Answer with Step-by-step explanation:
We are given that a real vector space
V=[tex]R^4[/tex]
a.We have to prove that dim V =4
Let (a,b,c,d) is an element of V
Suppose that four elements
(1,0,0,0),(0,1,0,0),(0,0,1,0) and (0,0,0,1)
a(1,0,0,0)+b(0,1,0,0)+c(0,0,1,0)+d(0,0,0,1)
=(a,b,c,d)
All four elements are independent and can span each element of V.
Hence, dimension of V=4
b.We are given that (1,1,0,0),(0,1,1,0,(0,0,1,1)
We have to prove that these elements can span V
If number of elements are less than dimension of then the elements can not span V.
[tex]\left[\begin{array}{cccc}1&1&0&0\\0&1&1&0\\0&0&1&1\end{array}\right][/tex]
Rank of matrix =3
Hence, rank is less than the dimension of V.Therefore, given elements can not span V.
c.We are given that (1,-1,0,0),(0,1,-1,0),(0,0,1,-1),(-1,0,0,1)
We have to show that these elements are linearly independent.
[tex]\left[\begin{array}{cccc}1&-1&0&0\\0&1&-1&0\\0&0&1&-1\\-1&0&0&1\end{array}\right][/tex]
Every row or column is not a linear combination of other rows or columns.
Therefore, these elements are linearly independent.
d.We are given that [tex]v_1,v_2,v_3,v_4\in V[/tex] and span[tex](v_1,v_2,v_3,v_4)=V[/tex]
We have to prove that given vectors are linear independent.
Let [tex]v_1=(1,0,0,0),v_2=(0,1,0,0),v_3=(0,0,1,0),v_4=(0,0,0,1)[/tex]
(a,b,c,d)=a(1,0,0,0)+b(0,1,0,0)+c(0,0,1,0)+d(0,0,0,1)
All four vectors are linearly independent because any element is not a linear combination other elements .
Hence, every four vectors [tex]v_1,v_2,v_3,v_4\in V[/tex] are linearly independent.
