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Math 790 Test 3 (Solutions) Satya Mandal Fall 05 Each Problem 10 points Due on: October 2, 2005 I like short proofs and elmentary proof. Unless otherwise stated, F is a field and V, W are two vector sapces over F.
- Let V, W be two vector spaces over F and let T : V → W be a set theoretic map. Prove that the following are equivalent:
(a) For u, v ∈ V and c, d ∈ F we have T (cu + dv) = cT (u) + dT (v) in W. (b) For u, v ∈ V and c ∈ F we have T (u + v) = T (u) + T (v) and T (cu) = cT (u) in W. (c) For u, v ∈ V and c ∈ F we have T (cu + v) = cT (u) + T (v) in W. (Recall, T is said to be a linear transformation if one of (or all) the above conditions are satisfied.) Solution. ((a) ⇒(b)): We have T (cu + dv) = cT (u) + dT (v). Take c = d = 1, we get T (u + v) = T (u) + T (v). Now take d = 0, we get T (cu) = cT (u). Therefore (b) is established. ((b) ⇒(c)): Using the additive part of the hypothesis, we have T (cu + v) = T (cu) + T (v). Using T (cu) = cT (u), we get T (cu + v) = T (cu) + T (v) = cT (u) + T (v). Hence, (c) is established.
((c) ⇒(a)): From the hypothesis in (c), we have T (cu + dv) = cT (u) + T (dv). Also taking v = 0 we get T (cu) = cT (u) for any c ∈ F and u ∈ V. So, we have
T (cu + dv) = cT (u) + T (dv) = cT (u) + dT (v)
and (a) is established.
- Let V, W be two vector spaces over F and let T : V → W be a linear transformation. Assume that dim(V ) = dim(W ) = n is finite. Prove that the following statements are equivalent: (a) T is invertible. (b) If e 1 , e 2 ,... , em ∈ V (here m ≤ n,) are linearly independent in V then the images T (e 1 ), T (e 2 ),... , T (em) are linearly independent in W. (c) T is onto. Solution. ((a) ⇒(b)): Suppose e 1 , e 2 ,... , em ∈ V are lin- early independent. We will prove that T (e 1 ), T (e 2 ),... , T (em) are linearly independent. Suppose c 1 T (e 1 ) + c 2 T (e 2 ) + · · · + cmT (em) = 0 for some ci ∈ F. Since T is linear, we have T (c 1 e 1 + · · · + cmem) = c 1 T (e 1 ) + c 2 T (e 2 ) + · · · + cmT (em) = 0. By (a), T is invertible and hence one to one. Therefore c 1 e 1 + · · · + cmem = 0. By linear indpendence of e 1 ,... , em, we have ci = 0. Therefore (b) is established. ((b) ⇒(c)): Suppose e 1 , e 2 ,... , en is a basis of V. By (b), T (e 1 ), T (e 2 ),... , T (en) are linearly independent. Since dim(W ) = n, it follows that T (e 1 ), T (e 2 ),... , T (en) is a basis of W. Therefore T (V ) = T (Span(e 1 ,... , en)) = Span(T (e 1 ), T (e 2 ),... , T (en)) = W. Hence T is onto and (c) is established. ((c) ⇒(a)): We need to show, T is one to one. We have nullity(T ) + rank(T ) = n. Since T is onto, rank(T ) = n. Hence, nullity(T ) = 0. So, null space of T is zero and T (v) = 0 implies v = 0. Hence T is one to one. So, (a) is established.
- Give the examples as follows:
(a) Give an example of a linear operator T : V → V such that T 2 = 0 but T 6 = 0. (b) Give two linear operator T, U : V → V such that T U = 0 but U T 6 = 0.
- Let V, W be two finite dimensional vector spaces over F. Assume dim V = n and dim W = m. Let Mm,n be the set of all m×n matrices with entries in F. Let E = {e 1 , e 2 ,... , en} be a basis of V and E′^ = {≤ 1 , ≤ 2 ,... , ≤m} be a basis of W. (a) For a linear transformation T : V → W define the matrix of T with respect to E and E′. (b) Prove that the map f : L(V, W ) → Mm,n such that f (T ) = matrix of T with respect to E and E′ is an isomorphism. (Try to understand the following diagram. Here A is the matrix of T.) V T^ // iso ≤ ≤
W iso ≤ ≤ Fn^ A^ //Fm Solution. We will prove only (b). Let me comment that to prove that f is ’isomorphism’, there are two general methods. First method proves that the map f is one to one and onto. Alternately, you can define a map g in the opposite direction and prove that f g = Id and gf = Id. I will write a proof in using this alternatve method. Define a map g : Mm,n → L(V, W ) as follows: For A ∈ Mm,n define T ∈ L(V, W ) by the equa- tion:
(T (e 1 ),... , T (en)) = (≤ 1 , ≤ 2 ,... , ≤m)A and let g(A) = T.
Note g is linear and gf = IdL(V,W ) and f g = IdMm,n. So, g is the inverse of f and the proof is complete.
f (T T −^1 ) = f (T −^1 T ) = f (Id) = In. By (a) f (T )f (T −^1 ) = f (T −^1 )f (T ) = In. Therefore f (T −^1 ) is the inverse of f (T ).
(⇒ ): Write f (T ) = A. Suppose A is ivertible. Let B be the inverse of A. Since f is onto, f (U ) = B for some U ∈ Mn,n. So, f (T U ) = f (T )f (U ) = AB = In. Since f is one to one, T U = Id. Similarly, U T = Id and therefore, T is invertible. This completes the proof.
- Let V be a finite dimensional vector space over F with dim(V ) = n. Let E = {e 1 ,... , en} and E′^ = {≤ 1 ,... , ≤n} be two basis of V. Let T ∈ L(V, V ) be linear operator. Let (e 1 ,... , en) = (≤ 1 ,... , ≤n)P for some n × n matrix. (a) Prove that P is an invertible matrix. (b) Let A be the matrix of T with respect to E and B be the matrix of T with respect to E′. Prove that B = P AP −^1. Solution. Proof of (a): There is also a matrix Q such that (≤ 1 ,... , ≤n) = (e 1 ,... , en)Q. Combining these two, we get (e 1 ,... , en) = (e 1 ,... , en)QP. Therefore QP = In and similarly, P Q = In. So, P is invert- ible. Proof of (b): Apply T to the equation: (e 1 ,... , en) = (≤ 1 ,... , ≤n)P. We get (T (e 1 ),... , T (en)) = (T (≤ 1 ),... , T (≤n))P. We also have (T (e 1 ),... , T (en)) = (e 1 ,... , en)A and (T (≤ 1 ),... , T (≤n)) = (≤ 1 ,... , ≤n)B. Therefore (e 1 ,... , en)A = (≤ 1 ,... , ≤n)BP = (e 1 ,... , en)P −^1 BP. Comparing, we get A = P −^1 BP. This completes the proof.
