let T be a linear operator on a finite-dimensional vector space V with the disti

Question

let T be a linear operator on a finite-dimensional vector space V with the distinct eigenvalues 1 , 2 ,...,k and corresponding multiplicities m1 , m2 , ... , mk suppose that is a basis for V such that [T] is an upper triangular matrix. prove that the diagonal entries of [T] are 1 , 2 , ...,k and that each i occurs mi times (1 i k)

Explanation / Answer

Section 5.1: Defn 1. A linear operator T : V ? V on a finite-dimensional vector space V is called diagonalizable if there is an ordered basis ß for V such that ß[T]ß is a diagonal matrix. A square matrix A is called diagonalizable if LA is diagonalizable. Note 1. If A =B [T]B where T = LA then to find ß such that ß[I]B B[T]B B[I]ß =ß [T]ß is diagonal is the same thing as finding an invertible S such that S-1AS is diagonal. Defn 2. Let T be a linear operator on V , a non-zero v ? V is called an eigenvector of T if ?? ? F, such that T(v) = ?v. The scalar ? is called the eigenvalue corresponding to the eigenvector v. Let A ? Mn(F),v ? Fn,v ? 0 is called an eigenvector of A if v is an eigenvector of LA. That is, Av = ?v. Theorem 5.1. A linear operator T on a finite dimensional vector space V is diagonalizable if and only if there is an ordered basis ß for V consisting of eigenvectors of T . Furthermore, if T is diagonalizable, ß = {v1, v2, . . . , vn} is an ordered basis of eigenvectors of T , and for D =ß [T]ß, D is a diagonal matrix and Dj,j is the e-val corresponding to vj,1 = j = n. Proof. If T is diagonalizable, then ?? ?1 0 0 ··· 0 0 ?? ß [ T ] ß = ?? 0 ? 2 0 · · · 0 0 ?? Example 1. ··· 0 0 0 ··· 0 ?n for basis ß = {v1,v2,...,vn}, which means that for all i, T(vi) = ?ivi. So each ?i is an eigenvalue corresponding to the eigenvector vi. Conversely, if there exists an ordered basis ß = {v1,v2,...,vn} such that each vi is an e-vctr of T then there exist ?1,?2,...,?n such that T (vi) = ?ivi and so ???1 00···00?? ß [ T ] ß = ?? 0 ? 2 0 · · · 0 0 ?? ?? cosa -sina ?? B[T]B = sina cosa , ··· 0 0 0 ··· 0 ?n whereB = {(1, 0), (0, 1)}, If a = p/2, T has no eigenvectors because the linear transformation is “rotation by a”. But if a = p, then it does have e-vectors (1, 0) and (0, 1). But this transformation is invertible. (Rotate by the negative angle) So invertible and diagonalizable are not the same. 1 2 Theorem 5.2. A ? Mn(F). ? is an e-val of A if and only if det(A - ?I) = 0. Proof. ? is an e-val of A ??v?=0suchthatAv=?v. ??v?=0suchthatAv-?v=0. ??v?=0suchthat (A-?In)v=0. ?det(A-?In)=0. Defn 3. For matrix A ? Mn(F), f(t) = det(A - tIn) is the characteristic polynomial of A. Defn 4. Let T be a linear operator on an n-dimensional vector space V with ordered basis ß. We define the characteristic polynomial f(t) of T to be the characteristic polynomial of A=ß [T]ß. Thatis,f(t)=det(A-tIn). Note 2. Similar matrices have the same characteristic polynomials, since if B = S-1AS, det(B - tIn) = det(S-1AS - tIn) = det(S-1AS - tS-1S) = det(S-1(A - tIn)S) = det(S-1) det(A - tIn) det(S) = 1 det(A - tIn) det(S) = det(A - tIn) det(S ) So that characteristic polynomials are “similarity invariants”. If B1 and B2 are 2 bases of V then B1[T]B1 is similar to B2[T]B2 and so we see that the definition of characteristic polynomial of T, does not depend on the basis used in the representation. So, we may say f(t) = det(T - tIn). Theorem 5.3. Let A ? Mn(F) (1) The characteristic polynomial of A is a polynomial of degree n with leading coefficient (-1)n . (2) A has at most n distinct eigenvalues. Proof. First, we will prove (1). We need a slightly stronger statement for our induction statement to be of use. If B is a square n × n matrix such that for some permutation ? ? Sn, and some subset K ? [n], (B)i,j =bi,?(i) -t, i (B)i,j = bi,j, i (B)i,j = bi,j, i ?K,?(i)=j ?K,j?=?(i) ??K where for all i,j ? [n], bi,j is a scalar, then det(B) is a polynomial in t of degree |K|. Furthermore, if |K| = n and ? = id, the leading coefficient is (-1)n. In this case, the entries on the diagonal of B are of the form bi,i - t. The proof is by induction. The base case is for is for n = 1. A = [a1,1], B = A-tI1, det(B) = a1,1 - t, which is a polynomial of degree 1 and has leading coefficient (-1)1 and if B = A, we have that det(B) = a1,1 which is a polynomial of degree 0. Assume n > 1 and the theorem is true for n - 1 × n - 1 matrices. Assume B satisfies the hypothesis of the statement. We compute det(B) by expanding on row 1. det(B) = ??n (-1)1+i(B)1,i|B(1|i)| i=1 We see that for each i, by induction, B(1|i) is an n - 1 square matrix with |K| or |K| - 1 entries of the form bi,j - t. If there exists an i ? [n], such that (B)1,i is of the form b1,i - t then B(1|i) has |K| - 1 entries of the form br,s - t and satisfies the induction hypothesis. det(B(1|i)) is therefore a polynomial of degree |K| - 1 and also for j ?= i, B(1|j) has |K| - 1 entries of the form br,s -t and satisfies the induction hypothesis. det(B(1|i)) is therefore a polynomial of degree |K|-1. Weseeinthiscasethatdet(B)isapolynomialofdegree1+|K|-1=|K|. If additionally, |K| = n and ? = id, then b1,1 is of the form b1,1 - t and B(1|1) has all of its diagonal entries of the form bi,i - t and we see that its leading coefficient is (-1)n-1 by induction and so the leading coefficient of det(B) is (-1)(-1)n-1 = (-1)n. And if for all i ? [n], (B)1,i is of the form b1,i then for all i ? [n], B(1|i) has |K| or entries of the form br,s - t and satisfies the induction hypothesis. So we see that in this case, det(B) is a polynomial of degree |K|. Now (2) follows by the fact from algebra that a polynomial of degree n over a field can have at most n roots. Theorem 5.4. Let T be a linear operator on a vector space V and let ? be an e-val of T . A vectorv?V isane-vctrofT correspondingto?ifandonlyifv?=0andv?N(T-?In). Proof. ? is an e-val of T ??v?=0suchthatT(v)=?v. ??v?=0suchthatT(v)-?v=0. ??v?=0such that(T-?In)(v)=0. ??v?=0suchthatv?N(T-?In). Section 5.2. Theorem 5.5. Let T be a linear operator on a vector space V and let ?1,?2,...,?k be distinct e-vals of T . If v1, v2, . . . , vk are e-vctrs of T such that for all i ? [k], ?i corresponds to vi, then {v1, v2, . . . , vk} is a linearly independent set. Proof. The proof is by induction on k. Let k = 1. {v1} is a linearly independent set. Assume k > 1 and the theorem holds for k - 1 distinct e-vals and e-vctrs. Now suppose ?1,?2,...,?k be distinct e-vals of T and v1,v2,...,vk are e-vctrs of T such that for all i ? [k], ?i corresponds to vi. We wish to show {v1, v2, . . . , vk} is a linearly independent set. Let a1v1+a2v2+···+akvk=0 (1) for some scalars a1,...,ak. Applying T - ?kI to both sides of the equation, we obtain, (T -?kI)(a1v1 +a2v2 +···+akvk)=0 a1(T -?kI)(v1)+a2(T -?kI)(v2)+···+ak(T -?kI)(vk))=0 (2) ?i ? [k - 1], ai(T - ?kI)(vi) = ai(T(vi) - ?kI(vi)) = ai(?ivi - ?kvi) = ai(?i - ?k)vi 3 4 and ak(T - ?kI)(vk) = ak(T(vk) - ?kI(vk)) = ak(?kvk - ?kvk) = ak(?k - ?k)vk So (2) becomes a1(?1 -?k)v1 +a2(?2 -?k)v2 +···+ak-1(?k-1 -?k)vk-1 =0 By induction, {v1, v2, . . . , vk-1} is a linearly independent set and so for all i ? [k - 1], ai(?i -?k)=0. But,since?i -?k ? 0,itmustbethatai =0. Now looking back at equation (1), we have that akvk = 0. But since vk is not the zero vector, it must be that ak = 0 as well. Therefore, {v1, v2, . . . , vk} is a linearly independent set. Cor 1. Let T be a linear operator on an n-dimensional vector space V . If T has n distinct e-vals, then T is diagonlizable. Proof. Suppose ?1, ?2, . . . , ?n are distinct e-vals of T with corresponding e-vctrs v1, v2, . . . , vn. By Theorem 5.5, {v1, v2, . . . , vn} is a linearly independent set. By Theorem 5.1, T is diag- onalizable. Defn 5. A polynomial f(t) in P(F) splits over F if there are scalars c,a1,...,an such that f (t) = c(t - a1 )(t - a2 ) · · · (t - an ). Theorem 5.6. The characteristic polynomial of any diagonalizable linear operator splits. Proof. Let T be a diagonalizable linear operator on V . Suppose ß is a basis of V such that D =ß [T]ß is diagonal. ???1 00···00?? D = ?? 0 ? 2 0 · · · 0 0 ?? ··· 0 0 0 ··· 0 ?n f(t) is the characteristic polynomial so ?? ? 1 - t 0 0 · · · 0 0 ?? f(t)=det(D-tI)=?? 0 ?2-t 0 ··· 0 0 ?? ··· 0 0 0 ··· 0 ?n-t =(?1 -t)(?2 -t)···(?n -t). Defn 6. Let ? be an e-val of a linear operator (or matrix) with characteristic polynomial f(t). The algebraic multiplicity (or just multiplicity) of ? is the largest positive integer k for which(t-?)k isafactoroff(t). Defn 7. Let T be a linear operator on a vector space V and let ? be an e-val of T. Define E? = {x ? V |T(x = ?x} = N(T-?I). The set E? is called the eigenspace of T corresponding to ?. The eigenspace of a matrix A ? Mn(F ) is the e-space of LA. Fact 1. E? is a subspace. Proof. Let a ? F, x,y ? E?. Then T(ax+y) = aT(x)+T(y) = a?x+?y = ?(ax+y). Theorem 5.7. Let T be a linear operator on a finite dimensional vector space V , and let ? be an e-val of T having multiplicity m. Then 1 = dim(E?) = m. Proof. Let {v1, v2, . . . , vp} be a basis of E?. Extend it to a basis ß = {v1, v2, . . . , vp, vp+1, . . . , vn} of V . Let A = [T ]ß , then since ?i = p, T(vi) = ?vi. So, ???Ip B?? A=0C ?? (? - t)Ip B ?? A - tI = 0 C - tIn-p Expanding on the 1st column, we see that det(A - tIn) = (? - t)p det(C - tIn-p) = (? - t)pq(t). So the multiplicity of ? is greater than or equal to p and the dim(E?) = p. Lemma 1. Let T be a linear operator on a vector space V and let ?1, ?2, . . . , ?k be distinct e-vals of T. For each i = 1,2,...,k, let vi ? E?i. If v1+v2+···vk = 0, then ?i ? [k],vi = 0. Proof. Renumbering if necessary, suppose for 1 = i = p, vi ?= 0 and for p+1 = i = k, vi = 0. Then, v1 + v2 + · · · , vp = 0. But this contradicts Theorem 5.5. Thus,?i?[k],vi =0. Theorem 5.8. Let T be a linear operator on a vector space V and let ?1,?2,...,?k be distinct e-vals of T. For each i = 1,2,...,k, let Si ? E?i, be a finite linearly independent set.ThenS=S1?S2?···?Sk isalinearlyindependentsubsetofV. Proof. For all i suppose Si = {vi,1, vi,2, . . . , vi,ni }. Then S={vi,j :1=i=k,1=j=ni}. Supposethereexists{ai,j}suchthat k ni ???? ai,jvi,j = 0. For each i, let wi = ni ai,j vi,j . Then wi ? E? , for all i and w1 + w2 + · · · + wk = 0. ?? i=1 j=1 j=1 i So, by the lemma, wi = 0, ?i. But each Si is linearly independent, so for all j, ai,j = 0. Theorem 5.9. Let T be a linear operator on a finite dimensional vector space V such that the characteristic polynomial of T splits. Let ?1,?2,...,?k be distinct e-vals of T. Then (1) T is diagonalizable if and only if the multiplicity of ?i is equal to dim(E?i ), ?i. (2) If T is diagonalizable and ßi is an ordered basis for E?i , for each i, then ß = ß1 ? ß2 ? · · · ? ßk is an ordered basis for V consisting of eigenvectors of T . Proof. For all i ? [k], let mi be the multiplicity of ?i, di = dim(E?i ), and dim(V ) = n. We will show (2) first. Suppose T is diagonalizable. let ßi be a basis for E?i , ?i ? [k]. We know ß = {ß1 ? ß2 ? ··· ? ßk} is a linearly independent set by Theorem 5.8. By Theorem 5.1, there is a basis ? of V such that ? consists of eigenvectors of T . Let x ? V . x?Span(?),sox=a1v1 +a2v2 +···+anvn wherev1,v2,...,vn areeigenvectorsofT. 5 6 Each vi ?Span(ßj), for some j ? [k]. So it can be expressed as a linear combination of vectors in ßj. Thus x ?Span(ß) and we have span(()ß) = V . Now we show (1). (?:)Weknowdi =mi,?i. But since by (2), ß is a basis, we have n=d1 +d2 +···+dk =m1 +m2 +···+mk =n Thus, by the “squeeze principle” we have d1 +d2 +···+dk =m1 +m2 +···+mk and m1 -d1 +m2 -d2 +···+mk -dk0 But?i?[k],mi -di =0andsomi =di. (?:) Suppose ?i, mi = di. We know m1 +m2 +···+mk = n since T splits and by Theorem 5.3 f(t) has degree n. Thus, we know d1 +d2 +···+dk = n and if ?i, ßi is an ordered basis for E?i , by Theorem 5.8 ß = ß1 ? ß2 ? · · · ? ßk is linearly independent. And, since |ß| = n, by Corollary 2, (b) to Theorem 1.10, ß is a basis of V . Then by Theorem 5.1, T is diagonalizable. Note 3. Test for diagonalization. A linear operator T on a vector space V of dimension n is diagonalizable if and only if both of the following hold. (1) The characteristic polynomial splits. (2) For each ?, eigenvalue of T, the multiplicity of ? equals the dimension of E?. Notice that E? = {x|(T?I)(x) = 0} = N(T - ?I) and n = nullity(T - ?I) + rank(T - ?I). So, dim(E?) = nullity(t - ?I) = n - rank(T - ?I). Proof. Assume T is diagonalizable. By Theorem 5.6, the characteristic polynomial of T splits. Now by Theorem 5.9, for each ?, eigenvalue of T, the multiplicity of ? equals the dimension of E?. Now assume (1) and (2) hold. Then by Theorem 5.9 again, T is diagonalizable. Defn 8. Let W1,W2,...,Wk be subspaces of a vector space V. The sum is: ??k i=1 Wi ={v1 +v2 +···+vk :vi ?Wi,?i?[k]} Fact 2. The sum is a subspace. Defn 9. Let W1,W2,...,Wk be subspaces of a vector space V. We call V the direct sum of W1,W2,...,Wk, written V = W1 ? W2 ? · · · ? Wk IfV isthesumofW1,W2,...,Wk and?j?[k],Wj n?? Wi ={0}. Example 2. V = R4 W1 = {(a,b,0,0)|a,b ? R} W2 = {(0,0,c,0)|c ? R} W3 = {(0,0,0,d)|d ? R} Theorem 5.10. Let W1, W2, . . . , Wk be subspaces of a finite-dimensional vector space V . Tfae i ? j (1) V = W1 ? W2 ? · · · ? Wk (2) V = ??ki=1 Wi and ?i, vi ? Wi if v1 + v2 + · · · vk = 0 then vi = 0, ?i (3) Each vector v ? V can be written uniquely as v = v1 +v2 +···vk, where ?i ? [k],vi ?Wi. (4) If ?i ? [k],?i is an ordered basis for Wi then ?1 ??2 ?···??k is an ordered basis for V. (5) For each i ? [k] there is an ordered basis ?i for Wi such that ?1 ??2 ?···??k is an ordered basis for V . Proof. (1)?(2): Supposefori?[k],v ?W v +v +···+v =0. Let1=i=k. Then ??ii12k?? vi =- j? ivj. Butsinceby(1),V =W1?W2?···?Wk,weknowthatvi =- j?=ivj =0. (2) ? (3): By (2), each vector v ? V can be written as v = v1 +v2 +···vk, where ?i ? [k],vi ? Wi. To show uniqueness, suppose that v = v1+v2+···vk and v = w1+w2+···wk, where?i?[k],vi,wi ?Wi. Then we have v1 +v2 +···vk =w1 +w2 +···wk v1 -w1 +v2 -w2 +···+vk -wk =0. oreachi?[k],vi -wi ?Wi,soby(2),vi -wi =0andwehavethatvi =wi. (3) ? (4): Let i ? [k] and ?i = {wi,1,wi,2,...,wi,ni} be an ordered basis for Wi. By (3), we know that ?1 ? ?2 ? · · · ? ?k spans V . To show linear independence, we suppose for some {ai,j :1=i=k,1=j =ni}, Notice that for each i ? [k], So by uniqueness, we have that must be that ?j ? [ni], ai,j = 0. ??k ?? i=1 ?1 ??2 ?···??k spans V, we have that V = Wj n ??ki=1,i? j Wi. Since v ? Wj , we have that ??k i i=1 But also, k ni ???? ai,jwi,j =0 ni ai,jwi,j ? Wi. We also have ?? i=1 j=1 j=1 0=0 ni ai,jwi,j = 0. Now since ?i is linearly independent, it j=1 (4) ? (5): We know that for each i ? [k], Wi has a finite basis, ?i. Thus ?1 ??2 ?···??k is an ordered basis for V by (4). (5) ? (1): Let i ? [k] and ?i = {wi,1,wi,2,...,wi,n } be an ordered basis for Wi. Since Wi. Let j ? [k] and consider v ? v = aj,1wj,1 + aj,2wj,2 + · · · + aj,nj wj,nj v= ??k i=1,i?=j xi 7 8 for some vectors xi ? Wi, which are linear combinations of the vectors in ?i. We have that xi =0 and hence all of the coefficients of vectors in ?1 ? ?2 ? · · · ? ?k are zero. This implies that v = 0. Theorem 5.11. A linear operator T on a finite-dimensional vector space V is diagonalizable if and only if V is the direct sum of the eigenspaces of T. Proof. Let ?1, ?2, . . . , ?k be the distinct eigenvalues of T . (?) Let T be diagonalizable. Then ?i ? [k], let ?i be an ordered basis of E?i . By Theorem 5.9, ?1 ??2 ?···??k is an ordered basis for V. By Theorem 5.10, V is the direct sum of E?i ’s. ??k (?) If V = i=1 E?i. Choose a basis ?i for each E?i. By Theorem 5.10, ?1 ??2 ?···??k is an ordered basis for V . Since there is a basis for V of E-vcts of T , T is diagonalizable, by Theorem 5.1. Section 5.3 - Skip. Section 5.4 - Invariant subspaces and the Cayley-Hamilton Theorem. Defn 1. Let T be a linear operator on a vector space V . A subspace W of V is called a T-invariantsubspaceofV if(W)?W. Defn 2. If T is a linear operator on V and W is a T-invariant subspace of V, then the restriction TW of T to W is a mapping from W to W and it follows that TW is a linear operator on W. aj,1wj,1 +aj,2wj,2 +···+aj,njwj,nj - Lemma 2. Exercise 21 from Section 4.3. If M ? Mn(F) can be expressed as ??A B?? M=OC, where A ? Mr(F), C ? Ms(F), s + r = n, and O is the all s × r matrix of all zeros. Proof. The proof is by induction on r. If r = 1, we form det M be expanding on column 1. Then det M = a1,1M(1|1) = det A · det C. Now assume for all such matrices M where A is r-1×r-1. Again, we expand on column 1 of M. det M = a1,1 det M (1|1) - a2,1 det M (2|1) + · · · (-1)r+1 det M (r|1) For each i, M(i|1) has the form ?? A(i|1) B* ?? OC, where B* is a submatrix of B. By induction, det M(i|1) = det A(i|1) det C. So, we have: detM =a1,1detA(1|1)detC-a2,1detA(2|1)detC+···(-1)r+1detA(r|1)detC = det C(a1,1 det A(1|1) - a2,1 det A(2|1) + · · · (-1)r+1 det A(r|1)) = det C det A ??k i=1,i?=j Theorem 5.21. Let T be a linear operator on a finite-dimensional vector space V , and let W be a T-invariant subspace of V. Then the caracteristic polynomial of TW divides the characteristic polynomial of T. Proof. Choose an ordered basis ? = {v1,v2,...,vk} for W, and extend it to an ordered basis ß = {v1,v2,...,vk,vk+1,...,vn} for V. Let A = [T]ß and B1 = [TW]?. Observe that A can be written in the form Let f (t) be the characteristic polynomial of T and g(t) the characteristic polynomial of TW . Then O B3 ??B-tI B ?? f(t)=det(A-tIn)=det 1 k 2 ??B B?? A=12. O B3 - tIn-k The following was presented by N.Vankayalapati. He provided a handout. by the Lemma. Thus g(t) divides f(t). Defn 3. Let T be a linear operator on a vector space V and let x be a nonzero vector in V . The subspace W = span({x,T(x),T2(x),...}) is called the T-cyclic subspace of V generated by x. Theorem 5.22. Let T be a linear operator on a finite-dimensional vector space V , and let W denote the T -cyclic subspace of V generated by a nonzero vector v ? V . Let k = dim(W ). Then (a) {v,T(v),T2(v),...,ak-1Tk-1(v)Tk(v)} is a basis for W. (b) If a0v + a1T (v) + · · · + T k-1(v) = 0, then the characteristic polynomial of TW is f(t)=(-1)k(a0 +a1t+···+ak-1tk-1 +tk). The following was presented by J. Stockford. He provided a handout. Theorem 5.23. (Cayley-Hamilton). Let T be a linear operator on a finite-dimensional vector space V , and let f(t) be the characteristic polynomial of T. Then f(T) = T0, the zero transformation. The following was presented by Q. Ding. He provided a handout. Cor 1. (Cayley-Hamilton Theorem for Matrices). Let A be an n × n matrix, and let f (t) be the characteristic polynomial of A. Then f(A) = O, the zero matrix. We did not cover the following theorems. Theorem 5.24. Let T be a linear operator on a finite-dimensional vector space V , and suppose that V = W1 ? W2 ? ··· ? Wk, where Wi is a T-invariant subspace of V for each i (1 = i = k). Suppose that fi(t) is the characteristic polynomial of TWi (1 = i = k). Then f (t)f ? (t)· · ? ·f ? (t) is the characteristic polynomial of T . 12k =g(t)·det(B3 -tIn-k) 9 10 Defn 4. Let B1 ? Mm(F), and let B2 ? Mn(F). We define the direct sum of B1 and B2, denoted B1 ?B2 as the (m+n)×(m+n) matrix A such that ?? (B1)i,j Ai,j =?(B2)(i-m),(j-m) 0 for 1 = i,j = m form+1=i,j=n+m otherwise If B1,B2,...,Bk are square matrices with entries from F, then we define the direct sum of B1,B2 ...,Bk recursively by B1 ?B2 ?···?Bk =(B1 ?B2 ?···?Bk-1)?Bk Of A=B1 ?B2 ?···?Bk, then we often write ?B1 O···O? A=?O B2 ··· O? ? . . . ? O O ··· Bk Theorem 5.25. Let T be a linear operator on a finite-dimensional vector space V , an d letW1,W2,...,Wk beT invariantsubspacesofV suchthatV =W1?W2?···?Wk. For each i, let ßi be an ordered basis for Wi, and let ß = ß1 ?ß2 ?···?ßk. Let A = [T]ß and Bi=[TWi]ßi foreachi.ThenA=B1?B2?···?Bk.

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