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 $$(1, 1, -1)$$.

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