Topics in Abstract Algebra/Printable version

Topics in Abstract Algebra

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Non-commutative rings

A ring is not necessarily commutative but is assumed to have the multiplicative identity.

Proposition. Let be a simple ring. Then: every morphism is either zero or an isomorphism. (Schur's lemma)

Theorem (Levitzky). Let be a right noetherian ring. Then every (left or right) nil ideal is nilpotent.

Commutative algebra

The set of all prime ideals in a commutative ring is called the spectrum of and denoted by . (The motivation for the term comes from the theory of a commutative Banach algebra.)

Spec AEdit

The set of all nilpotent elements in forms an ideal called the nilradical of . Given any ideal , the pre-image of the nilradical of is an ideal called the radical of and denoted by . Explicitly, if and only if for some .

Proposition A.14. Let .

  • (i)
  • (ii)

Proof. Routine.

Exercise. A ring has only one prime ideal if and only if its nilradical is maximal.

Exercise. Every prime ideal in a finite ring is maximal.

Proposition A.2. Let be a ring. If every principal ideal in is prime, then is a field.
Proof. Let . Since is in , which is prime, . Thus, we can write . Since is prime, is a domain. Hence, .

Lemma. Let . Then is prime if and only if


Proof. () Clear. () Let be the image of in . Suppose is a zero-divisor; that is, for some . Let , and . Since , and is strictly larger than , by the hypothesis, . That is, .

Theorem A.11 (multiplicative avoidance). Let be a multiplicative system. If is disjoint from , then there exists a prime ideal that is maximal among ideals disjoint from .
Proof. Let be a maximal element in the set of all ideals disjoint from . Let and be ideals strictly larger than . Since is maximal, we find and . By the definition of , ; thus, . By the lemma, is prime then.

Note that the theorem applies in particular when contains only 1.

Exercise. A domain A is a principal ideal domain if every prime ideal is principal.

A Goldman domain is a domain whose field of fractions is finitely generated as an algebra. When is a Goldman domain, K always has the form . Indeed, if , let . Then .

Lemma. Let be a domain with the field of fractions , and . Then if and only if every nonzero prime ideal of contains .
Proof. () Let , and . If is disjoint from , then, by the lemma, there is a prime ideal disjoint from , contradicting the hypothesis. Thus, contains some power of , say, . Then and so are invertible in () If is a nonzero prime ideal, it contains a nonzero element, say, . Then we can write: , or ; thus, .

A prime ideal is called a Goldman ideal if is a Goldman domain.

Theorem A.21. Let be a ring and . Then is the intersection of all minimal Goldman ideals of A containing
Proof. By the ideal correspondence, it suffices to prove the case . Let . Let . Since is not nilpotent (or it will be in ), by multiplicative avoidance, there is some prime ideal not containing . It remains to show it is a Goldman ideal. But if is a nonzero prime, then since collapses to zero if it is disjoint from . By Lemma, the field of fractions of is obtained by inverting and so is a Goldman ideal. Hence, the intersection of all Goldman ideals reduces to zero.

In some rings, Goldman ideals are maximal; this will be discussed in the next section. On the other hand,

Lemma. Let . Then is a Goldman ideal if and only if it is the contraction of a maximal ideal in .

Theorem. The following are equivalent.

  1. For any , is the intersection of all maximal ideals containing .
  2. Every Goldman ideal is maximal.
  3. Every maximal ideal in contracts to a maximal ideal in .

Proof. Clear.

A ring satisfying the equivalent conditions in the theorem is called a Hilbert-Jacobson ring.

Lemma. Let be domains such that is algebraic and of finite type over . Then is a Goldman domain if and only if is a Goldman domain.
Proof. Let be the fields of fractions of and , respectively.

Theorem A.19. Let be a Hilbert-Jacobson ring. Then is a Hilbert-Jacobson ring.
Proof. Let be a Goldman ideal, and . It follows from Lemma something that is a Goldman domain since it is contained in a , a Goldman domain. Since is a Hilbert-Jacobson ring, is maximal and so is a field and so is a field; that is, is maximal.


To do:
Explain why is a field (or point to a location where it can be understood why it is so...).

Theorem A.5 (prime avoidance). Let be ideals, at most two of which are not prime, and . If , then for some .
Proof. We shall induct on to find that is in no . The case being trivial, suppose we find such that for . We assume ; else, we're done. Moreover, if for some , then the theorem applies without and so this case is done by by the inductive hypothesis. We thus assume for all . Now, ; if not, since is prime, one of the ideals in the left is contained in , contradiction. Hence, there is in the left that is not in . It follows that for all . Finally, we remark that the argument works without assuming and are prime. (TODO: too sketchy.) The proof is thus complete.

An element p of a ring is a prime if is prime, and is an irreducible if either or is a unit..

We write if , and say divides . In a domain, a prime element is irreducible. (Suppose . Then either or , say, the former. Then , and . Canceling out we see is a unit.) The converse is false in general. We have however:

Proposition. Suppose: for every and , whenever (1) is the only principal ideal containing . Then every irreducible is a prime.
Proof. Let be an irreducible, and suppose and . Since implies and , there is a such that . But then and so ( is an irreducible.) Thus, .

Theorem A.16 (Chinese remainder theorem). Let . If , then

is exact.

The Jacobson radical of a ring is the intersection of all maximal ideals.

Proposition A.6. is in the Jacobson radical if and only if is a unit for every .
Proof. Let be in the Jacobson radical. If is not a unit, it is in a maximal ideal . But then we have: , which is a sum of elements in ; thus, in , contradiction. Conversely, suppose is not in the Jacobson radical; that is, it is not in some maximal ideal . Then is an ideal containing but strictly larger. Thus, it contains , and we can write: with and . Then , and would cease to be proper, unless is a non-unit.

Note that the nilradical is contained in the Jacobson radical, and they coincide in particular if prime ideals are maximal (e.g., the ring is a principal ideal domain). Another instance of this is:

Exercise. In , the nilradical and the Jacobson radical coincide.

Theorem A.17 (Hopkins). Let A be a ring. Then the following are equivalent.

  1. A is artinian
  2. A is noetherian and every prime ideal is maximal.
  3. is finite and discrete, and is noetherian for all maximal ideal .

Proof. (1) (3): Let be prime, and . Since is artinian (consider the short exact sequence), the descending sequence stabilizes eventually; i.e., for some unit u. Since is a domain, is a unit then. Hence, is maximal and so is discrete. It remains to show that it is finite. Let be the set of all finite intersections of maximal ideals. Let be its minimal element, which we have by (1). We write . Let be an arbitrary maximal ideal. Then and so by minimality. Thus, for some i. (3) (2): We only have to show is noetherian.

A ring is said to be local if it has only one maximal ideal.

Proposition A.17. Let be a nonzero ring. The following are equivalent.

  1. is local.
  2. For every , either or is a unit.
  3. The set of non-units is an ideal.

Proof. (1) (2): If is a non-unit, then is the Jacobson radical; thus, is a unit by Proposition A.6. (2) (3): Let , and suppose is a non-unit. If is a unit, then so are and . Thus, is a non-unit. Suppose are non-units; we show that is a non-unit by contradiction. If is a unit, then there exists a unit such that . Thus either or is a unit, whence either or is a unit, a contradiction. (3) (1): Let be the set of non-units. If is maximal, it consists of nonunits; thus, where we have the equality by the maximality of .

Example. If is a prime ideal, then is a local ring where is its unique maximal ideal.

Example. If is maximal, then is a local ring. In particular, is local for any maximal ideal .

Let be a local noetherian ring.

A. Lemma

  • (i) Let be a proper ideal of . If is a finite generated -module, then .
  • (ii) The intersection of all over is trivial.

Proof: We prove (i) by the induction on the number of generators. Suppose cannot be generated by strictly less than generators, and suppose we have that generates . Then, in particular,

where are in ,

and thus

Since is not a unit, is a unit; in fact, if is not a unit, it belongs to a unique maximal ideal , which contains every non-units, in particular, , and thus , which is nonsense. Thus we find that actually x_2, ..., x_n generates ; this contradicts the inductive hypothesis.

An ideal is said to be primary if every zero-divisor in is nilpotent. Explicitly, this means that, whenever and , . In particular, a prime ideal is primary.

Proposition. If is primary, then is prime. Conversely, if is maximal, then is primary.
Proof. The first part is clear. Conversely, if is maximal, then is a maximal ideal in . It must be unique and so is local. In particular, a zero-divisor in is nonunit and so is contained in ; hence, nilpotent.

Exercise. prime primary.

Theorem A.8 (Primary decomposition). Let be a noetherian ring. If , then is a finite intersection of primary ideals.
Proof. Let be the set of all ideals that is not a finite intersection of primary ideals. We want to show is empty. Suppose not, and let be its maximal element. We can write as an intersection of two ideals strictly larger than . Indeed, since is not prime by definition in particular, choose and such that . As in the proof of Theorem A.3, we can write: where is the set of all such that . By maximality, . Thus, they are finite intersections of primary ideals, but then so is , contradiction.

Proposition. If is indecomposable, then the set of zero divisors is a union of minimal primes.

Integral extensionEdit

Let be rings. If is a root of a monic polynomial , then is said to be integral over . If every element of is integral over , then we say is integral over or is an integral extension of . More generally, we say a ring morphism is integral if the image of is integral over . By replacing with , it suffices to study the case , and that's what we will below do.

Lemma A.9. Let . Then the following are equivalent.

  1. is integral over .
  2. is finite over .
  3. is contained in an -submodule of that is finite over .

Proof. (1) means that we can write:

Thus, spans . Hence, (1) (2). Since (2) (3) vacuously, it remains to show (3) (1). Let be generated over by . Since , we can write

where . Denoting by the matrix , this means that annihilates . Hence, by (3). Noting is a monic polynomial in we get (1).

The set of all elements in B that are integral over A is called the integral closure of A in B. By the lemma, the integral closure is a subring of containing . (Proof: if and are integral elements, then and are contained in , finite over .) It is also clear that integrability is transitive; that is, if is integral over and is integral over , then is integral over .

Proposition. Let be an integral extension where is a domain. Then

  • (i) is a field if and only if is a field.
  • (ii) Every nonzero ideal of has nonzero intersection with .

Proof. (i) Suppose is a field, and let . Since and is integral over , we can write:

Multiplying both sides by we see . For the rest, let . We have an integral equation:


Since is a domain, if is the minimal degree of a monic polynomial that annihilates , then it must be that . This shows that , giving us (ii). Also, if is a field, then is invertible and so is .

Theorem (Noether normalization). Let be a finitely generated -algebra. Then we can find such that

  1. is integral over .
  2. are algebraically independent over .
  3. are a separating transcendence basis of the field of fractions of if is separable over .

Exercise A.10 (Artin-Tate). Let be rings. Suppose is noetherian. If is finitely generated as an -algebra and integral over , then is finitely generated as an -algebra.

Exercise. A ring morphism (where is an algebraically closed field) extends to (Answer:

Noetherian ringsEdit

Exercise. A ring is noetherian if and only if every prime ideal is finitely generated. (See T. Y. Lam and Manuel L. Reyes, A Prime Ideal Principle in Commutative Algebra for a systematic study of results of this type.)

The next theorem furnishes many examples of a noetherian ring.

Theorem A.7 (Hilbert basis). is a noetherian ring if and only if is noetherian.
Proof. By induction it suffices to prove is noetherian. Let . Let be the set of all coefficients of polynomials of degree in . Since , there exists such that


For each , choose finitely many elements of whose coefficients generate . Let be an ideal generated by for all . We claim . It is clear that . We prove the opposite inclusion by induction on the degree of polynomials in . Let , the leading coefficient of and the degree of . Then . If , then

In particular, if , then has degree strictly less than that of and so by the inductive hypothesis . Since , then. If , then and the same argument shows .

Exercise. Let be the ring of continuous functions . is not noetherian.

Let be a noetherian local ring with . Let . Then is called an ideal of definition if is artinian.


The local ring is said to be regular if the equality holds in the above.

Theorem. Let be a noetherian ring. Then .
Proof. By induction, it suffices to prove the case .

Theorem. Let be a finite-dimensional -algebra. If is a domain with the field of fractions , then .
Proof. By the noether normalization lemma, is integral over where are algebraically independent over . Thus, . On the other hand, .

Theorem. Let be a domain with (ACCP). Then is a UFD if and only if every prime ideal of height 1 is principal.
Proof. () By Theorem A.10, contains a prime element . Then

where the second inclusion must be equality since has height 1. () In light of Theorem A.10, it suffices to show that is a GCD domain. (TODO: complete the proof.)

Theorem. A regular local ring is a UFD.

Theorem A.10 (Krull's intersection theorem). Let be a proper ideal. If is either a noetherian domain or a local ring, then .

Theorem A.15. Let . If is noetherian,

for some .

In particular, the nilradical of is nilpotent.
Proof. It suffices to prove this when . Thus, the proof reduces to proving that the nilradical of A is nilpotent. Since is nilpotent, we have finitely many nilpotent elements that spans . The power of any linear combination of them is then a sum of terms that contain the high power of some if we take the sufficiently high power. Thus, is nilpotent.

Proposition A.8. If is noetherian, then is noetherian.

Corollary. If is noetherian, then is noetherian.

Zariski topologyEdit

Given , let . (Note that .) It is easy to see

, and .

It follows that the collection of the sets of the form includes the empty set and and is closed under intersection and finite union. In other words, we can define a topology for by declaring to be closed sets. The resulting topology is called the Zariski topology. Let , and write .

Proposition A.16. We have:

  • (i) is quasi-compact.
  • (ii) is canonically isomorphic to .

Proof. We have: .

Exercise. Let be a local ring. Then is connected.

Corollary. is a closed surjection.

Theorem A.12. If is noetherian for every maximal ideal and if is finite for each , then is noetherian.

Integrally closed domainEdit

Lemma A.8. In a GCD domain, if , then .

Proposition A.9. In a GCD domain, every irreducible element is prime.
Proof. Let be an irreducible, and suppose . Then . If , is a unit, the case we tacitly ignore. Thus, by the lemma, , say, is a nonunit. Since is irreducible, and so .

In particular, in a polynomial ring that is a GCD domain, every irreducible polynomial is a prime element.

Theorem (undefined: ACC). Let A be a ring that satisfies the ascending chain conditions on principal ideals (example: noetherian ring). Then every x in is a finite product of irreducibles.

Theorem A.10. Let A be a domain. The following are equivalent.

  1. Every nonzero nonunit element is a finite product of prime elements.
  2. (Kaplansky) Every nonzero prime ideal contains a prime element.
  3. is a GCD domain and has (ACC) on principal ideals.

Proof. (3) (2): Let . If is nonzero, it then contains a nonzero element x, which we factor into irreducibles: . Then for some . Finally, irreducibles are prime since is a GCD domain. (2) (1): Let be the set of all products of prime elements. Clearly, satisfies the hypothesis of Theorem A.11 (i.e., closed under multiplication). Suppose, on the contrary, there is a nonzero nonunit . It is easy to see that since , and are disjoint. Thus, by Theorem A.11, there is a prime ideal containing and disjoint from . But, by (2), contains a prime element ; that is, intersects , contradiction. (1) (3): By uniqueness of factorization, it is clear that is a GCD domain.

A domain satisfying the equivalent conditions in the theorem is called a unique factorization domain or a UFD for short.

Corollary. If is a UFD, then is a UFD. If A is a principal ideal domain, then is a UFD.

Theorem A.13 (Nagata criterion). Let A be a domain, and a multiplicatively closed subset generated by prime elements. Then is a UFD if and only if is a UFD.

Field theory

Basic definitionsEdit

Let be a field extension; i.e., is a subfield of a field . Then has a k-algebra structure; in particular, a vector space structure. A transcendental element is an element that is not integral; in other words, is transcendental over if and only if is (isomorphic to) the polynomial ring in one variable. The situation can be phrased more abstract as follows. Given an element x in an extension and an indeterminate , we have the exact sequence:

by letting and the kernel of that map. Thus, is transcendental over if and only if . Since is a PID, when nonzero, is generated by a nonzero polynomial called the minimal polynomial of , which must be irreducible since is a domain and so is prime. (Note that if we replace by , say, then it is no longer a PID; therefore the kernel is no longer principal. So, in general, if a subset is such that is a polynomial ring where members of are variables, then is said to be algebraically independent; By convention, the empty set is algebraically independent, just as it is linearly independent.) Finally, as a custom, we call an integral field extension an algebraic extension.

When has finite dimension over , the extension is called finite extension. Every finite extension is algebraic. Indeed, if is transcendental over , then is a "polynomial ring" and therefore is an infinite-dimensional subspace of and L must be infinite-dimensional as well.

Exercise. A complex number is called an algebraic number if it is integral over . The set of all algebraic numbers is countable.

A field is called algebraically closed if it admits no nontrivial algebraic field extension. (A field is always an algebraic extension of itself, a trivial extension.) More concretely, a field is algebraically closed if every root of a polynomial over that field is already in that field. It follows from the Axiom of Choice (actually equivalent to it) that every field is a subfield of some algebraically closed field.

Separable extensionsEdit

A field extension is said to be separable if it is separable as k-algebra; i.e., is reduced for all field extension . The next theorem assures that this is equivalent to the classical definition.

Theorem. A field is a separable algebraic over if and only if every irreducible polynomial has distinct roots (i.e., and its derivative have no common root.)

For the remainder of the section, denotes the characteristic exponent of a field; (i.e., if and otherwise.) If the injection

is actually surjective (therefore, an automorphism), then a field is called perfect. Examples: Fields of characteristic zero and finite fields are perfect. Imperfect fields are therefore rather rare; they appear in algebraic geometry, a topic in later chapters. We let be the union of adjoined with -th roots of elements in over all positive integers . is then called the perfect closure since there is no strictly smaller subfield of that is perfect.

Proposition. A -algebra is separable if and only if is reduced.

Proposition. The following are equivalent.

  • (i) A field is perfect.
  • (ii) Every finite extension is separable.
  • (iii) Every extension is separable.

Proof. Suppose (ii) is false; it is then necessary that and . Finally, if (iii) is false, then there is an extension such that is not reduced. Since is algebraic over by construction, it has a finite extension such that is not reduced. This falsifies (ii).

In particular, any extension of a perfect field is perfect.

Separable extensionsEdit

Let be a field extension, and be the characteristic exponent of (i.e., if has characteristic zero; otherwise, .) is said to be separable over if is a domain. A maximal separable extension is called the separable closure and denoted by .

A field is said to be perfect if its separable closure is algebraically closed. A field is said to be purely inseparable if it equals its separable closure. (As the reader would notice, the terminology so far is quite confusing; but it is historical.)

Lemma. An algebraic extension is separable if and only if the minimal polynomial of any element has no multiple root.
Proof. We may assume that the extension is finite.

Proposition. A field is perfect if and only if either (i) its characteristic is zero or (ii) is an automorphism of
Proof. First suppose . Let be an irreducible polynomial. If and have a common root, then, since is irreducible, must divide and so since . On the other hand, if , then


Thus, a field of characteristic is perfect.

Corollary. A finite field is perfect.

Proposition. Let be a finite extension. Then is separable over if and only if is separable over and is separable over .

Proposition. Every finite field extension factors to a separable extension followed by a purely inseparable extension. More precisely,

Exercise. (Clark p. 33) Let be a field of characteristic 2, , a root of , and . Then (i) is purely inseparable and is separable. (ii) There is no nontrivial purely inseparable subextension of K/F.

Theorem (Primitive element). Let be a finite extension, where (but not necessarily ) are separable over . Then for some .
Proof. It suffices to prove the case (TODO: why?) Let be the minimal polynomials of .

Theorem. Let be a finitely generated field extension. Then the following are equivalent.

  1. is separable over .
  2. has a separating transcendence basis over .
  3. is a domain.

Transcendental extensionsEdit

Theorem (undefined: Lüroth) (Lüroth). Any subfield of containing but not equal to is a pure transcendental extension of .

Let be a field extension of degree . An element defines a -linear map:


We define

Proposition. Let be finite field extensions. Then

  • (i)
  • (ii)

Theorem A.8 (Hilbert 90). If is a finite Galois extension, then


Corollary. Let is a cyclic extension, and generate . If such that , then

for some .

A. Theorem A field extension is algebraic if and only if it is the direct limit of its finite subextensions.

A field extension is said to be Galois if

Here, we used the notation of invariance:

(In particular, when is a finite extension, is a Galois extension if and only if .) When is Galois, we set , and call the Galois group of .

A. Theorem A field extension is Galois if and only if it is normal and separable.

Integrally closed domainEdit

A domain is said to be integrally closed if equals the integral closure of in the field of fractions.

Proposition. GCD domains and valuation domains are integrally closed.
Proof. Suppose is integral over ; i.e.,


We may assume . It follows:


and so . Since by Lemma A.8, we have that is a unit in , and thus . The case of valuation domains is very similar.

Proposition. "integrally closed" is a local property.

Proposition. Let be a domain. The following are equivalent.

  1. Every finitely generated submodule of a projective -module is projective.
  2. Every finitely generated nonzero ideal of is invertible.
  3. is a valuation domain for every prime ideal .
  4. Every overring of is the intersection of localizations of .
  5. Every overring of is integrally closed.

A domain satisfying any/all of the equivalent conditions in the proposition is called the Prüfer domain. A notherian Prüfer domain is called a Dedekind domain.

Proposition A.10. Let be an integrally closed domain, and a finite extension of . Then is integral over if and only if its minimal polynomial in is in .

A Dedekind domain is a domain whose proper ideals are products of prime ideals.

A. Theorem Every UFD that is a Dedekind domain is a principal ideal domain.
Proof: Let be a prime ideal. We may assume is nonzero; thus, it contains a nonzero element . We may assume that is irreducible; thus, prime by unique factorization. If is prime, then we have . Thus, every prime ideal is principal.

Theorem Let A be an integral domain. Then A is a Dedekind domain if and only if:

  • (i) A is integrally closed.
  • (ii) A is noetherian, and
  • (iii) Every prime ideal is maximal.

A. Theorem Let A be a Dedekind domain with fraction field K. Let L be a finite degree field extension of K and denote by S the integral closure of R in L. Then S is itself a Dedekind domain.

A Lemma Let be an integral domain. Then is a Dedekind domain if and only if every localization of is a discrete valuation ring.

Lemma Let be a noetherian ring. Then every ideal contains a product of nonzero prime ideals.
Proof: Let be the set of all ideals that do not contain a product of nonzero prime ideals. If the lemma is false, is nonempty. Since is noetherian, has a maximal element . Note that is not prime; thus, there are such that but and . Now, . Since both and are strictly larger than , which is maximal in , and are both not in and both contain products of prime ideals. Hence, contains a product of prime ideals.

A local principal ideal domain is called a discrete valuation ring. A typical example is a localization of a Dedekind domain.

Henselian ringsEdit


Linear algebra

The Moore-Penrose inverseEdit

Inverse matrices play a key role in linear algebra, and particularly in computations. However, only square matrices can possibly be invertible. This leads us to introduce the Moore-Penrose inverse of a potentially non-square real- or complex-valued matrix, which satisfies some but not necessarily all of the properties of an inverse matrix.

Definition. Let be an m-by-n matrix over a field and be an n-by-m matrix over , where is either , the real numbers, or , the complex numbers. Recall that refers to the conjugate transpose of . Then the following four criteria are called the Moore–Penrose conditions for :

  1. ,
  2. ,
  3. ,
  4. .

We will see below that given a matrix , there exists a unique matrix that satisfies all four of the Moore–Penrose conditions. They generalise the properties of the usual inverse.

Remark. If is an invertible square matrix, then the ordinary inverse satisfies the Moore-Penrose conditions for . Observe also that if satisfies the Moore-Penrose conditions for , then satisfies the Moore-Penrose conditions for .

Basic properties of the Hermitian conjugateEdit

We assemble some basic properties of the conjugate transpose for later use. In the following lemmas, is a matrix with complex elements and n columns, is a matrix with complex elements and n rows.

Lemma (1). For any -matrix ,
Proof. The assumption says that all elements of A*A are zero. Therefore,

Therefore, all equal 0 i.e. .

Lemma (2). For any -matrix ,
Proof. :

Lemma (3). For any -matrix ,
Proof. This is proved in a manner similar to the argument of Lemma 2 (or by simply taking the Hermitian conjugate).

Existence and uniquenessEdit

We establish existence and uniqueness of the Moore-Penrose inverse for every matrix.

Theorem. If is a -matrix and and satisfy the Moore-Penrose conditions for , then .
Proof. Let be a matrix over or . Suppose that and are Moore–Penrose inverses of . Observe then that

Analogously we conclude that . The proof is completed by observing that then

Theorem. For every -matrix there is a matrix satisfying the Moore-Penrose conditions for
Proof. The proof proceeds in stages.

is a 1-by-1 matrix

For any , we define:

It is easy to see that is a pseudoinverse of (interpreted as a 1-by-1 matrix).

is a square diagonal matrix

Let be an n-by-n matrix over with zeros off the diagonal. We define as an n-by-n matrix over with as defined above. We write simply for .

Notice that is also a matrix with zeros off the diagonal.

We now show that is a pseudoinverse of :

is a general diagonal matrix

Let be an m-by-n matrix over with zeros off the main diagonal, where m and n are unequal. That is, for some when and otherwise.

Consider the case where . Then we can rewrite by stacking where is a square diagonal m-by-m matrix, and is the m-by-(nm) zero matrix. We define as an n-by-m matrix over , with the pseudoinverse of defined above, and the (nm)-by-m zero matrix. We now show that is a pseudoinverse of :

  1. By multiplication of block matrices, so by property 1 for square diagonal matrices proven in the previous section,.
  2. Similarly, , so
  3. By 1 and property 3 for square diagonal matrices, .
  4. By 2 and property 4 for square diagonal matrices,

Existence for such that follows by swapping the roles of and in the case and using the fact that .

is an arbitrary matrix

The singular value decomposition theorem states that there exists a factorization of the form


is an m-by-m unitary matrix over .
is an m-by-n matrix over with nonnegative real numbers on the diagonal and zeros off the diagonal.
is an n-by-n unitary matrix over .[1]

Define as .

We now show that is a pseudoinverse of :

This leads us to the natural definition:

Definition (Moore-Penrose inverse). Let be a -matrix. Then the unique -matrix satisfying the Moore-Penrose conditions for is called the Moore-Penrose inverse of .

Basic propertiesEdit

We have already seen above that the Moore-Penrose inverse generalises the classical inverse to potentially non-square matrices. We will now list some basic properties of its interaction with the Hermitian conjugate, leaving most of the proofs as exercises to the reader.

Exercise. For any -matrix ,

The following identities hold:

  1. A+ = A+ A+* A*
  2. A+ = A* A+* A+
  3. A = A+* A* A
  4. A = A A* A+*
  5. A* = A* A A+
  6. A* = A+ A A*

Proof of the first one: and imply that . □

The remaining identities are left as exercises.

Reduction to the Hermitian caseEdit

The results of this section show that the computation of the pseudoinverse is reducible to its construction in the Hermitian case. It suffices to show that the putative constructions satisfy the defining criteria.

Proposition. For every -matrix ,
Proof. Observe that

Similarly, implies that i.e. .

Additionally, so .

Finally, implies that .

Therefore, .

Exercise. For every -matrix ,


We now turn to calculating the Moore-Penrose inverse for a product of two matrices,

Proposition. If has orthonormal columns i.e. , then for any -matrix of the right dimensions,