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Linear Algebra (Honors)

by: Kaylin Wehner

Linear Algebra (Honors) MATH 115AH

Kaylin Wehner
GPA 3.55


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This 4 page Class Notes was uploaded by Kaylin Wehner on Friday September 4, 2015. The Class Notes belongs to MATH 115AH at University of California - Los Angeles taught by Staff in Fall. Since its upload, it has received 32 views. For similar materials see /class/177813/math-115ah-university-of-california-los-angeles in Mathematics (M) at University of California - Los Angeles.


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Date Created: 09/04/15
MATH 115AH INNER PRODUCT SPACES AND THE SPECTRAL THEOREM NB These notes are not really intended as a substitute for the text they should be viewed as a supplement They do not address general bilinear forms at all rather7 llve tried to take the most direct path to spectral theory and so have only treated inner products Some of this material is not covered in Artin7s book I know you re swamped with work already but thinking a little about the exercises might not be such a bad idea 1 Inner Product Spaces An inner product on a real vector space V is just a positive de nite symmetric bilinear form ltgtVgtltVgtRi The standard example you should keep in mind is the usual dot product on the Euclidean space R lt117 H71n7y17 7yngt 11y1 Inyn An inner product on a complex vector space V is a positive de nite hermitiansymmetric sesquilinear form ltgtVgtltVgtC This means lt satis es 0 Sesquilinearity lt is linear in the rst variable lt139U1 A2v27 70gt A1ltv17 10gt A2ltv27 70gt and conjugatelinear in the second variable ltU7 lwl A2w2gt KW w1gt MW 102 o Hermitian symmetry lt1 wgt ltw7vgti 0 Positive de niteness lt1 11gt 2 0 for all u E V with equality if and only if v 0 An inner product space is a vector space V over R or C equipped with an inner product People sometimes say hermitian inner product and hermitian inner product space to single out the complex case The standard inner product on C is lt21 i zn 101 21131 i i znwni An inner product space carries a natural notion of size namely7 the norm of v E V is given by HUN 1 ltv7vgtA You should check that this corresponds to the usual notion of length in R if lt is the dot product mentioned above NB Norm is a general concept which I won7t de ne here not every norm arises from an inner product II Orthogonality Two vectors 1 w in an inner product space V are orthogonal if 1 wgt 0 Given a subspace W C V7 we de ne the orthogonal complement WL y Wiv6Vlltvwgt0VwEWi So Wi is the set of all vectors in V that are orthogonal to everything in Wi Exercise 1 Wi is a subspace of V Exercise 2 V W 69 Wi This last exercise tells us that every 1 E V can be written uniquely as v w u7 where w E W and u 6 Wii So we can de ne the orthogonal projection 7r V a W by 7rv wi 7r is a surjective linear transformation with ker7r Wk Thought Exercise Just so you donlt forget your group theory WT is the usual projection homo morphism from V to the quotient group VWii III Orthonormal Bases A asis R v17i i i 7vn of an inner product space V is called orthonormal if ltvi7vj 6177 where tilj is the Kronecker deltai What this means is that the vjls all have norm length 1 and are pairwise orthogonali The standard example you should keep in mind is the standard basis e17iu7en 0 R73 The following result is crucial Theorem 1 GramSchmidt Given a basis v17 i i i 7vn of an inner product space V there exists an orthonormal basis 1717 i i 7 7377 such that Spani17 i i i 7ij Spanv17i i i 7 vj for all i S j S n Know the proof of this at all costs Its pretty geometrically intuitive think about orthogonal com plementsi IV Adjoints et al The Matrix Viewpoint Recall that a real matrix A 6 MAR is symmetiic if A A i It turns out that to de ne a useful analogue for complex matrices7 the transpose is inadequate We introduce Aquot A 7 the adjoint ofiA E The adjoint obeys most of the same properties of the transpose7 except now AA AA for A E C A matrix A 6 MAC is called selfadjoint also hermitian7 hermitian symmetiic if A A Note that selfadjoint means the same thing as symmetric when we re talking about real matricesi A real matrix A 6 MAR is orthogonal if At A li For the complex analogue7 A 6 MAC is unitary if A A li In each case7 of course7 we re implicitly requiring that A is invertible Once again7 unitary and orthogonal mean the same thing when we7re talking about real matricesi V Adjoints et al The Operator Viewpoint Take V to be a nitedimensional inner product space real or complex Given an operator T E EndV7 we de ne the adjoint of T to be the unique operator T E EndV that satis es Tv7w v7Tw V v7w E V You should be worried at this point Why does T exist7 and why is it unique The following exercises will give you the answer to this7 and they will also give the connection to the matrix viewpoint above I suppose this is optional material Artin doesn7t even bother to de ne what an adjoint operator is You can skip ahead to the de nition of a selfadjointhermitian operator if youlre feeling lazy Exercise 3 In the special case where V R or V C and is the standard inner product7 verify by computation that LA LAJ Exercise 4 Now take V to be a general inner product space over K R or C7 and let R v17 i i i 7 on be an orthonormal basis of Vi Let e17iu7en be the standard basis of Kni Show that the map V a K given by vj ej is an isometric isomorphism7 iiei7 an isomorphism that satis es ltv7wgtltlt1gtv7 39wgt7 where the inner product on the right hand side is the standard inner product on Kni Exercise 5 Draw an appropriate commutative diagram and think about the previous two exercises while staring at it If A is the matrix of T with respect to R then welll get T 1 o Ly o in Now we make the natural de nitions An operator T E EndV is selfadjoint Artin says hermitian ifT T ie if Tvwgt vTugt V 1110 6 V T E EndV is unitary if T is invertible with T T li Exercise 6 T is unitary if and only if it preserves the inner product TvTwgt vwgt for all 1110 6 Vi Artin takes this to be the de nition of a unitary operator The following theorem is a consequence of the rst three exercises in this section Theorem 2 Let V be a nitedimensional complex resp real inner product space T E EndV and A the matrix ofT with respect to an orthonormal basis Then A is selfadjoint resp symmetric and only ifT is selfadjoint and A is unitary resp orthogonal and only if T is unitary Finally do this Exercise 7 Show that if T is a selfadjoint operator then the eigenvalues of T are real Hint Look at Tvvgt for suitable vi OK here it comes i i i VI The Spectral Theorem for SelfAdjoint Operators Complex Case Theorem 3 SPEC C Let V be a nitedimensional complex inner product space and let T E EndV be a selfadjoint operator Then V has an orthonormal basis consisting of eigenvectors of T By Theorem 2 above and Exercise 7 this is equivalent to Theorem 4 SPEC C Matrix form Let A 6 Mn C be a selfadjoint matrix Then A is unitarily diagonalizalzlequot ie there is a unitary matrix P E GLnC such that PAP PAP 1 is diagonal with real entries Before moving on to the proof lets say a couple words about why this is so great We re making a pretty strong assumption viz that our operator is selfadjointi But the only diagonalization result we had before required T to have n distinct eigenvalues this is no longer required with the Spectral Theoremi Moreover we donlt just get any old basis of eigenvectors we get an orthonormal one Of course keep in mind that we can t do any of this without an inner product structure on our space or thonormal77 wouldnlt mean anything without one Proof of Theorem 3 Say dimV ni By the Fundamental Theorem of Algebra the charac teristic polynomial of T has a root A E C by Exercise 7 A is actually reali Let 1 be a A eigenvector for T Replacing v with HziH if necessary we can assume 1 Let W Spanvi then dimW n 7 1 and V Spanv EB Wi If we can show W is Tinvariant well be done by induction on n dimV as follows The theorem is trivial when n 1 Suppose we know the theorem for all vector spaces of dimension n 7 1 then in particular it holds for Wi So if W is Tinvariant we can view the restriction of T as an element of EndW and we can nd an orthonormal basis vb i i i 11771 for W consisting of eigenvectors of i Then 1 v1 i i i 11771 is an orthonormal basis of V consisting of eigenvectors o i So now we just show W is Tinvariant ie we show that Tu E W for all w E W which is equivalent to wEWgt Twvgt Oi But this is a small matter w E W gt ltwvgt 0 gt ltTwvgt ltwTvgt 5xltwvgt 0 since T is selfadjointl D VII The Spectral Theorem for SelfAdjoint Operators Real Case Theorem 5 SPEC R Let V be a nitedimensional real inner product space T E EndV a selfadjoint operator Then V has an orthonormal basis of eigenvectors of T Exercise 8 Formulate and prove the matrix version of the real Spectral Theoreml Proof of Theorem 5 The only possible dif culty in copying the proof of Theorem 3 would be in nding an eigenvector to start with But we can be cheap about this By GramSchmidt choose some orthonormal basis of V and let A 6 Mn R be the matrix ofT with respect to this basis By Theorem 2 A is a symmetric matrix So LA is a selfadjoint operator on C cfl Exercise 3 LA has real eigenvalues by Exercise 7 but these are just the eigenvalues of A and hence those of T as well So we7ve proved the existence of a real eigenvalue and hence the existence of an eigenvector The rest of the proof follows exactly as above D VIII The Spectral Theorem for Normal Operators To wrap up welll present the beall endall result of nitedimensional spectral theoryl Unless the professor tells you otherwise you aren t responsible for knowing it Let V be an inner product spacer An operator T E EndV is normal if it commutes with its adjoint ie T Tquot T Tl NB Selfadjoint operators are normal Theorem 6 The Spectral Theorem Let V be a nitedimensional inner product space T E EndV a normal operator Then V has an orthonormal basis of eigenvectors of T Proof The main idea of the proof is the same as in the selfadjoint case nd an eigenvector show the orthogonal complement is Tinvariant and apply induction on n dimVl Doing the rst two things is considerably harder in this case however First et R TT S 7T7T Then R and S are selfadjoint and we have T R iSl Since TTquot T T we see that RS SRl Now R has an eigenvector as usual by the Fundamental Theorem of Algebra say Rv Au I claim the eigenspace ERA kerR 7 AI is S invariant z e ERA gt R 7 ANSI 5R7 Am 50 0 gt Sz 6 ER So we can view 5 as an element of EndER and hence S has an eigenvector z in ER But this means I is a simultaneous eigenvector for R and 5 say RI Ar and Sr uzl Moreover I is an eigenvector for T Tz R1 iSz A iuzl Without loss of generality 1 now all we have to do is show that SpanzL is Tinvariantl To this end note that Tquot RiS Rquot Rquot 7 i5


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