Consider the set of vectors T in P, T-{I.- -21-1,1 +5,4 l,3° + 3 -4-1, 21, + 3r-
ID: 3115923 • Letter: C
Question
Consider the set of vectors T in P, T-{I.- -21-1,1 +5,4 l,3° + 3 -4-1, 21, + 3r-4t + 3}, find the basis of the vector space spanned by T. Compare dim(span(T)) and dim(P3). Can you add or remove an appropriate number of polynomial vectors in order to form a basis of the "parent" vector space (i.e. P3). Explicitly write the basis you find. Provide a printout of your MATLAB work The basis of span(T) is dim(span(7))- di dim(P3)- Explictly write the basis of Ps by adding or removing an appropriate number of polynomial vectors to span(T):Explanation / Answer
We have T = { t3-t2-2t-1, t3+5t2+1, 3t3+3t2-4t-1, 2t3+3t2-4t+3}. Let A =
1
1
3
2
-1
5
3
3
-2
0
-4
-4
-1
1
-1
3
It may be observed that the entries in the columns of A are the coeffients of t3,t2,t and the scalar multiple of 1 in the vectors in T.To find the basis of span(T), we will reduce A to its RREF as under:
Add 1 times the 1st row to the 2nd row
Add 2 times the 1st row to the 3rd row
Add 1 times the 1st row to the 4th row
Multiply the 2nd row by 1/6
Add -2 times the 2nd row to the 3rd row
Add -2 times the 2nd row to the 4th row
Multiply the 3rd row by -3/5
Add -10/3 times the 3rd row to the 4th row
Add -5/6 times the 3rd row to the 2nd row
Add -2 times the 3rd row to the 1st row
Add -1 times the 2nd row to the 1st row
Then the RREF of A is
1
0
2
0
0
1
1
0
0
0
0
1
0
0
0
0
It may now be observed that the 1st,2nd and the 4th column of A are linearly independent and that the 3rd column is a linear combination of the first 2 columns. Hence a basis for span(T) is { t3-t2-2t-1, t3+5t2+1, 2t3+3t2-4t+3} or, {t3,t2,t}.
Thus dim (span(T)) = 3.
The standard basis for P3 is {t3,t2,t,1} so than dim(P3) = 4.
If 1 is added to the basis {t3,t2,t} for span(T), we will get the basis for the parent vector space, i.e. P3.
1
1
3
2
-1
5
3
3
-2
0
-4
-4
-1
1
-1
3