Download Sol - Math TAMU

HW 7
4.1 #3. (2 pts.) Verify that the set of polynomials f in P3 such that f (2) = 0 forms a vector
space with the usual operations on 3-vectors.
Sol. Let V denote the collection of polynomials f ∈ P3 such that f (2) = 0. First we
check that V is closed under vector addition and scalar multiplication. Suppose f , g ∈ V and
c ∈ R . The addition f + g is a polynomial whose evaluation at x = 2 satisfies
(f + g)(2) = f (2) + g(2) = 0 + 0 = 0.
We also have
(cf )(2) = c f (2) = c 0 = 0
(1 pt. so far)
From elementary calculation of polynomials, for f , g, h ∈ V and a, b ∈ R, we check that
(1) f + g = g + h
(2) f + (g + h) = f + g + h = (f + g) + h
(3) ∃0 (the polynomial with 0 value globally) so that f + 0 = 0 + f = f
(4) ∀f , ∃! − f so that f + (−f ) = (−f ) + f = 0
(5)
(6)
(7)
(8)
a(f + g) = af + ag
(a + b)f = af + bf
(ab)f = a(bf )
1f = f (another 1 pt. )
4.1 #6. (2 pts.) Prove that the set of nonsingular n×n matrices under the usual operations
is not a vector space.
Sol. If such collection V is a vector space then V contains a zero vector, i.e., zero matrix
0 of size n × n. However the matrix 0 is singular, and so 0 ∈
/ V. This implies that V can
not be a vector space.
4.1 #12. (2 pts.) Let V be a vector space. Prove that the identity element for vector
addition in V is unique.
Sol. Suppose V contains two identity elements v1 and v2 . Then for a vector w ∈ V we
have w = w + v1 and w = w + v2 . By taking the difference we further have
0 = w − w = (w + v1 ) − (w + v2 ) = v1 − v2 ,
and hence v1 = v2 .
4.2 #4. (2 pts.) Show that the set of vectors of the form [a, b, 0, c, a − 2b + c] in R5 forms
a subspace of R5 under the usual operations.
1
Sol. Let V denote the set of vectors of the above form. By rewriting
[a, b, 0, c, a − 2b + c] = a[1, 0, 0, 0, 1] + b[0, 1, 0, 0, −2] + c[0, 0, 0, 1, 1],
(1 pt. so far) we see that V is the set of linear combinations of three vectors [1, 0, 0, 0, 1], [0, 1, 0, 0, −2]
and [0, 0, 0, 1, 1]. That is V = Span(S) where S consists of those three vectors. As discussed
in the proof of Thm 4.5 (section 4.3) one can check that Span(S) is closed under addition
and scalar multiplication. (another 1 pt. )
4.2 #9. (2 pts.) Show that the set W of solutions to the differential equation y 00 +2y 0 −9y = 0
is a subspace of the vector space of all twice-differentiable real-valued functions defined on
R.
Sol. We first show that W is nonempty. Observe that the 2nd order equation determines
a quadratic equation (the corresponding characteristic equation)
x2 + 2x − 9 = 0.
√
The Quadratic Formula implies that x = −1 ± 10. One can verify that
√
y1 = e(−1+
10)x
√
and
y2 = e(−1−
10)x
are solutions of the given differential equation as follows.
√
√
√
√
y10 = (−1 + 10)e(−1+ 10)x , y100 = (−1 + 10)2 e(−1+ 10)x
o
n
√
√
√
√
y100 + 2y10 − 9y = (−1 + 10)2 + 2(−1 + 10) − 9 e(−1+ 10)x = 0e(−1+ 10)x = 0
(similar calculation shows that y2 is also a solution) Thus the collection W contains at least
two elements y1 and y2 . (1 pt. so far)
To show W is a subspace it remains that W is closed under addition and scalar multiplication. Let y = f and y = g be elements of W. Then f + g ∈ W since
(f + g)00 + 2(f + g)0 − 9(f + g) = {f 00 + 2f 0 − 9f } + {g00 + 2g0 − 9g} = 0 + 0 = 0.
For a scalar c ∈ R we have
(cf )00 + 2(cf )0 − 9(cf ) = c(f 00 + 2f 0 − 9f ) = c0 = 0.
Therefore W is a vector space. (another 1 pt. )
Note that the solution set W is spanned by S = {y1 , y2 }, i.e., every solution of the
differential equation can be written by a linear combination of y1 and y2 . This provides a
fundamental method for one to solve a homogeneous differential equations of degree n.
2