠀}}_1& {\stackrel{鈫�}{蠀}}_2& 鈰�& {\stackrel{鈫�}{蠀}}_m\\ |& |& & |\end{array}\right]$