DS-GA 3001: PROBLEM SESSION 2 SOLUTIONS VLADIMIR KOBZAR

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1 DS-GA 3: PROBLEM SESSION SOLUTIONS VLADIMIR KOBZAR Note: In this discussion we allude to the fact the dimension of a vector space is given by the number of basis vectors. We will shortly establish this result formally. Question. Consider the non-empty set of functions # V : p : R Ñ R ˇˇ + ppxq a k x k for a i P R, x P R Define an addition operation ` : V ˆV Ñ V and a scalar multiplication operation : R ˆ V Ñ V such that the triple pv, `, q is a real vector space. Find a basis for this vector space, and deduce its dimension. Soln: The numbering below corresponds to the numbering of the vector space axioms in the previous lecture. ř [ a ] Define : R ˆ V Ñ V by pλpqpxq λppxq. Then λppxq n pλa kqx k where ppxq ř n a kx k, so λp P V. [ b ] Define ` : V Ñ V by pp ` p qpxq p pxq ` p pxq. Then a kx k and p pxq ` p pxq ř n pa k ` a k qx k where p pxq ř n p pxq ř n a kx k. Therefore, p ` p P V. [] pp ` p qpxq ř n pa k ` a k qx k ř n p qpxq Thus p ` p p ` p for any p, p P V. [] For p 3 defined by p 3 pxq ř n a 3kx k, we have pp ` pp ` p 3 qqpxq a k x k ` p n p pa k ` a k qx k pp ` a k x k ` a 3k x k q ÿ a k x k ` a k x k q ` a 3k x k ppp ` p q ` p 3 qpxq Thus p ` pp ` p 3 q pp ` p q ` p 3 for any p, p, p 3 P V. [3] Define by pxq. By the Fundamental Theorem of Algebra, an n degree polynomial with more than n roots is identically zero. Therefore P V is well defined by pxq ř n xk, and we Date: September, 6.

2 VLADIMIR KOBZAR have pp ` qpxq a k x k ` Therefore, p ` p for any p P V. [4] We have x k ppxq pp ` p q pqpxq a k x k ` p q a k x k Therefore, p ` p qp for any p P V. [5] We have p pqpxq Therefore, p p for any p P V. [6] We have c pc pqpxq c pc a k x k ppxq a k x k q pc c q Therefore, c pc pq pc c qp for any p P V. [7] We have ppc ` c qpqpxq pc ` c q c a k x k a k x k pc c qppxq a k x k ` c a k x k c ppxq ` c ppxq Therefore, pc ` c q p c p ` c p for any c, c P R and p P V. [8] We have cpp ` p qpxq cp n c a k x k ` ÿ a k x k ` c a k x k q a k x k cp pxq ` cp pxq Therefore, c pp ` p qp c p ` c p for any c P R and p, p P V. By the construction of V, the set of monomials b tx k u n is a spanning set. By the Fundamental Theorem of Algebra, an n degree ř polynomial with more than n roots is identically zero. Therefore n a k x k for all x P R if and only if a k for ď k ď n. Thus, b is a basis of V.

3 DS-GA 3: PROBLEM SESSION SOLUTIONS 3 Question. Let V : R nˆn, the space of nˆn matrices. Show that V is a real vector space. Show [argue] that the dimension of V is n P N. (Sketch of proof) The closure under addition and scalar multiplication and the other vector space properties follow from the canonical addition and scalar multiplication ` : V Ñ V by A ` A where the ij-th entry of the sum is given by pa ` A q ij a ij ` a ij and a ij and a ij are the ij-th entries of A and A respectively, and : R ˆ V Ñ V by λa where the ij-th entry of the is given by pλaq ij λa ij where a ij are the ij-th entries of A. The set of n ˆ n matrices E ij that have all zero entries except for the ij-th entry, where ď i, j ď n are linearly independent and span V. There are n such matrices. Question 3. Let V : CpR, Rq be the set of all continuous functions on R with range in R. Show that V is a real vector space. What can you say about the dimension of V? Let W : C pr, Rq be the set of all differentiable functions whose derivative is continuous. Show that W is a subspace of V : CpR, Rq. Soln: [ a ] Define : R ˆ V Ñ V by pλfqpxq λfpxq. The closure of V under scalar multiplication follows from the fact that continuous functions remain continuous under scaling. [ b ] Define ` : V ˆ V Ñ V by pf ` gqpxq fpxq ` gpxq. The closure under addition follows from the fact that sums of continuous functions are continuous. [] We have pf ` gqpxq fpxq ` gpxq gpxq ` fpxq pg ` fqpxq Thus f ` g g ` f for any f, g P V. [] For h P V, we have pf ` pg ` hqqpxq fpxq ` pgpxq ` hpxqq pfpxq ` gpxqq ` hpxq ppf ` gq ` hqpxq Thus f ` pg ` hq pf ` gq ` h for any f, g, h P V. [3] Define by pxq, which is trivially a continuous function. We have, pf ` qpxq fpxq ` fpxq Therefore, f ` f for any f P V. Note that if pxq for any x, then the above equality will not hold, and therefore, P V is well-defined. [4] We have pf ` p q fqpxq fpxq ` p qfpxq Therefore, f ` p qf for any f P V.

4 4 VLADIMIR KOBZAR [5] We have Therefore, f f for any f P V. [6] We have p fqpxq fpxq c pc fqpxq c pc fpxqq pc c q fpxq pc c qfpxq Therefore, c pc fq pc c qf for any f P V. [7] We have ppc ` c qfqpxq pc ` c q fpxq c fpxq ` c fpxq c fpxq ` c fpxq Therefore, pc ` c q f c f ` c f for any c, c P R and f P V. [8] We have cpf ` f qpxq cpf pxq ` f pxqq cf pxq ` cf pxq Therefore, c pf ` f qp c f ` c f for any c P R and f, f P V. Since any monomial is continuous, and as shown above, the set of monomials is linearly independent. Given any finite dimension k, we can find l linearly independent monomials in V such that l ą k. Thus, V cannot be finite-dimensional. [ a ] For C pr, Rq functions, the fact that f is differentiable implies that there exists f : R Ñ R such that for every a P R, there exists pλfq on R defined by pλfq paq λf paq. The continuity of f implies that λf is continuous on R. This confirms that pλfq P W. [ b ] The fact that f and g are differentiable implies that there exists f, g : R Ñ R such that for every a P R, we have pf ` gq on R defined by pf ` gq paq f paq ` g paq. The existence of pf ` gq implies that pf ` gq is continuous on R. This confirms that pf ` gq P W. Question 4. Let T : R 4 Ñ R 3 given by M 4 since the remain- Find a basis for the kernel and range of T. Soln: Range M is spanned by and ing columns are linearly dependent. Solving the linear system given by 4 x x x 3 x 4 yields, for example, the following two linearly independent vectors,

5 which span ker M: DS-GA 3: PROBLEM SESSION SOLUTIONS 5 and (We will study Gaussian elimination as a systematic method to solve systems of linear equations.) Question 5. Find the range R T and the nullspace N T of the following linear mappings T : U Ñ V and give their dimension. What would be the matrix representation of T in the canonical basis? (a) U V R n, T px, x, x 3,..., x n q px `x, x `x, x 3,..., x n q Soln: Vectors in the form pa, a,,..., q, and only such vectors, will be mapped to zero. Therefore, the nullspace will be spanned by p,,,..., q and dim R T n (b) U V # R n, T px, x, x 3,..., x n q px `x, x x, x 3,..., x n q a ` a Soln: ðñ a a. Thus dim N T and a a dim R T n Question 6. (a) Find the change of basis matrix U from the canonical basis S in R to the basis S tv, v u with ˆ ˆ 4 v, v 3 5 Soln: ˆx x S ˆ 4 ˆx x v ` x v 3 5 Thus the change of basis matrix U from S to S is given by: U ˆ ˆ 5 3 x

6 6 VLADIMIR KOBZAR ˆ (b) Use U SÑS to compute the coordinates of the vector u with respect to the basis S. Soln: 5 ˆ ˆ ˆ 3 (c) We say that a linear transformation is given with respect to a single basis if both the domain and the range are coordinatized using the same basis. Let T be a linear mapping given by ˆ ˆ x x x T x x ` x with respect to the canonical basis S. Find the matrix representation of T with respect to S. Soln: T is represented by the following matrix in S: ˆ T Therefore, ˆ T UT U 5 ˆ ˆ ˆ