In the case of groups, we could only form a quotient group if the subgroup we were modding out by as sufficiently “nice” (i.e., was a normal subgroup); likewise, in the case of rings we required that a subring be nice (an ideal) in order to form a quotient ring.  With modules however, we can always form the quotient module with a submodule.  We can do this since, as is an abelian group under addition, all subgroups are normal and we can form the quotient group .  This is naturally an abelian group, so in order to turn this into an -module we have to define multiplication of elements in and , which we do in the most obvious way: .  To check that this is in fact an -module, we simply verify that the three axioms of a module hold.

Finally, if is a unitary module, then so too is : .

We say a map between two -modules is a homomorphism if is a group homomorphism with the additional property that for each , .  As you would expect, the kernel of this homomorphism is a submodule of , and the image is a submodule of .

Supposing that , we define to be the smallest submodule of containing , which is naturally the intersection of all submodules containing

If is a finite set, we may write in place of , and in the event that is a singleton, we say that is a cyclic submodule of .  We call the elements of the generating set of the submodule, and call the elements of the generators of this submodule.

To be more precise, given a ring and an abelian group (written additively), we say that is a (left) -module if there exists a ring homomorphism .  (Recall that if is an abelian group, the collection of endomorphisms of forms a ring under piecewise addition and function composition.)  This says simply that given an and an , we can define the multiplication of by as .  Given any and any , the following are immediate consequences of our definition.

If in addition the ring has an identity, we may also require that for all .  We call such a module unitary.  Though this isn’t strictly required, we will assume that all modules we deal with are unitary if the ring has identity.  Notice that we only perform multiplication on the left.  If we define multiplication on the right, we have a right -module.  We will assume that all of our modules are left module unless otherwise specified.

The three properties listed above aren’t simply consequences of our definition of a module, but can actually be used as an alternative definition.  That is, if we have an abelian group , a ring , and some multiplication satisfying the above properties, we then have a module.  To see this, define for each a -endomorphism by .  Notice that the second property above guarantees that is indeed an endomorphism: .  Now defining a map as , properties one and three guarantee that this is in fact a ring homomorphism.

As stated earlier, a module can be thought of as a generalization of a vector space.  In fact, if our ring is a field, then any -module is simply a vector space over .  A module can also be thought of as a generalization of an abelian group, in the sense that every abelian group is in fact a -module.  Suppose that is an abelian group.  For each positive , define as .  If , define .  Finally, if , define as the inverse (in ) of .

Notice that every ring can be viewed as a module over itself; is an -module where scalar multiplication is simply the ring’s usual multiplication.  Additionally, if is a subring of , then can be viewed as an -module as the three properties of an -module are satisfied by usual multiplication in the ring.  Similarly, if is left ideal of , then is a left -module; if is a right ideal, then it’s a right -module.

As we may pull our scalars from a ring instead of a field, we can treat some more “interesting” objects are scalars.  For instance, suppose that is an -vector space and a linear transformation.  Then , the set of all polynomials with coefficients from , forms a ring.  For any with , we define where means the -fold composition of with itself.  For any we define as .  Notice this satisfies the properties of a -module.

Thus this module, which we’ll denote , has polynomials as its scalars.

Notice that if is an eigenvalue of with associated eigenvector we have the following.

Of course, satisfies this equation, but that’s a trivial solution.  For any other, non-trivial, solution we’d require that is non-singular, and so .  Thus if is an eigenvalue of , we must have .

Now suppose that is such that .  Then there is a non-trivial solution to , so , and is an eigenvalue.  We’ve shown that is an eigenvalue of if and only if .  Furthermore, is a polynomial in (this is obvious if is , and inductively we can show that this is true for matrices).  This means that with the characteristic polynomial, the problem of finding eigenvalues is reduced to finding the roots of a polynomial.

As an example, suppose

Then the characterstic polynomial is

So we see that the eigenvalues are , , and (notice the last two are complex conjugates of one another).

Now, once we’ve found the eigenvalues, the next step is to find the eigenvectors.  Since

what we want is to find the nullspace of , since these are all the vectors that will take to zero.  In our particular example, for ,

Now we take the row-reduced echelon form of this matrix, since it shares the same null space:

This tells us that

So the eigenvectors associated with the eigenvalue are the multiples of .  We’d repeat the above process with to find the other eigenvectors.

Now since is a change of basis matrix, each of its columns gives the coordinates to a basis vector of some basis.  Let’s call that basis and let through be the elements of that basis.  Now, if we take the above equation and multiply by on the right, notice that

That is, the -th column of is equal to the -th column of , which is just times the -th column of .  Since each column of is just a linear combination of the columns of , though, we have

This means that when we plug in the -th column of to the linear transformation represented by , we get back a multiple of that column.  Calling the linear transformation , we have that

.

Vectors such as whose image under is just a multiple of the vector are called eigenvectors of .  That multiple, the above, is called an eigenvalue of .  These eigenvectors and eigenvalues are associated with a particular linear transformation, so when we talk about the eigenvectors and eigenvalues of a matrix, we really mean the eigenvectors and eigenvalues of the transformation represented by that matrix.  Notice that this means that eigenvalues are independent of the chosen basis; since similar matrices represent the same transformation just with respect to different bases, similar matrices have the same eigenvalues.

We assumed that was similar to a diagonal matrix above, but this isn’t always true.  If is similar to a diagonal matrix, say , then as we’ve just shown, the columns of are eigenvectors of .  Since these form the columns of a non-singular matrix, the eigenvectors of form a basis for the vector space.  Also, if the eigenvectors of form a basis, let’s take those basis vectors as columns of .

So a matrix is diagonalizable (similar to a diagonal matrix) if and only if its eigenvectors form a basis for the vector space.

Now, the matrix representation of this transformation with respect to the standard basis is clearly

But suppose we were to use a different basis for , like say .  We see that our transformation maps these basis vectors as follows:

Notice that with the vectors we have on both the left and the right above are the coordinates with respect to the standard basis.  We’d like to see what the matrix representing looks like with respect to the basis, so let’s convert the vectors on the right to -coordinates.  Recalling that a change of basis is simply a system of equations where the columns of the coefficient matrix are the coordinates of the basis vectors (and the inverse of this matrix if we want to go the other way), we have that

So with respect to our basis, the representation of is

We will denote this matrix as where the right-most subscript means that inputs are in -coordinates, and the left-most subscript means the outputs are in -coordinates as well.  Note that we could calculate by going from coordinates to standard coordinates, using the earlier matrix, then going back to -coordinates.  That is,

where refers to the -to-standard basis change of basis matrix.  For notational convenience, define the following

Then the above becomes

Any matrices and for which there exists a with satisfying this equation are called similar matrices.  Note that two matrices are similar if and only if they represent the same linear transformation, but with respect to different bases.  Also note that every non-singular matrix represents a change of basis matrix.  Similarity forms an equivalence relation on the set of square matrices / the set of linear transformations from a finite dimensional vector space to itself.

Specifically, suppose and are vector spaces over the field .  A function is called a linear transformation if for all scalars and for all vectors we have

Note that because of this linearity, a linear transformation is completely determined by how it maps the basis vectors of the domain.  Suppose that is a basis for V.  Let be any vector in with .  We then have

.

So if we know each , we can figure out where any other vector will be sent by .  This does not mean that is necessarily a basis for the range, .  It could be that , in which case these vectors are linearly dependent and can’t both be in the basis.  We do have that span , however, so as long as they’re linearly independent they’ll form a basis.

The main thing we want to notice about linear transformations for right now is that if both and are finite dimensional, then a linear transformation can be represented as a matrix.  Suppose that is -dimensional with the basis mentioned above, and that is -dimensional with basis .  Note that the properties of matrix multiplication tell us that any matrix defines a linear transformation from to :

Now suppose is any other linear transformation.  Suppose that the coordinate vector of with respect to the basis is

Now let .  We then have

Thus a linear transformation between finite dimensional vector spaces can be represented as a matrix.  Notice that the entries of our matrix depend on our particular chosen bases: if one basis were altered, the matrix would change, even though the transformation is the same.  We will denote the matrix representing with respect to the and bases as

if we delete the first row and second column we have

Let’s also note that if the first row of is all zeros except for the first entry, the determinant of is simply the determinant of multiplied by that first entry.  That is,

To see this, first note that we can zero out the entries in the first column below the by performing a sequence of elementary row operations that don’t change the determinant.  Now we clearly have

Supposing we can write as a product of elementary matrices, , to calculate its determinant, we can then obtain the matrix above by looking at the product

Each of these is then an elementary matrix whose determinant is the same as the determinant of the associated matrix, so we have our result.

Now, applying the linearity we discussed last time,

Notice that we can can zero out the elements in the first column below giving us

In general, for the -th column, we want to do a series of column swaps bringing the -th column to the front of the matrix, but keeping the other columns in order.  (For this reason a single column swap won’t work, since that permutes the remaining columns.)  Each time we swap columns, the determinant is multiplied by -1.  If we move the -th column to the left by swapping with column , then with column , and so on, we perform swaps.  Thus

So in total we have

We could also perform this expansion along another row, but we’d have to perform some row swaps to move that row to the top first.  Supposing we decide to expand along the -th row, we’ll perform swaps, each time multiplying the determinant by -1.  We’d multiply our result above, then by .  Distributing that across our sum though, we note that

So, if we expand along the -th row, our formula becomes

Each term of the sum with the factor removed is called a cofactor; The -th cofactor, which we’ll denote is

The procedure we’ve just described is known as the Laplace expansion (or cofactor expansion) for the determinant, and so now we have a more efficient way of calculating determinants.  (Notice that we could expand along a column instead of a row: just repeat the above procedure on the transpose of the matrix.)

If is singular, once we put it in row reduced form it must have a row of zeros. We can now break up into a product of elementary matrices, one of which will have be of the form . We know that this matrix will have determinant zero, so the product will be zero, and thus . Likewise, if , we can write as a product of elementary row matrices and one will be , so is singular. Now we know that a matrix is singular if and only if its determinant is zero.

Suppose now that the first row of can be written as for some vectors and . We wish to show that

First suppose that the rows through form a linearly dependent set. Then our matrix is singular so has determinant zero. The determinants on the right in the above equation are zero too, so our we have our result.

Suppose now that through are linearly independent. We can then extend these to a basis for by adding a vector, call it . Then there exist scalars for such that

Some simple manipulations from last time give us the following.

And likewise,

Now we combine these results,

Since we can swap rows without altering the determinant, this result holds for any rows.  A similar argument shows the result also holds for columns.

1. For any two matrices and ,
2. The determinant of is .

That’s it.  That’s all you need to define the determinant.  Of course, now we have to worry about whether a function with these properties even exists or not, and if so, is that function is unique?  Before answering either of those questions, though, we need to establish that every matrix can be written as a product of elementary matrices (where by an elementary matrix we mean one which results in an elementary row (or column) operation when multiplied on the left (or right), including zeroing out a row or column).  To see this, recall that for every matrix there exist non-singular matrices and such that

where .  In the above, represents a sequence of elementary row operations, and gives a sequence of elementary column operations; and are the products of elementary matrices.  Since these are non-singular we have

The middle matrix is clearly a product of matrices of the form .  If we agree to also call such matrices elementary, then every square matrix is a product of elementary matrices.

Supposing we have a function satisfying the properties given above, we can calculate the determinant of an elementary matrix pretty easily.  In the following, means we multiply the -th row by in the identity matrix; means we swap rows and ; means we add times row to row .

First we’ll look at scalar multiples of a particular row:

(these are scalars in a field, so they commute)

Using similar identities we can show and .  Now we know the determinants for all elementary matrices.  Since every square matrix is a product of elementary matrices, and we can split the determinant up across a product of matrices, we can even calculate the determinant of any square matrix based solely on the two properties given earlier.  (Note this isn’t necessarily an efficient way to calculate the determinant, just a possible way.)

At this point it shouldn’t seem too surprising that the two properties above are all we need to define a determinant: every square matrix is a product of elementary matrices, and we can “massage” elementary matrices into a form whose determinant we can calculate.  In coming posts we’ll see how to expand this to define the Laplace / cofactor expansion for the determinant; the relationship between determinants and inverses; and Cramer’s rule, which tells us how to compute a single coordinate in the solution vector to a system of equations.

When we write down a vector like we implicitly mean that these are the coordinates to use with the vectors in the standard basis; these are the coefficients we multiply the basis vectors by to get describe our vector.

But how would we find the appropriate coordinates if we were to use as our basis?  Let’s suppose our coefficients are .  Then what we want to do is find the values of the such that

So what we have is a system of equations:

Notice that since the columns of this matrix form a basis for , the matrix is invertible, and so we can easily solve for the .

In general, if we have a basis , we will write

to mean that the are the coefficients of the vectors in our basis, and will let be the matrix converts the coordinates for the standard basis to coordinates in our basis.  Likewise, will be the matrix which converts coordinates from the basis to coordinates with the standard basis.

Generalizing on the argument above we see that we can calculate by simply using our basis vectors as our columns.  For we take the inverse of this matrix.

Now suppose we have two bases, and .  The matrix which will take coordinates with respect to the basis and convert them into coordinates for the basis is denoted .  One way to calculate is to convert our -coordinates into standard coordinates, and then convert those into -coordinates:

Alternatively, we could note that

(with the 1 in the -th spot)

(recall is the notation I use for the -th column of )

So the -th column of is simply -th vector from our basis, but in -coordinates.