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Math 314/814: Matrix Theory Dr. S. Cooper, Fall 2008
Homework Solutions – Week of November 11
Section 6.2:
(3) Let c 1 , c 2 , c 3 , c 4 be scalars such that
c 1
−^1
(^) + c 2
(^) + c 3
(^) + c 4
−^1
Comparing entries of the matrices, this leads to the linear system of equations: −c 1 + 3c 2 − c 4 = 0 c 1 + 2c 3 = 0 − 2 c 1 + c 2 − 3 c 3 − c 4 = 0 2 c 1 + c 2 + c 3 + 7c 4 = 0 We form the associated augmented matrix and row-reduce:
^ −→
^.
We see that there is a free variable, and so we conclude that the given matrices are linearly dependent. To find a dependence relation we need to solve the system. Let c 4 = t. Then c 3 = 2t, c 2 = −t and c 1 = − 4 t. Letting t = 1 and subbing in for the scalars in the initial matrix equation we see that −^1 − 1 7
−^1
−^1
(5) Suppose we have scalars c 1 and c 2 such that c 1 (x) + c 2 (1 + x) = 0 + 0x. By rearranging this is equivalent to c 2 + (c 1 + c 2 )x = 0 + 0x. Comparing coefficients we see that c 2 = 0 and hence c 1 = 0. Thus, the given polyno- mials are linearly independent.
(6) Suppose we have scalars c 1 , c 2 and c 3 such that
c 1 (1 + x) + c 2 (1 + x^2 ) + c 3 (1 − x + x^2 ) = 0 + 0x + 0x^2. By rearranging this is equivalent to
(c 1 + c 2 + c 3 ) + (c 1 − c 3 )x + (c 2 + c 3 )x^2 = 0 + 0x + 0x^2.
Comparing coefficients we obtain the linear system of equations c 1 + c 2 + c 3 = 0 c 1 − c 3 = 0 c 2 + c 3 = 0 We form the associated augmented matrix and row-reduce:
Since there are no free variables, we have only the trivial solution c 1 = c 2 = c 3 = 0. Therefore, the given polynomials are linearly independent.
(10) The only scalars which solve the linear combination
c 1 (1) + c 2 (sin x) + c 2 (cos x) = 0 are c 1 = c 2 = c 3 = 0. Thus, the given functions are linearly independent.
(11) Using the trig identity sin^2 x + cos^2 x = 1, we have the linear combination
1 − sin^2 x − cos^2 x = 0.
Therefore, the given functions are linearly dependent. One possible linear dependence relation is 1 = sin^2 x + cos^2 x.
(17) (a) The set {u, +v, v + w, u + w} is linearly independent. To see this, suppose we have scalars c 1 , c 2 and c 3 such that
c 1 (u, +v) + c 2 (v + w) + c 3 (u + w) = 0.
We form the associated augmented matrix and row-reduce:
^ −→
^.
Thus, c 1 = c 2 = c 3 = c 4 = 0. That is, the matrices in B are linearly independent. Since dim(M 22 ) = 4 and B consists of 4 linearly independent matrices in M 22 , Theorem 6.10 (c) says that B is a basis for M 22.
(22) Suppose that c 1 , c 2 , c 3 are scalars such that
c 1 x + c 2 (1 + x) + c 3 (x − x^2 ) = 0 + 0x + 0x^2. Rearranging we see that this equation is equivalent to c 2 + (c 1 + c 2 + c 3 )x − c 3 x^2 = 0 + 0x + 0x^2. Comparing coefficients we see immediately that c 2 = c 3 = 0. This implies that c 1 = 0. Thus, the polynomials in B are linearly independent. Since dim(P 2 ) = 3 and B consists of 3 linearly independent polynomials in P 2 , Theorem 6.10 (c) says that B is a basis for P 2.
(23) Suppose that c 1 , c 2 , c 3 are scalars such that
c 1 (1 − x) + c 2 (1 − x^2 ) + c 3 (x − x^2 ) = 0 + 0x + 0x^2. Rearranging we see that this equation is equivalent to (c 1 + c 2 ) + (−c 1 + c 3 )x + (−c 2 − c 3 )x^2 = 0 + 0x + 0x^2. Comparing coefficients this gives the linear system c 1 + c 2 = 0 −c 1 + c 3 = 0 −c 2 − c 3 = 0 We form the augmented matrix and row-reduce:
Since there are free variables, we see that the polynomials in B are not linearly inde- pendent. In fact, if we solve the system for c 1 , c 2 , c 3 , we see that
x − x^2 = −(1 − x) + (1 − x^2 ).
Therefore, B is not a basis for P 2.
(27) We want to find scalars c 1 , c 2 , c 3 and c 4 such that
c 1
(^) + c 2
(^) + c 3
(^) + c 4
Comparing corresponding entries, this gives the linear system:
c 1 + c 2 + c 3 + c 4 = 1 c 2 + c 3 + c 4 = 2 c 3 + c 4 = 3 c 4 = 4 We form the augmented matrix and row-reduce:
^ −→
^.
We conclude that c 1 = c 2 = c 3 = −1 and c 4 = 4. Thus,
[A]B =
^.
(29) We want to find scalars c 1 , c 2 and c 3 such that
c 1 (1) + c 2 (1 + x) + c 3 (−1 + x^2 ) = 2 − x + 3x^2.
Rearranging this is equivalent to the equation
(c 1 + c 2 − c 3 ) + c 2 x + c 3 x^2 = 2 − x + 3x^2.
which is true if and only if a^ (a^ +^ b) c (c + d)
(a^ +^ c)^ (b^ +^ d) c d
Comparing entries of the matrices, we must have c = 0 and a = d. So,
V =
a^ b 0 a
(^) : a, d ∈ R
a
(^) + b
(^) : a, b ∈ R
= span
We now verify that the above two matrices that span V are also linearly independent. Suppose we have scalars c 1 and c 2 such that
c 1
(^) + c 2
Comparing entries we see that c 1 = c 2 = 0. Thus,
B =
is a basis for V. Thus, dim(V ) = 2.
(45) We need to find a polynomial f (x) in P 2 which makes the set {1 + x, 1 + x + x^2 , f (x)} linearly independent. Let f (x) = 1. If c 1 , c 2 , c 3 are scalars such that c 1 (1 + x) + c 2 (1 + x + x^2 ) + c 3 (1) = 0 = 0x + ox^2 then c 1 + c 2 + c 3 = 0 c 1 + c 2 = 0 c 2 = 0 This system has only the trivial solution c 1 = c 2 = c 3 = 0. Thus, the set B = {1 + x, 1 + x + x^2 , 1 } is linearly independent. Since dim(P 2 ) = 3 and B consists of three linearly independent polynomials in P 2 , Theorem 6.10 (c) implies that B is a basis for P 2.
(51) Since we already have a spanning set for the subspace, we simply need to throw away the polynomials which depend on the others. The first two polynomials are not scalar multiples of one another and so { 1 − x, x − x^2 } is a linearly independent set. Observe that the third and fourth polynomials are linear combinations of the first two, i.e.,
1 − x^2 = (1 − x) + (x − x^2 ) 1 − 2 x + x^2 = (1 − x) − (x − x^2 )
and so we throw away the third and fourth given polynomials. Therefore, { 1 − x, x − x^2 } is a linearly independent set of vectors that spans the given subspace. We conclude that { 1 − x, x − x^2 } is a basis for the subspace in question. (53) Since we already have a spanning set for the subspace, we simply need to throw away the functions which depend on the others. The first two functions are not scalar multiples of one another and so {sin^2 x, cos^2 x} is a linearly independent set. Observe that the third function is a linear combination of the first two via the trig identity cos 2x = cos^2 x − sin^2 x and so we throw away the third function. Therefore, {sin^2 x, cos^2 x} is a linearly independent set of vectors that spans the given subspace. We conclude that {sin^2 x, cos^2 x} is a basis for the subspace in question.
Section 6.3:
(2) (a) Observe that x = 5
Thus, [x]B =
Similarly, x = − 7
(3) (a) Observe that
x =
Thus,
[x]B =
Similarly,
x =
−^1
Thus,
[x]C =
(b) We need to find the coordinate vector of each basis vector in B with respect to C. We have (^)
and (^)
and (^)
Thus, by definition,
PC←B =
(c) We have
PC←B[x]B =
= [x]C^.
(d) We need to find the coordinate vector of each basis vector in C with respect to B. We have (^)
and (^)
and (^)
Thus, by definition,
PB←C =
(e) We have
PB←C [x]C =
= [x]B.
(9) (a) We have
A = 4
Thus,
[A]B =
^.
So, by definition,
PC←B =
^.
(c) We have
PC←B[A]B =
^ =
^ = [A]C^.
(d) We need to find the coordinate vector of each basis vector in C with respect to B. We have 1 2 0 − 1
and (^) 2 1 1 0
and (^) 1 1 0 1
and (^) 1 0 0 1
So, by definition,
PB←C =
^.
(e) We have
PB←C [A]C =
^ =
^ = [A]B.
(11) (a) Observe that f (x) = 2(sin x + cos x) − 5 cos x. Thus, [f (x)]B =
Similarly, f (x) = 2(sin x + cos x) − 3 cos x. Thus, [f (x)]C =
(b) We need to find the coordinate vector of each basis vector in B with respect to C. We have sin x + cos x = 1(sin x) + 1(cos x) and cos x = 0(sin x) + 1(cos x). Thus, by definition, PC←B =
(c) We have
PC←B[f (x)]B =
(^) = [f (x)]C.
(d) We need to find the coordinate vector of each basis vector in C with respect to B. We have sin x = 1(sin x + cos x) − (cos x) and cos x = 0(sin x + cos x) + cos x. Thus, by definition, PB←C =
To find the x′y′-coordinates, we need to find the matrix PC←B. We see that 1 0
= − √^1
− √^1
and (^)
0 1
= √^1
− √^1
Thus,
PC←B =
−^1 /
(a) We calculate
C
−^1 /
−^1 /
(b) Let x be the vector whose x′y′-coordinates are (4, −4). We want to find [x]B. We calculate
[x]B = PB←C [x]C = (PC←B)−^1
−^1 /
