Mathematics 3: Algebra Workshop 4 Working with linear maps The aim of this workshop is to work with linear maps. (1) Let F be a field, and define the linear map T : F 3 → F 3 by x1 x1 1 1 0 T x2 = B x2 , where B = 0 1 1 . x3 x3 1 0 1 Find the rank and nullity of T when (a) F = R; (b) F = F2. (a) Call the matrix A. Its columns are linearly independent over R, so the kernel contains only the 0 vector, and hence the nullity is 0. Also, as R3 is 3-dimensional, these columns span R3. Hence the rank of T is 3 (the dimension of the image of T ). 1 (b) Over F2, however, the columns of A sum to 0, so that 1 is in 1 the kernel of T . So the nullity of T is 1, and the rank is2. [This illustrates the importance of the particular field over which a vector space is defined: its properties of the space can depend strongly on the field.] (2) Changing bases. Please hand your answer to this Q at the week 5 workshop. (a) For a linear map T : Rn → Rm having matrix A with respect to the standard bases of Rn and Rm, what is its matrix when the basis of Rn is taken to be the columns of a nonsingular matrix B ∈ Rn×n, and the basis of Rm is taken to be the columns of a nonsingular matrix H ∈ Rm×m? (see Monday’s lecture). (b) Define T : R3 → R2 by x1 x1 1 0 1 T x2 = A x2 , where A = . 0 1 0 x3 x3 1 2 Now take the columns of B in Q1 as a new basis of R3, and a new basis of R2 to be the columns of 1 1 H = . 0 1 What is the new matrix A′ of T with respect to these bases? 1 ′ (c) Now the checking part: Write the vector 2 as a linear combination c1b1 + 1 ′ ′ c2b2 + c3b3 of the columns b1, b2, b3 of B (easy to do by ‘eyeballing’). Calculate ′ c1 ′′ 1 ′ ′ c1 ′′ ′′ A c2, and call the answer ′′ . Finally, check that A 2 = c1h1 + c2h2, ′ c2 c3 1 where h1 and h2 are the columns of H. (a) From lectures, the matrix wrt the new bases is H−1AB. (b) Easy to calculate that −1 1 −1 H = . 0 1 and then that ′ −1 2 0 0 A = H AB = . 0 1 1 ′ ′ ′ ′′ ′′ (c) Easy to find that c1 =0,c2 =1,c3 =1, and hence that c1 =0, c2 = 1 ′′ ′′ 2 2. Then can check that both A 2 and c h + c h are equal to . 1 1 2 2 2 1 (3) (Revision?) Show that a linear map T : U → V between vector spaces U, V (over some field F ) is an injection iff its kernel is {0}. Suppose the kernel is 0. If T (x) = T (x′) then, by linearity, T (x − x′)=0, so that x−x′ is in ker(T ), and hence is 0. So x = x′, showing that T is injective. Conversely, if T is injective, then, since T (0)=0, there can be no other vectors mapped to 0 by T . So the kernel consist of 0 only. (4) For a field F , let ψ : F n → F n be a linear map. (a) If ψ is a surjection, show that ψ is an injection (and so a bijection). (b) If ψ is a injection, show that ψ is surjection (and so a bijection). (a) We apply the Rank-Nullity Theorem. If ψ is surjective, the rank of its image is n, and so, by the Rank-Nullity Theorem, the rank of its 3 kernel is n − n = 0. Hence its kernel consists of the 0 vector only, making ψ an injection. (b) We apply the Rank-Nullity Theorem again. If ψ is injective, the rank of its kernel is 0, and so, by the Rank-Nullity Theorem, the rank of its image is n−0= n. Hence its image consists of all of F n, making ψ a surjection. [Compare with the ‘Discrete Maths’ fact that, for a function f from an n-element set to itself: if f is an injection then it’s a bijection, and if f is a surjection then it’s a bijection.] n n (5) For a field F , let φ : F → F be a linear map, and let b1, b2,...,bn be a basis for F n. n (a) If φ is also a bijection , show that φ(b1),φ(b2),...,φ(bn) is also a basis for F . n (b) Conversely, show that if φ(b1),φ(b2),...,φ(bn) is also a basis for F , then φ must be a bijection. (a) Assume φ is also a bijection. We need to show that φ(b1),φ(b2),...,φ(bn) are linearly independent, and that they span F n. n Linear independence: Suppose that j=1 cjφ(bj)=0. Then, by linearity, n n φ( j=1 cjbj) = 0, so j=1 cjbj is in theP kernel of φ. But, since φ is n injective,P its kernelP is 0, so j=1 cjbj =0. But then, since b1, b2,...,bn is a basis, these vectors are linearlyP independent, and so all of c1,...,cn are 0. Spanning: Take any x ∈ F n. As φ is surjective, x = φ(x′) for some ′ n n x ∈ F . But since b1, b2,...,bn is a basis, these vectors span F , and ′ n so x = j=1 cjbj, say. Applying φ, we have P n n ′ x = φ(x )= φ( cjbj)= cjφ(bj), Xj=1 Xj=1 as req’d. (b) Assume that φ(b1),φ(b2),...,φ(bn) is also a basis. We must show that φ is injective and surjective. n Injectivity. Take j=1 cjbj in the kernel of φ, and apply φ. Using n linearity, get j=1 cjPφ(bj)=0. As φ(b1),φ(b2),...,φ(bn) are linearly independent, all of c1,...,cn Pmust be 0. So the kernel of φ is 0 and φ is injective. Surjectivity. By assumption, every vector x ∈ F n can be written as n n x = j=1 cjφ(bj). But this is equal, by linearity, to φ( j=1 cjbj). Hence x isP in the image of φ. P.