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Lecture 19: Polar and singular valuedecompositions; generalized eigenspaces; the

decomposition theorem (1)

Travis Schedler

Thurs, Nov 17, 2011 (version: Thurs, Nov 17, 1:00 PM)

Goals (2)

I Polar decomposition and singular value decomposition

I Generalized eigenspaces and the decomposition theorem

Read Chapter 7, begin Chapter 8, and do PS 9.

Warm-up exercise (3)

(a) Let T be an invertible operator on a f.d. i.p.s. and setP :=

√T ∗T and S := TP−1. Show that S is an isometry.

Recall P is positive, so

T = SP

is a polar decomposition (i.e., S is an isometry and Ppositive).

(b) Now suppose T = 0. Show that polar decompositionsT = SP are exactly

T = S0

for every isometry S , i.e., we have always P = 0 but S can beanything.

One-dimensional analogue: Either z ∈ C is invertible, in which casez = (z/|z |)|z | = sp or else z is zero, in which case z = s · 0 for anys of absolute value one.

Solution to warm-up exercise (4)

(a) S∗S = (TP−1)∗TP−1 = (P−1)∗T ∗TP−1 = P−1P2P−1 = I .Here we used that P∗ = P and hence (P−1)∗ = P−1 as well.

(b) Since isometries are invertible, 0 = SP for S an isometryimplies P = S−10 = 0. On the other hand clearly S0 = 0 forall S .

Polar decomposition and SVD (5)Proposition: every complex number z is expressible as z = r · e iθ,where r ≥ 0 and θ ∈ [0, 2π). (Unique if z nonzero).

Equivalently: z = s · r , for |s| = 1 and r = |z | =√

z · z ≥ 0.

TheoremLet V be a f.d. i.p.s. and T ∈ L(V ). Then there is an expressionT = SP, for S an isometry and P positive.P is unique and P =

√T ∗T .

Moreover, S is unique iff T is invertible.

Corollary (Singular Value Decomposition (SVD))

There exists orthonormal bases (e1, . . . , en) and (f1, . . . , fn) of Vsuch that Tei = si fi , for si ≥ 0 the singular values.Moreover, (e1, . . . , en) is an orthonormal eigenbasis of T ∗T witheigenvalues s2i .

Proof: Let (e1, . . . , en) be an orthonormal eigenbasis of T ∗T ands1, . . . , sn the square roots of the eigenvalues. When si 6= 0, setfi := s−1i Tei . Then extend the resulting fi to an orthonormaleigenbasis.

Polar decomposition and SVD (5)Proposition: every complex number z is expressible as z = r · e iθ,where r ≥ 0 and θ ∈ [0, 2π). (Unique if z nonzero).Equivalently: z = s · r , for |s| = 1 and r = |z | =

√z · z ≥ 0.

TheoremLet V be a f.d. i.p.s. and T ∈ L(V ). Then there is an expressionT = SP, for S an isometry and P positive.P is unique and P =

√T ∗T .

Moreover, S is unique iff T is invertible.

Corollary (Singular Value Decomposition (SVD))

There exists orthonormal bases (e1, . . . , en) and (f1, . . . , fn) of Vsuch that Tei = si fi , for si ≥ 0 the singular values.Moreover, (e1, . . . , en) is an orthonormal eigenbasis of T ∗T witheigenvalues s2i .

Proof: Let (e1, . . . , en) be an orthonormal eigenbasis of T ∗T ands1, . . . , sn the square roots of the eigenvalues. When si 6= 0, setfi := s−1i Tei . Then extend the resulting fi to an orthonormaleigenbasis.

Polar decomposition and SVD (5)Proposition: every complex number z is expressible as z = r · e iθ,where r ≥ 0 and θ ∈ [0, 2π). (Unique if z nonzero).Equivalently: z = s · r , for |s| = 1 and r = |z | =

√z · z ≥ 0.

TheoremLet V be a f.d. i.p.s. and T ∈ L(V ). Then there is an expressionT = SP, for S an isometry and P positive.P is unique and P =

√T ∗T .

Moreover, S is unique iff T is invertible.

Corollary (Singular Value Decomposition (SVD))

There exists orthonormal bases (e1, . . . , en) and (f1, . . . , fn) of Vsuch that Tei = si fi , for si ≥ 0 the singular values.Moreover, (e1, . . . , en) is an orthonormal eigenbasis of T ∗T witheigenvalues s2i .

Proof: Let (e1, . . . , en) be an orthonormal eigenbasis of T ∗T ands1, . . . , sn the square roots of the eigenvalues. When si 6= 0, setfi := s−1i Tei . Then extend the resulting fi to an orthonormaleigenbasis.

Polar decomposition and SVD (5)Proposition: every complex number z is expressible as z = r · e iθ,where r ≥ 0 and θ ∈ [0, 2π). (Unique if z nonzero).Equivalently: z = s · r , for |s| = 1 and r = |z | =

√z · z ≥ 0.

√T ∗T .

Moreover, S is unique iff T is invertible.

Corollary (Singular Value Decomposition (SVD))

√z · z ≥ 0.

√T ∗T .

Moreover, S is unique iff T is invertible.

Corollary (Singular Value Decomposition (SVD))

Uniqueness of polar decomposition (6)

I If T = SP, then T ∗T = P∗S∗SP = P∗P = P2, soP =

√T ∗T . Thus P is unique (positive operators have unique

positive square roots; see the slides for Lecture 18 or Axler).

I If T is invertible, S = TP−1 so S is unique.

I Conversely, if T is not invertible, neither is P, and we canreplace S by SS ′ where S ′ is an isometry such that S ′v = vfor all eigenvectors v of nonzero eigenvalue. So then S is notunique.

Existence of polar decomposition (7)

I Set P :=√

T ∗T .

I range(P) is P-invariant and P is an isomorphism there (it hasan eigenbasis with nonzero eigenvalues).Define thus P|−1range(P) : range(P)→ range(P).

Consider S1 := TP|−1range(P) : range(P)→ range(T ).

S∗1S1 = I , so 〈u, v〉 = 〈S1u, S1v〉 for all u, v ∈ range(P).

I Recall: null(P) = null(T ). So dim range(P) = dim range(T ).Thus S1 takes an on. basis (e1, . . . , em) of range(P) to anon. basis (f1, . . . , fm) of range(T ).

I Extend (ei ) and (fi ) to on. bases of V and extend S1 toS ∈ L(V ) by S(ei ) = fi when i > m.

I So S takes an on. basis to another on. basis, i.e., it is anisometry.

I Finally, T = SP, since it is true on range(P) andnull(T ) = null(P) = range(P)⊥ = Span(em+1, . . . , en).

Nonuniqueness of SVD (8)

I Note: the SVD is not unique, even if T is invertible: theorthonormal eigenbasis (ei ) of T ∗T is not unique. (e.g., onecan reorder them and multiply by ±1, at the least.)

I On the other hand, the polar decomposition is unique iff T isinvertible.

I Example: T = TA, A =

(0 −21 0

).

I We can guess that A = SP =

(0 −11 0

)(1 00 2

). So this is

the answer (unique since A, equivalently P, is invertible).

I For SVD we could have e1 =

(10

), e2 =

(01

), s1 = 1, s2 = 2,

f1 =

(01

), f2 =

(−10

).

I But we could also swap everything: s1 with s2, e1 with e2, andf1 with f2. Or we could take e1 to −e1 (hence f1 to −f1)and/or e2 to −e2 (hence f2 to −f2).

Nonuniqueness of SVD (8)

I Note: the SVD is not unique, even if T is invertible: theorthonormal eigenbasis (ei ) of T ∗T is not unique. (e.g., onecan reorder them and multiply by ±1, at the least.)

I On the other hand, the polar decomposition is unique iff T isinvertible.

I Example: T = TA, A =

(0 −21 0

).

I We can guess that A = SP =

(0 −11 0

)(1 00 2

). So this is

the answer (unique since A, equivalently P, is invertible).

I For SVD we could have e1 =

(10

), e2 =

(01

), s1 = 1, s2 = 2,

f1 =

(01

), f2 =

(−10

).

I But we could also swap everything: s1 with s2, e1 with e2, andf1 with f2. Or we could take e1 to −e1 (hence f1 to −f1)and/or e2 to −e2 (hence f2 to −f2).

Nonuniqueness of SVD (8)

I Note: the SVD is not unique, even if T is invertible: theorthonormal eigenbasis (ei ) of T ∗T is not unique. (e.g., onecan reorder them and multiply by ±1, at the least.)

I On the other hand, the polar decomposition is unique iff T isinvertible.

I Example: T = TA, A =

(0 −21 0

).

I We can guess that A = SP =

(0 −11 0

)(1 00 2

). So this is

the answer (unique since A, equivalently P, is invertible).

I For SVD we could have e1 =

(10

), e2 =

(01

), s1 = 1, s2 = 2,

f1 =

(01

), f2 =

(−10

).

I But we could also swap everything: s1 with s2, e1 with e2, andf1 with f2. Or we could take e1 to −e1 (hence f1 to −f1)and/or e2 to −e2 (hence f2 to −f2).

Nonuniqueness of SVD (8)

I On the other hand, the polar decomposition is unique iff T isinvertible.

I Example: T = TA, A =

(0 −21 0

).

I We can guess that A = SP =

(0 −11 0

)(1 00 2

). So this is

the answer (unique since A, equivalently P, is invertible).

I For SVD we could have e1 =

(10

), e2 =

(01

), s1 = 1, s2 = 2,

f1 =

(01

), f2 =

(−10

).

Nonuniqueness of SVD (8)

I On the other hand, the polar decomposition is unique iff T isinvertible.

I Example: T = TA, A =

(0 −21 0

).

I We can guess that A = SP =

(0 −11 0

)(1 00 2

). So this is

the answer (unique since A, equivalently P, is invertible).

I For SVD we could have e1 =

(10

), e2 =

(01

), s1 = 1, s2 = 2,

f1 =

(01

), f2 =

(−10

).

Nonuniqueness of SVD (8)

I On the other hand, the polar decomposition is unique iff T isinvertible.

I Example: T = TA, A =

(0 −21 0

).

I We can guess that A = SP =

(0 −11 0

)(1 00 2

). So this is

the answer (unique since A, equivalently P, is invertible).

I For SVD we could have e1 =

(10

), e2 =

(01

), s1 = 1, s2 = 2,

f1 =

(01

), f2 =

(−10

).

Computing SVD and polar decomposition (9)

I The best way to compute these is to do SVD first; then let Pbe the operator with eigenvectors (ei ) and eigenvalues (si ),and let S be the isometry Sei = fi for all i .

I To compute SVD, given T , compute first T ∗T .

I Then find the eigenvalues of T ∗T (2× 2 case: characteristicpolynomial: for A =M(T ) in an orthonormal basis, these arethe roots of x2 − tr(AtA)x + det(AtA).)

I Find the eigenspaces and an on. eigenbasis (ei ) of T ∗T .

I Next, set P :=√

T ∗T , by taking the nonnegative square rootof the eigenvalues. These eigenvalues are the si .

I Finally, let fi := s−1i Tei for the nonzero si ; for the remaining fijust extend the ones we get to an orthonormal basis.

Computing SVD and polar decomposition (9)

I The best way to compute these is to do SVD first; then let Pbe the operator with eigenvectors (ei ) and eigenvalues (si ),and let S be the isometry Sei = fi for all i .

I To compute SVD, given T , compute first T ∗T .

I Then find the eigenvalues of T ∗T (2× 2 case: characteristicpolynomial: for A =M(T ) in an orthonormal basis, these arethe roots of x2 − tr(AtA)x + det(AtA).)

I Find the eigenspaces and an on. eigenbasis (ei ) of T ∗T .

I Next, set P :=√

T ∗T , by taking the nonnegative square rootof the eigenvalues. These eigenvalues are the si .

I Finally, let fi := s−1i Tei for the nonzero si ; for the remaining fijust extend the ones we get to an orthonormal basis.

Computing SVD and polar decomposition (9)

I The best way to compute these is to do SVD first; then let Pbe the operator with eigenvectors (ei ) and eigenvalues (si ),and let S be the isometry Sei = fi for all i .

I To compute SVD, given T , compute first T ∗T .

I Then find the eigenvalues of T ∗T (2× 2 case: characteristicpolynomial: for A =M(T ) in an orthonormal basis, these arethe roots of x2 − tr(AtA)x + det(AtA).)

I Find the eigenspaces and an on. eigenbasis (ei ) of T ∗T .

I Next, set P :=√

T ∗T , by taking the nonnegative square rootof the eigenvalues. These eigenvalues are the si .

I Finally, let fi := s−1i Tei for the nonzero si ; for the remaining fijust extend the ones we get to an orthonormal basis.

Computing SVD and polar decomposition (9)

I To compute SVD, given T , compute first T ∗T .

I Find the eigenspaces and an on. eigenbasis (ei ) of T ∗T .

I Next, set P :=√

T ∗T , by taking the nonnegative square rootof the eigenvalues. These eigenvalues are the si .

Computing SVD and polar decomposition (9)

I To compute SVD, given T , compute first T ∗T .

I Find the eigenspaces and an on. eigenbasis (ei ) of T ∗T .

I Next, set P :=√

T ∗T , by taking the nonnegative square rootof the eigenvalues. These eigenvalues are the si .

Computing SVD and polar decomposition (9)

I To compute SVD, given T , compute first T ∗T .

I Find the eigenspaces and an on. eigenbasis (ei ) of T ∗T .

I Next, set P :=√

T ∗T , by taking the nonnegative square rootof the eigenvalues. These eigenvalues are the si .

Example (10)

I Example from before: A =

(0 −21 0

).

I First, AtA =

(1 00 4

).

I Next, an eigenbasis is e1 =

(10

)and e2 =

(01

)with

eigenvalues 1 and 4.I So P =

√AtA has the same eigenbasis, with eigenvalues

s1 =√

1 = 1 and s2 =√

4 = 2.

I Then f1 = s−11 Ae1 = 1−1(

01

)=

(01

). Also

f2 = s−12 Ae2 = 2−1(−20

)=

(−10

).

I Now P =

(1 00 2

)and S =

(0 −11 0

), as desired.

I In general: P = (e1e2)

(s1 00 s2

)(e1e2)−1 and

S = (f1f2)(e1e2)−1.

Example (10)

I Example from before: A =

(0 −21 0

).

I First, AtA =

(1 00 4

).

I Next, an eigenbasis is e1 =

(10

)and e2 =

(01

)with

eigenvalues 1 and 4.I So P =

√AtA has the same eigenbasis, with eigenvalues

s1 =√

1 = 1 and s2 =√

4 = 2.

I Then f1 = s−11 Ae1 = 1−1(

01

)=

(01

). Also

f2 = s−12 Ae2 = 2−1(−20

)=

(−10

).

I Now P =

(1 00 2

)and S =

(0 −11 0

), as desired.

I In general: P = (e1e2)

(s1 00 s2

)(e1e2)−1 and

S = (f1f2)(e1e2)−1.

Example (10)

I Example from before: A =

(0 −21 0

).

I First, AtA =

(1 00 4

).

I Next, an eigenbasis is e1 =

(10

)and e2 =

(01

)with

eigenvalues 1 and 4.

I So P =√

AtA has the same eigenbasis, with eigenvaluess1 =

√1 = 1 and s2 =

√4 = 2.

I Then f1 = s−11 Ae1 = 1−1(

01

)=

(01

). Also

f2 = s−12 Ae2 = 2−1(−20

)=

(−10

).

I Now P =

(1 00 2

)and S =

(0 −11 0

), as desired.

I In general: P = (e1e2)

(s1 00 s2

)(e1e2)−1 and

S = (f1f2)(e1e2)−1.

Example (10)

I Example from before: A =

(0 −21 0

).

I First, AtA =

(1 00 4

).

I Next, an eigenbasis is e1 =

(10

)and e2 =

(01

)with

eigenvalues 1 and 4.I So P =

√AtA has the same eigenbasis, with eigenvalues

s1 =√

1 = 1 and s2 =√

4 = 2.

I Then f1 = s−11 Ae1 = 1−1(

01

)=

(01

). Also

f2 = s−12 Ae2 = 2−1(−20

)=

(−10

).

I Now P =

(1 00 2

)and S =

(0 −11 0

), as desired.

I In general: P = (e1e2)

(s1 00 s2

)(e1e2)−1 and

S = (f1f2)(e1e2)−1.

Example (10)

I Example from before: A =

(0 −21 0

).

I First, AtA =

(1 00 4

).

I Next, an eigenbasis is e1 =

(10

)and e2 =

(01

)with

eigenvalues 1 and 4.I So P =

√AtA has the same eigenbasis, with eigenvalues

s1 =√

1 = 1 and s2 =√

4 = 2.

I Then f1 = s−11 Ae1 = 1−1(

01

)=

(01

). Also

f2 = s−12 Ae2 = 2−1(−20

)=

(−10

).

I Now P =

(1 00 2

)and S =

(0 −11 0

), as desired.

I In general: P = (e1e2)

(s1 00 s2

)(e1e2)−1 and

S = (f1f2)(e1e2)−1.

Example (10)

I Example from before: A =

(0 −21 0

).

I First, AtA =

(1 00 4

).

I Next, an eigenbasis is e1 =

(10

)and e2 =

(01

)with

eigenvalues 1 and 4.I So P =

√AtA has the same eigenbasis, with eigenvalues

s1 =√

1 = 1 and s2 =√

4 = 2.

I Then f1 = s−11 Ae1 = 1−1(

01

)=

(01

). Also

f2 = s−12 Ae2 = 2−1(−20

)=

(−10

).

I Now P =

(1 00 2

)and S =

(0 −11 0

), as desired.

I In general: P = (e1e2)

(s1 00 s2

)(e1e2)−1 and

S = (f1f2)(e1e2)−1.

Example (10)

I Example from before: A =

(0 −21 0

).

I First, AtA =

(1 00 4

).

I Next, an eigenbasis is e1 =

(10

)and e2 =

(01

)with

eigenvalues 1 and 4.I So P =

√AtA has the same eigenbasis, with eigenvalues

s1 =√

1 = 1 and s2 =√

4 = 2.

I Then f1 = s−11 Ae1 = 1−1(

01

)=

(01

). Also

f2 = s−12 Ae2 = 2−1(−20

)=

(−10

).

I Now P =

(1 00 2

)and S =

(0 −11 0

), as desired.

I In general: P = (e1e2)

(s1 00 s2

)(e1e2)−1 and

S = (f1f2)(e1e2)−1.

Generalized eigenvectors (11)

Goal: Although not all f.d. vector spaces admit an eigenbasis, overF = C they always admit a basis of generalized eigenvectors.

DefinitionA generalized eigenvector v of T eigenvalue λ is one such that, forsome m ≥ 1, (T − λI )mv = 0.

Examples:

I m = 1 above if and only if v is an (ordinary) eigenvector.

I If T is nilpotent, then all vectors are generalized eigenvectorsof eigenvalue zero. So, even though it does not have aneigenbasis, every basis is a basis of generalized eigenvectors!

DefinitionLet V (λ) be the generalized eigenspace of eigenvalue λ: the spanof all generalized eigenvectors of eigenvalue λ.

Note that V (λ) is T -invariant, since (T − λI )mv = 0 implies(T − λI )mTv = T (T − λI )mv = 0.

Generalized eigenvectors (11)

Goal: Although not all f.d. vector spaces admit an eigenbasis, overF = C they always admit a basis of generalized eigenvectors.

DefinitionA generalized eigenvector v of T eigenvalue λ is one such that, forsome m ≥ 1, (T − λI )mv = 0.

Examples:

I m = 1 above if and only if v is an (ordinary) eigenvector.

I If T is nilpotent, then all vectors are generalized eigenvectorsof eigenvalue zero. So, even though it does not have aneigenbasis, every basis is a basis of generalized eigenvectors!

DefinitionLet V (λ) be the generalized eigenspace of eigenvalue λ: the spanof all generalized eigenvectors of eigenvalue λ.

Note that V (λ) is T -invariant, since (T − λI )mv = 0 implies(T − λI )mTv = T (T − λI )mv = 0.

Generalized eigenvectors (11)

Goal: Although not all f.d. vector spaces admit an eigenbasis, overF = C they always admit a basis of generalized eigenvectors.

DefinitionA generalized eigenvector v of T eigenvalue λ is one such that, forsome m ≥ 1, (T − λI )mv = 0.

Examples:

I m = 1 above if and only if v is an (ordinary) eigenvector.

I If T is nilpotent, then all vectors are generalized eigenvectorsof eigenvalue zero. So, even though it does not have aneigenbasis, every basis is a basis of generalized eigenvectors!

DefinitionLet V (λ) be the generalized eigenspace of eigenvalue λ: the spanof all generalized eigenvectors of eigenvalue λ.

Note that V (λ) is T -invariant, since (T − λI )mv = 0 implies(T − λI )mTv = T (T − λI )mv = 0.

Generalized eigenvectors (11)

Examples:

I m = 1 above if and only if v is an (ordinary) eigenvector.

Note that V (λ) is T -invariant, since (T − λI )mv = 0 implies(T − λI )mTv = T (T − λI )mv = 0.

Generalized eigenvectors (11)

Examples:

I m = 1 above if and only if v is an (ordinary) eigenvector.

Note that V (λ) is T -invariant, since (T − λI )mv = 0 implies(T − λI )mTv = T (T − λI )mv = 0.

Generalized eigenvectors (11)

Examples:

I m = 1 above if and only if v is an (ordinary) eigenvector.

Note that V (λ) is T -invariant, since (T − λI )mv = 0 implies(T − λI )mTv = T (T − λI )mv = 0.

Generalized eigenvectors (11)

Examples:

I m = 1 above if and only if v is an (ordinary) eigenvector.

Note that V (λ) is T -invariant, since (T − λI )mv = 0 implies(T − λI )mTv = T (T − λI )mv = 0.

The decomposition theorem (12)

TheoremLet V be f.d., F = C, and T ∈ L(V ). Then V is the direct sum ofits generalized eigenspaces: V =

⊕λ V (λ).

First step:

LemmaSuppose that λ 6= µ. Then V (λ) ∩ V (µ) = {0}.

Proof.

I Suppose v ∈ V (λ) ∩ V (µ) is nonzero. Let (T − λI )mv = 0but not (T − λI )m−1v . So u := (T − λI )m−1v is a nonzeroeigenvector of eigenvalue λ.

I Now let m′ ≥ 1 be such that (T − µI )m′v = 0. Then also

(T − µI )m′u = (T − µI )m

′(T − λI )m−1v =

(T − λI )m−1(T − µI )m′v = 0.

I But, (T − µI )m′u = (λ− µ)m

′u 6= 0, since λ 6= µ. This is a

contradiction.

The decomposition theorem (12)

TheoremLet V be f.d., F = C, and T ∈ L(V ). Then V is the direct sum ofits generalized eigenspaces: V =

⊕λ V (λ).

First step:

LemmaSuppose that λ 6= µ. Then V (λ) ∩ V (µ) = {0}.

Proof.

I Suppose v ∈ V (λ) ∩ V (µ) is nonzero. Let (T − λI )mv = 0but not (T − λI )m−1v . So u := (T − λI )m−1v is a nonzeroeigenvector of eigenvalue λ.

I Now let m′ ≥ 1 be such that (T − µI )m′v = 0. Then also

(T − µI )m′u = (T − µI )m

′(T − λI )m−1v =

(T − λI )m−1(T − µI )m′v = 0.

I But, (T − µI )m′u = (λ− µ)m

′u 6= 0, since λ 6= µ. This is a

contradiction.

The decomposition theorem (12)

TheoremLet V be f.d., F = C, and T ∈ L(V ). Then V is the direct sum ofits generalized eigenspaces: V =

⊕λ V (λ).

First step:

LemmaSuppose that λ 6= µ. Then V (λ) ∩ V (µ) = {0}.

Proof.

I Suppose v ∈ V (λ) ∩ V (µ) is nonzero. Let (T − λI )mv = 0but not (T − λI )m−1v . So u := (T − λI )m−1v is a nonzeroeigenvector of eigenvalue λ.

I Now let m′ ≥ 1 be such that (T − µI )m′v = 0. Then also

(T − µI )m′u = (T − µI )m

′(T − λI )m−1v =

(T − λI )m−1(T − µI )m′v = 0.

I But, (T − µI )m′u = (λ− µ)m

′u 6= 0, since λ 6= µ. This is a

contradiction.

Lemma on generalized eigenspaces (13)

LemmaV (λ) = null(T − λI )dimV .

I.e., if v is a generalized eigenvector of eigenvalue λ, we can takem = dim V before: (T − λI )dimV v = 0.

Proof.

I Let Ui := (T − λI )iV (λ).

I Since V (λ) is T -invariant (hence (T − λI )-invariant),U0 ⊇ U1 ⊇ · · · .

I However, if Ui = Ui+1, then (T − λI ) is injective on Ui . Since(T − λI ) is nilpotent, this implies Ui = {0}.

I So U0 ) U1 ) · · · ) Um = {0}, and dim Ui ≤ dim V (λ)− i .Hence m ≤ dim V (λ) ≤ dim V , and (T − λI )dimV v = 0 forall v ∈ V (λ).

Lemma on generalized eigenspaces (13)

LemmaV (λ) = null(T − λI )dimV .

I.e., if v is a generalized eigenvector of eigenvalue λ, we can takem = dim V before: (T − λI )dimV v = 0.

Proof.

I Let Ui := (T − λI )iV (λ).

I Since V (λ) is T -invariant (hence (T − λI )-invariant),U0 ⊇ U1 ⊇ · · · .

I However, if Ui = Ui+1, then (T − λI ) is injective on Ui . Since(T − λI ) is nilpotent, this implies Ui = {0}.

I So U0 ) U1 ) · · · ) Um = {0}, and dim Ui ≤ dim V (λ)− i .Hence m ≤ dim V (λ) ≤ dim V , and (T − λI )dimV v = 0 forall v ∈ V (λ).

One more lemma (14)

LemmaV = (T − λI )dimV ⊕ range(T − λI )dimV =V (λ)⊕ range(T − λI )dimV .

Proof.

I Since the dimensions are equal, we need to show just that(T − λI )dimV ∩ range(T − λI )dimV = {0}.

I Let v ∈ (T − λI )dimV ∩ range(T − λI )dimV . Writev = (T − λI )dimV u.

I Since (T − λI )2 dimV u = (T − λI )dimV v = 0, also u is ageneralized eigenvector of eigenvalue λ.

I But, by the last lemma, then (T − λI )dimV u = 0, sov = 0.

One more lemma (14)

LemmaV = (T − λI )dimV ⊕ range(T − λI )dimV =V (λ)⊕ range(T − λI )dimV .

Proof.

I Since the dimensions are equal, we need to show just that(T − λI )dimV ∩ range(T − λI )dimV = {0}.

I Let v ∈ (T − λI )dimV ∩ range(T − λI )dimV . Writev = (T − λI )dimV u.

I Since (T − λI )2 dimV u = (T − λI )dimV v = 0, also u is ageneralized eigenvector of eigenvalue λ.

I But, by the last lemma, then (T − λI )dimV u = 0, sov = 0.

Proof of the decomposition theorem (15)

⊕λ V (λ).

I Proof: By induction on dim V . Let λ be an eigenvalue of T ,so V (λ) 6= {0}.

I Write V = V (λ)⊕ range(T − λI )dimV .I Since dim range(T − λI )dimV < dim V , the ind. hyp. shows

that range(T − λI ) is the direct sum of the generalizedeigenspaces of T |range(T−λI )dim V .

I To conclude, we claim that for µ 6= λ,V (µ) ⊆ range(T − λI )dimV . Thus V (µ) is a generalizedeigenspace of T |range(T−λI )dim V .

I For this, we show that (T − λI )dimV V (µ) = V (µ).I First, V (µ) is T -invariant, so (T − λI )dimV V (µ) ⊆ V (µ).

We only need to show (T − λI )dimV is injective on V (µ).I This means that V (µ) ∩ null(T − λI )dimV = {0}. But this is

V (µ) ∩ V (λ) = {0}, by the lemmas.

Proof of the decomposition theorem (15)

⊕λ V (λ).

I Proof: By induction on dim V . Let λ be an eigenvalue of T ,so V (λ) 6= {0}.

I Write V = V (λ)⊕ range(T − λI )dimV .I Since dim range(T − λI )dimV < dim V , the ind. hyp. shows

that range(T − λI ) is the direct sum of the generalizedeigenspaces of T |range(T−λI )dim V .

I To conclude, we claim that for µ 6= λ,V (µ) ⊆ range(T − λI )dimV . Thus V (µ) is a generalizedeigenspace of T |range(T−λI )dim V .

I For this, we show that (T − λI )dimV V (µ) = V (µ).I First, V (µ) is T -invariant, so (T − λI )dimV V (µ) ⊆ V (µ).

We only need to show (T − λI )dimV is injective on V (µ).I This means that V (µ) ∩ null(T − λI )dimV = {0}. But this is

V (µ) ∩ V (λ) = {0}, by the lemmas.

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