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Solutions – Assignment 2
- Suppose that a v 1 1 (^) + a v 2 2 (^) + ... + a vn n + bu = 0
. Then − bu = a v 1 1 (^) + a v 2 2 (^) + ... + a vn n
But u ∉Span
B. The only way that we can make these two facts coincide is if
we cannot divide by b. Thus we know that − b = 0. This results in the equation a v 1 1 (^) + a v 2 2 (^) + ... + a vn n = 0
. Since we know that the vectors in B are linearly
independent, this means that all of the coefficients a 1 (^) , a 2 ,..., an must be zero. We already have that b = 0 , and so we can conclude that the expanded set of vectors,
{ v 1^ ,^ v 2^ ,...,^ vn^ , u }
, is also linearly independent.
- (Bonus) This is a result of one of the limitations of the definition of the span of the set of vectors. Recall that the span of these vectors is the set of all linear combinations of a finite subset of these vectors. That means that any sequence in this span can only have a finite number of non-zero entries. Since there is no such limitation on elements of the vector space V, this set of vectors cannot form a basis for V.
3) To check whether B is a basis, we must check two of three things
a) Linear independence b) Spanning c) Number of vectors
a) Suppose that 0 0 1 1 1 3 1 1 2 1 0 0 0 4 2 0 1 1 0 2 2 3 2 4 2
a b c d
a b c d a b c d b c a c d
^ − −
=^ +^ +^ +
+^ −^ +^ +^ −^ +
−^ −^ +
This gives a system of linear equations:
2 0 3 0 2 0 4 2 0
a b c d a b c d b c a c d
^ +^ −^ +^ =
We solve this using the matrix 1 1 1 2 1 3 1 1 0 2 1 0 4 0 1 2
Which row reduces to the identity. Thus the system is consistent and has a unique solution: all coefficients must be zero. Thus, the vectors are linearly independent.
b) Suppose that
1 1 1 3 1 1 2 1 0 4 2 0 1 1 0 2 2 3 2 4 2
w x a b c d y z a b c d a b c d b c a c d
−^ −
=^ +^ +^ +
−^ −
+^ −^ +^ +^ −^ +
−^ −^ +
This leads to a system of linear equations similar to those above (except that the solutions aren’t 0). However, since the matrix row reduces to give the identity, we know that the system is consistent regardless of the values for w, x, y, and z. Thus, the set of matrices spans the space of 2x2 matrices.
c) There are four matrices in B and we know from class that the dimension of
the vector space of 2x2 matrices is 4 (4=2x2). Thus B contains the correct
number of vectors.
The truly astute student might notice another approach to this. Using the standard
basis of this space of matrices, the coordinates of the vectors in B are actually
the columns in the matrix (**). This is not a coincidence, checking the arithmetic, you’ll notice that this must always happen. Since the matrix row reduces to the identity, the column vectors form a basis for R^4. This is sufficient to prove that
B is a basis.
- First, we note that all four polynomials do actually have a root at x = 1. Indeed, 1 2 3 2 2 3 2 4 3 2
p p x p x p X
Thus, these polynomials are all vectors in V. Now suppose that ap 1 (^) ( ) x + bp 2 (^) ( ) x + cp 3 (^) ( ) x + dp 4 ( ) x = q x ( ) (1)
for some polynomial q(x) (of degree at most three which has a root at x = 1). Then we have 2 3 2 2 3 2 3 2
q x a x x b x x x c x x d x x x b d x a b c d x a b c d x b c
Since we are working with linear independence, let us consider the situation where q x ( ) = 0. (i.e. s 0 (^) = 0, s 1 (^) = 0, and s 2 = 0 ) This gives us the following solution to our system of equations: 17 2 8
a d b d c d
^ = −
Therefore, from our original equation (1), we get
0 = q x ( ) = − 17 dp 1 (^) ( ) x − 2 dp 2 (^) ( ) x + 8 dp 3 (^) ( ) x + dp 4 ( ) x. Dividing through by d and putting p 4 (^) ( ) x on the other side of the equation tells us that p 4 (^) ( ) x is a linear combination of the first three polynomials. Indeed, 1 2 3 2 3 2 2 3 2 4
p x p x p x x x x x x x x x x x p x
Since p 4 (^) ( ) x ∈ Span { p 1 (^) ( ), x p 2 (^) ( ), x p 3 ( ) x }, we can safely remove it from B without changing the span. Furthermore, since we can see that the first three polynomials are linearly independent (the matrix only had 1 free variable), the collection B ' := (^) { p 1 (^) ( ), x p 2 (^) ( ), x p 3 ( ) x }forms a basis for V.
a) Let
= ^ ^ ^
B. It is easily seen that B is a linearly
independent set of vectors. Since B contains three vectors and V is three
dimensional, we can conclude that B is a basis for V.
b) The change of basis matrix is 1 1 0 1 0 1 2 0 0
A
= ^
Indeed, we can check that 1 1 0 : 1 0 1 2 0 0
u a b c
a b a c a
= ^ ^ + ^ ^ + ^
= ^ +
whose coordinates in terms of B are
a u b c
= ^
B^ can be gotten by matrix
multiplication:
a b u a c a a b c A u
= ^ +
= ^ ^ ^
B
- For ease of notation, I will define ( ) : sin( ) sin(2 ) ( ) : 2 cos( ) 3sin( ) ( ) : cos( ) sin(2 ).
f x x x g x x x h x x x
a) We wish to solve sin( ) x = af ( ) x + bg (^) ( x (^) ) + ch x ( )for the coefficients a, b, and c. So we have that sin( ) x = ( a − 3 )sin( ) b x + ( a − c )sin(2 ) x + (2 b + c ) cos( ) x Therefore, we see the system of equations
3 1 0 2 0
a b a c b c
^ −^ =
