## Linear Maps between vector spaces: Additional Definitions and Properties of Linear Maps

### Matrix Transpose

Let be a linear map. If is the matrix representation after choosing particular orthonormal basis sets for the underlying spaces, then, the transpose of or , is a map whose columns are the rows of .

In component form, this means:

The above definition relies on components. Another equivalent but more convenient definition is as follows.

Let be a linear map. Then, is the unique linear map that satisfies:

Any of the above two definitions can be used to show the following facts about the transpose of square matrices.

Notice that since the determinant of a square matrix is the same whether we consider the rows or the columns, then:

For example

### Matrix Inverse

Let be a linear map. The following are all equivalent:

- is invertible
- The rows of the matrix representation of are linearly independent
- The kernel of the contains only the zero vector

In this case, the inverse of is denoted and satisfies:

Notice that is unique, because if there is another matrix such that , then .

Notice also that if such that and , then, .

If the linear maps and are invertible, then it is easy to show that is also invertible and:

#### Matrix Inverse in

Consider the matrix:

Then, the inverse of can be shown to be:

Try it out, input the values of the matrix and press evaluate to calculate its inverse.

#### Matrix Inverse in

Consider the matrix:

If , and , then, the inverse of can be shown to be:

Try it out, input the values of the matrix and press evaluate to calculate its inverse.

### Invariants

Consider with the two orthonormal basis sets and with a coordinate transformation matrix such that .

Clearly, the components of vectors and the matrices representing linear operators change according to the chosen coordinate system (basis set). Invariants are functions of these components that do not change whether or is chosen as the basis set.

The invariants usually rely on the fact that .

#### Vector Invariants

##### Vector Norm

A vector has the representation with components when is the basis set. Alternatively, it has the representation with components when is the basis set.

The norm of the vector is an invariant since it is equal whether we use or .

The norm of when is the basis set:

The norm of is also equal to the norm of :

##### Vector Dot Product

Similarly, the dot product between two vectors is invariant:

#### Matrix Invariants in

We will restrict our discusion of invariants when the underlying space is . A linear operator has the matrix representation with components when is the basis set.

Alternatively, it has the representation with components when is the basis set. The following are some invariants of the matrix :

##### First Invariant, Trace

The trace of or is defined as:

is invariant for if we consider the components in :

It is straight forward from the definition to show that :

The above definition for the first invariant depends on the components in a given coordinate system. Another definition according to P. Chadwick that is independent of a coordinate system is given as follows:

where, and are three arbitrary linearly independent vectors. Use the components of and to verify that the two definitions are equivalent.

##### Second Invariant

The second invariant is defined as:

Clearly, since is invariant, so is :

Another definition for the second invariant according to P. Chadwick that is independent of a coordinate system is given as follows:

where, and are three arbitrary linearly independent vectors. Use the components of and to verify that the two definitions are equivalent.

##### Third Invariant, the Determinant

The third invariant is defined as the determinant of the matrix ;

Clearly, is invariant:

Another definition for the third invariant according to P. Chadwick that is independent of a coordinate system is given as follows:

where, and are three arbitrary linearly independent vectors. Use the components of and to verify that the two definitions are equivalent.

The trace (first invariant) and determinant (third invariant) of a matrix are related as follows:

##### Eigenvalues are Invariants

The eigenvalues of the matrices and are the same (why?).

It is worth mentioning that the three invariants mentioned above appear naturally in the characteristic equation of :

Input the components of a matrix in the following tool and three angles for coordinate transformation. The tool then calculates the three matrix invariants along with the eigenvalues and eigenvectors in both coordinate systems. As expected, the invariants and the eigenvalues are the same. However, the components of the eigenvectors are different. The vectors themselves are the same, but the components are different according to the relationship .

### Cayley-Hamilton Theorem

The Cayley-Hamilton Theorem is an important theorem in linear algebra that asserts that a matrix satisfies its characteristic equation. In other words, let . The eigenvalues of are those that satisfy:

where are polynomial expressions of the entries of the matrix . In particular, . Then, the Cayley-Hamilton Theorem asserts that:

The first equation is a scalar equation which is a polynomial expression of the variable . However, the second equation is a matrix equation in which the sum of the given matrices gives the 0 matrix. Without attempting a formal proof for the theorem, in the following we will show how the theorem applies to and .

#### Two Dimensional Matrices

Consider the matrix:

Therefore, the characteristic equation of is given by:

I.e.,

The matrix satisfies the characteristic equation as follows:

Where:

and

The following Mathematica code illustrates the above expressions.

View Mathematica Code:

M = {{M11, M12}, {M21, M22}} A = M.M - Tr[M] M + Det[M]*IdentityMatrix[2] FullSimplify[A]

#### Three Dimensional Matrices

Consider the matrix:

Therefore, the characteristic equation of is given by:

I.e.,

The matrix satisfies the characteristic equation as follows:

The above polynomial expressions in the components of the matrix equate to zero as illustrated using the following Mathematica code:

View Mathematica Code:

M = {{M11, M12,M13}, {M21, M22,M23},{M31, M32,M33}} I2=1/2*(Tr[M]^2-Tr[M.M]); A = M.M.M - Tr[M] M.M + I2*M-Det[M]*IdentityMatrix[3] FullSimplify[A]

One can show using induction that for , the matrix for can be written as a linear combination of , , and such that:

where , , and are functions of the invariants , , and .

Similarly, if is invertible, then, for can be written as:

where , , and are functions of the invariants , , and .