Topics in Abstract Algebra/Linear algebra

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The Moore-Penrose inverse[edit | edit source]

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 conjugate[edit | edit source]

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 uniqueness[edit | edit source]

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

where:

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 properties[edit | edit source]

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 case[edit | edit source]

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 ,


Products[edit | edit source]

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, .
Proof. Since , . Write and . We show that satisfies the Moore–Penrose criteria for .

Therefore, .

Exercise. If has orthonormal rows, then for any -matrix of the right dimensions, .


Another important special case which approximates closely that of invertible matrices is when has full column rank and has full row rank.

Proposition. If has full column rank and has full row rank, then .
Proof. Since has full column rank, is invertible so . Similarly, since has full row rank, is invertible so .

Write (using reduction to the Hermitian case). We show that satisfies the Moore–Penrose criteria.

Therefore, .

We finally derive a formula for calculating the Moore-Penrose inverse of .

Proposition. If , then .
Proof. Here, , and thus and . We show that indeed satisfies the four Moore–Penrose criteria.

Therefore, . In other words:

and, since

Projectors and subspaces[edit | edit source]

The defining feature of classical inverses is that What can we say about and ?

We can derive some properties easily from the more basic properties above:

Exercise. Let be a -matrix. Then and


We can conclude that and are orthogonal projections.

Proposition. Let be a -matrix. Then and are orthogonal projections
Proof. Indeed, consider the operator : any vector decomposes as

and for all vectors and satisfying and , we have

.

It follows that and . Similarly, and . The orthogonal components are now readily identified.

We finish our analysis by determining image and kernel of the mappings encoded by the Moore-Penrose inverse.

Proposition. Let be a -matrix. Then and .
Proof. If belongs to the range of then for some , and . Conversely, if then so that belongs to the range of . It follows that is the orthogonal projector onto the range of . is then the orthogonal projector onto the orthogonal complement of the range of , which equals the kernel of .

A similar argument using the relation establishes that is the orthogonal projector onto the range of and is the orthogonal projector onto the kernel of .

Using the relations and it follows that the range of P equals the range of , which in turn implies that the range of equals the kernel of . Similarly implies that the range of equals the range of . Therefore, we find,

Applications[edit | edit source]

We present two applications of the Moore-Penrose inverse in solving linear systems of equations.

Least-squares minimization[edit | edit source]

Moore-Penrose inverses can be used for least-squares minimisation of a system of equations that might not necessarily have an exact solution.

Proposition. For any matrix , where .
Proof. We first note that (stating the complex case), using the fact that satisfies and , we have

so that ( stands for the Hermitian conjugate of the previous term in the following)

as claimed.

Remark. This lower bound need not be zero as the system may not have a solution (e.g. when the matrix A does not have full rank or the system is overdetermined). If is injective i.e. one-to-one (which implies ), then the bound is attained uniquely at .


Minimum-norm solution to a linear system[edit | edit source]

The proof above also shows that if the system is satisfiable i.e. has a solution, then necessarily is a solution (not necessarily unique). We can say more:

Proposition. If the system is satisfiable, then is the unique solution with smallest Euclidean norm.
Proof. Note first, with , that and that . Therefore, assuming that , we have

Thus

with equality if and only if , as was to be shown.

An immediate consequence of this result is that is also the uniquely smallest solution to the least-squares minimization problem for all , including when is neither injective nor surjective. It can be shown that the least-squares approximation is unique. Thus it is necessary and sufficient for all that solve the least-squares minimization to satisfy . This system always has a solution (not necessarily unique) as lies in the column space of . From the above result the smallest which solves this system is .

Notes[edit | edit source]

  1. Some authors use slightly different dimensions for the factors. The two definitions are equivalent.

References[edit | edit source]

  • Ben-Israel, Adi; Greville, Thomas N.E. (2003). Generalized inverses: Theory and applications (2nd ed.). New York, NY: Springer. doi:10.1007/b97366. ISBN 978-0-387-00293-4.
  • Campbell, S. L.; Meyer, Jr., C. D. (1991). Generalized Inverses of Linear Transformations. Dover. ISBN 978-0-486-66693-8.
  • Nakamura, Yoshihiko (1991). Advanced Robotics: Redundancy and Optimization. Addison-Wesley. ISBN 978-0201151985.
  • Rao, C. Radhakrishna; Mitra, Sujit Kumar (1971). Generalized Inverse of Matrices and its Applications. New York: John Wiley & Sons. pp. 240. ISBN 978-0-471-70821-6.