Homework due Class 32
1. For which values of t the following 4-vectors form a basis of R4:
(1,t,3,4), (t, t, 3, 4t), (1,t, 3t, 4), (1, 1, 3, 4t) ?
From the webnotes, we need 4 linearly independent vectors to form a basis
for R4. We also know (from the webnotes) that four vectors are
linearly independent if their matrix of coefficients has a determinant not
equal to zero.
> A:=matrix(4,4,[1,t,1,1,t,t,t,1,3,3,3*t,3,4,4*t,4,4*t]);
> =matrix(4,4,[1,t,1,1,t,t,t,1,3,3,3*t,3,4,4*t,4,4*t]);
| A := |
[ 1 |
t |
1 |
1 ] |
| [ t |
t |
1 |
1 ] |
| [ 3 |
3 |
3t |
3 ] |
| [ 4 |
4t |
4 |
4t ] |
> det(A);
36 t3 - 36 t2 - 12 t4 + 12 t
> solve("=0,t);
0, 1, 1, 1
Thus, for all values of t not equal to 0 or 1, we have a basis of R4.
2. Find a basis in R4 containing the following vectors: (1,1,2,2),
(2,3,3,3).
To have a basis in R4 we need 4 linearly independent vectors.
We will try the set of vectors (1,1,2,2); (2,3,3,3); (0,0,1,0); (0,0,0,1)
For this set to be linearly independent (from the webnotes) we need the
determinant of its matrix of coefficients to be non-zero. So from maple
we get:
> A:=matrix([[1,2,0,0],[1,3,0,0],[2,3,1,0],[2,3,0,1]]);
> =matrix([[1,2,0,0],[1,3,0,0],[2,3,1,0],[2,3,0,1]]);
| A := |
[ 1 |
2 |
0 |
0 ] |
| [ 1 |
3 |
0 |
0 ] |
| [ 2 |
3 |
1 |
0 ] |
| [ 2 |
3 |
0 |
1 ] |
> det(A);
> t(A);
1
Thus, the determinant not zero, so the four vectors are linearly
independent and we have a basis of R4.
3. Find a basis of R3 contained in the following set of vectors: (1, 1,
1), (2, 2, 2), (3, 3, 3), (1, 2, 3), (2, 3,
4), (3, 2, 1).
From one of the theorems about basis in the web notes, we have:
2. If S is a linearly dependent set in an n-dimensional space V and
V=span(S) then by removing some elements of S
we can get a basis of V.
Also, we know that a basis of R3 needs three vectors from the webnotes.
But...
(1,1,1); (2,2,2); and (3,3,3) are all scalar multiples, so we can only
have one of these three in the basis of R3. (from linear independence)
(2,3,4) is a linear combination of (1,1,1) and (1,2,3); so it cannot be
in this basis. (from linear independence) ie. (1,1,1) + (1,2,3) = (2,3,4)
(3,2,1) is a linear combination of (1,1,1) and (1,2,3); so it cannot be
in this basis. (from linear independence) ie. 4*(1,1,1) +(-1)*(1,2,3) =
(3,2,1).
Thus, we only have 2 vectors left (1,1,1) and (1,2,3) and cannot (from
the web notes) form a basis of R3.
4. What is the dimension of the subspace of R4 spanned by the vectors
(1,2,3,4), (1,1,1,1), (3, 4, 5, 6), (5, 7, 9, 11)?
From a theorem about spans in the WebNotes, if S is a subset
of a vector space V and a is an element in S which is equal to a
linear combination of other elements of S, then S' (the
set obtained by removing a from S) is such that span(S)=span(S').
So let S={(1,2,3,4), (1,1,1,1), (3, 4, 5, 6), (5, 7, 9, 11)}. Then
notice that (3,4,5,6) equals a linear combination of other elements of S:
(3,4,5,6) = (1,2,3,4) + 2*(1,1,1,1).
Also notice that (5,7,9,11) is a linear combination of these same
two elements of S:
(5,7,9,11) = 2*(1,2,3,4) + 3*(1,1,1,1).
So let S' be the set obtained by removing (3,4,5,6) and (5,7,9,11) from S.
Then span(S')=span(S).
Since (1,1,1,1) and (1,2,3,4) are linearly independent (they are not
proportional), they form a basis of the subspace spanned by S', which is
identical to the subspace spanned by S. Since S' only contains two
elements and is the basis for the subset, this subset has dimension 2.