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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