Is the span of X is X? Does every vector space have finite basis?

Question a: Is the span of $\phi$ is $\phi$ ? justify. The vectors $u_1 = ( 2, -3 , 1) , u_2 = (1 , 4 , -2) , u_3 = (-8 , 12 , -4) , u_4 = ( 1 , 37 , -17) , u_5 = (-3 , -5 , 8)$ generate $R^3$ . Find a subset of the set $[u_1, u_2, u_3, u_4, u_5]$.

Solution:

First part:
NO, it is the trivial vector space that includes only the zero vector.

Second part:
The vectors $u_1 = ( 2, -3 , 1) , u_2 = (1 , 4 , -2), u_3 = (-8 , 12 , -4) , u_4 = ( 1 , 37 , -17) , u_5 = (-3, -5, 8)$ generate $R^3$ . To find the subset we have to search the linearly dependent set from $[u_1, u_2, u_3, u_4, u_5]$.

Let $a_1, a_2, a_3, a_4, a_5 \in R$.
Then the linear combination is
$a_1u_1 + a_2u_2 + a_3u_3 + a_4u_4 + a_5u_5 = 0$
$a_1\left[\begin{array}{cc} 2\\-3\\1 \end{array}\right] + a_2 \left[\begin{array}{cc} 1\\4\\-2 \end{array}\right] +a_3\left[\begin{array}{cc} -8\\12\\-4 \end{array}\right] +a_4\left[\begin{array}{cc} 1\\37\\-17 \end{array}\right] + a_5\left[\begin{array}{cc} -3\\5\\8 \end{array}\right] = \left[\begin{array}{cc} 0\\0\\0 \end{array}\right]$
The augmented matrix is,
$\left[\begin{array}{cc} 2&1&-8&1&-3|0\\-3&4&12&37&5|0\\1&-2&-4&-17&8|0 \end{array}\right]$

$\left[\begin{array}{cc} 2&\frac{1}{2} & -4 & \frac{1}{2} & \frac{-3}{2}|0\\ -3 & 4 & 12 & 37 & 5|0\\ 1 & -2 & -4 & -17 & 8|0 \end{array}\right]$ $\frac{1}{2}R_1 \Rightarrow R_2$

$\left[\begin{array}{cc} 1 & \frac{1}{2} & -4 & \frac{1}{2} & \frac{-3}{2}|0\\ 0 & \frac{11}{2} & 0 & \frac{77}{2} & \frac{-19}{2}|0\\ 0 & -\frac{-5}{2} & 8 & \frac{-35}{2} & \frac{19}{2}|0 \end{array}\right]$ $R_2 \Leftarrow R_2 +3R_1$ , $R_3 \Leftarrow R_3 -3 R_1$

$\left[\begin{array}{cc} 1 & \frac{1}{2} & -4 & \frac{1}{2} & \frac{-3}{2}|0\\ 0 & 1 & 0 & 7 & -209|0\\ 0 & -\frac{-5}{2} & 8 & \frac{-35}{2} &\frac{19}{2}|0 \end{array}\right]$ $R_2 \Leftarrow \frac{2}{11}R_2$

$\left[\begin{array}{cc} 1 & \frac{1}{2} & -4 & \frac{1}{2} & \frac{-3}{2}|0\\ 0 & 1 & 0 & 7 & -209|0\\ 0 & 0 & 8 & 0 & -513|0 \end{array}\right]$ $R_3 \Leftarrow R_3 +\frac{5}{2} R_2$

$\left[\begin{array}{cc} 1 & 0 & -4 & -3 & 103|0\\ 0 & 1 & 0 & 7 &-209|0\\ 0 & 0 & 8 & 0 & -513|0 \end{array}\right]$ $R_1 \Leftarrow R_1 -\frac{1}{2}R_2$

$\left[\begin{array}{cc} 1 & 0 & -4 & -3 & 103|0\\ 0 & 1 & 0 & 7 & -209|0\\ 0 & 0 & 1 & 0 & \frac{-513}{8}|0 \end{array}\right]$ $R_3 \Leftarrow \frac{1}{8}R_3$

$\left[\begin{array}{cc} 1 & 0 & 0 & -3 & \frac{-307}{2}|0\\ 0 & 1 & 0 & 7 & -209|0\\ 0 & 0 & 1 & 0 &\frac{-513}{8}|0 \end{array}\right]$ R_1 \Leftarrow R_1 + 4R_3 $$

Which shows that the subset of the set $[u_1, u_2, u_3, u_4, u_5]$ that is basis for $R^3$ are $[u_1, u_2, U_3]$

Question b: Does every vector space have finite basis? Justify. Prove that the set of solutions to the system of linear equations 2x – y + z = 0, 3x – 2y + Z = 0 is a subspace of $R^3$. Find the basis for this $R^3$.

Solution:\
First part
Every vector space have finite basis is false. The space c[0 , 1] or the space of all Polynomials has no finite basis, only infinite one.

Second part
To prove the set of linear equations $2x – y + z = 0 \\ 3x – 2y + z = 0$ is a subspace of $R^3$.
Now
Let us denote the solution set by S
Where $S = [a(1, 1, -1) : a \in R]$
$= [(a, a, -a): a \in R]$ is the solutions set of equations $2x – y + z = 0 3x – 2y + z = 0$.
Now
We show S is the subspace of $R^3$. Clearly S is subspace of $R^3$.
$V_{ss1}$ : Clearly $\vec{o}$ = $(0, 0, 0) \in R^3$ is a zero vector of $R^3$ .Now we have to show 0 is zero vector of S.
let $\vec{o} = (d, d, -d) \in s$
such that
$u + 0 = u$ $\forall u \in s$
Where $u = (a , a ,-a)$
$(a , a , -a) +(d , d , -d)$ = $(a , a , -a)$ $\Rightarrow (d, d,-d) = 0$ .
Hence $\vec{o}$ = $(0, 0, 0)$ is solutions to the equcation.
Hence $\vec{o}$ = $(0, 0, 0) \in S$
$V_{ss2}$ : Let $u = (a , a , -a)$ and $v = (b, b, -b)$ where $a, b \in R$ and $u , v \in S$
Then $u + v = (a , a ,-a) + (b , b , -b)$ $= (a + b , a + b , -a - b)$ $=(c , c , -c)$
where $c = a + b$ and $c \in R$. Hence $u + v \in S$
$V_{ss3}$ :Let $u = (a , a , -a)$ and $k \in F$
Such that
$ku = k(a, a , -a)$
$= (ka , kb , -kc)\in S$. Hence set of solution to the given equations is subspace. $2x – y + z = 0$
$3x – 2y + Z = 0$
By solving we get $x = z , z = -y$. Let $x = s , y = s , z = -s$
we have the solution would be $[ ( s , s , - s) = s ( 1 , 1 , -1) : s \in R ]$ and basis would be $(1, 1, -1)$.