Elementary Module Theory (IV): Linear Algebra

Throughout this article, a general ring is denoted R while a division ring is denoted D.

Dimension of a Vector Space

First, let’s consider the dimension of a vector space V over D, denoted dim(V). If W is a subspace of V, we proved earlier that any basis of W can be extended to give a basis of V, thus dim(W) ≤ dim(V).

Furthermore, we claim that if $\{v_i + W\}$ is a basis of the quotient space V/W, then the v_i‘s, together with a basis $\{w_j\}$ of W, form a basis of V:

If $\sum_i r_i v_i + \sum_j r_j' w_j = 0$ for some $r_i, r_j' \in D$ , its image in V/W gives $\sum_i r_i (v_i + W) = 0$ and thus each $r_i$ is zero. This gives $\sum_j r_j' w_j = 0$ ; since $\{w_j\}$ forms a basis of W, each $r_j' = 0.$ This proves that $\{v_i\} \cup \{w_j\}$ is linearly independent.
Let $v\in V$ . Its image v+W in V/W can be written as a linear combination $\sum_i r_i (v_i + W) = v+W$ for some $r_i \in R.$ Hence $v - \sum_i r_i v_i \in W$ and can be written as a linear combination of $\{w_j\}.$ So v can be written as a linear combination of $\{v_i\} \cup \{w_j\}.$

Conclusion: dim(W) + dim(V/W) = dim(V). Now if f : V → W is any homomorphism of vector spaces, the first isomorphism theorem tells us that V/ker(f) is isomorphic to im(f). Hence, dim(V) = dim(ker(f)) + dim(im(f)).

If V is finite-dimensional and dim(V) = dim(W), then:

(f is injective) iff (ker(f) = 0) iff (dim(ker(f)) = 0) iff (dim(im(f)) = dim(V)) iff (dim(im(f)) = dim(W)) iff (im(f) = W) iff (f is surjective).

Thus, (f is injective) iff (f is surjective) iff (f is an isomorphism).

For infinite-dimensional V and W, take the free vector spaces $V = W = D^{(\mathbf{N})}$ and let f : V → W take the tuple $(r_1, r_2, \ldots) \mapsto (0, r_1, r_2, \ldots).$ Then f is injective but not surjective.

Over a general ring, even if M and N are free modules, the kernel and image of f : M → N may not be free. This follows from the fact that a submodule of a free module is not free in general, as we saw earlier. Hence it doesn’t make sense to talk about dim(ker(f)) and dim(im(f)) for such cases.

In a Nutshell. The main results are:

for a D-linear map f : V → W, dim(V) = dim(ker(f)) + dim(im(f));

if dim(V) = dim(W), then f is injective iff it is surjective.

Matrix Algebra

Recall that an R-module M is free if and only if it has a basis $\{m_i\}_{i\in I}$ , in which case we can identify $R^{(I)} \cong M$ via $(r_i)_{i\in I}\mapsto \sum_{i\in I} r_i m_i.$ Let’s restrict ourselves to the case of finite free modules, i.e. modules with finite bases. If $M\cong R^a$ and $N\cong R^b,$ the group of homomorphisms is identified with $\text{Hom}(M, N)\cong R^{ab}$ in terms of b × a matrices in R.

Let’s make this identification a bit more explicit. Pick a basis $\{m_1, \ldots, m_a\}$ of M and $\{n_1, \ldots, n_b\}$ of N. We have:

$R^a \cong M, \ (r_1, \ldots, r_a) \mapsto \sum_{i=1}^a r_i m_i\$ and $\ R^b \cong N, (r_1, \ldots, r_b)\mapsto \sum_{j=1}^b r_j n_j.$

A module homomorphism f : M → N is expressed as a matrix as follows:

Example 1

Take R = R, the field of real numbers and $M = \{a + bx + cx^2 : a, b, c\in\mathbf{R}\}$ and $N = \{a + bx : a, b\in \mathbf{R}\}$ where x is an indeterminate here. The map f : M → N given by f(p(x)) = dp/dx is easily checked to be R-linear.

Pick basis {1, x x²} of M and {1, x} of N. Since f(1) = 0, f(x) = 1 and f(x²) = 2x, the resulting f takes $m_1 \mapsto 0, m_2\mapsto n_1, m_3 \mapsto 2n_2.$ Hence, the matrix corresponding to these bases is $\begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 2\end{pmatrix}.$

On the other hand, if we pick basis {1+x, –x, 1+x²} of M and basis {1+x, 1+2x} of N, then

$f(m_1) = f(1+x) = 1 = 2n_1 - n_2$ ;
$f(m_2) = -1 = -2n_1 + n_2$ ;
$f(m_3) = 2x = -2n_1 + 2n_2$

which gives the matrix representation $\begin{pmatrix} 2 & -2 & -2 \\ -1 & 1 & 2\end{pmatrix}.$

Example 2

Let M = {a + b√2 : a, b integers} which is a Z-module. Take f : M → M which takes z to (3-√2)z. It’s clear that f is a homomorphism of additive groups and hence Z-linear. Since the domain and codomain modules are identical (M), let’s pick a single basis.

If we pick {1, √2}, then

$f(m_1) = f(1) = 3-\sqrt 2 = 3m_1 - m_2$ ;
$f(m_2) = f(\sqrt 2) = -2 + 3\sqrt 2 = -2m_1 + 3m_2$

thus giving the matrix representation $\begin{pmatrix} 3 & -2 \\ -1 & 3\end{pmatrix}.$ Replacing the basis by {-1, 1+√2} would give us: $\begin{pmatrix} -4 & -1 \\ 1 & 2\end{pmatrix}.$

Thus, the matrix representation for f : V → W depends on our choice of bases for V and W. If V = W, then it’s often convenient to pick the same basis.

Dual Module

We saw earlier that $\text{Hom}(R, M) \cong M$ as an R-module isomorphism. What about Hom(M, R) then?

Definition. The dual module of left-module M is defined to be $M^* := \text{Hom}(M, R).$ This is a right R-module, via the following right action:

if $r\in R$ and $f:M\to R$ , then the resulting $f\cdot r$ takes $m\mapsto f(m)r$ .

From the universal property of direct sums and products, we see that:

$(\oplus_{i\in I} M_i)^* \cong \prod_{i\in I} M_i^*.$

Let’s check that we get a right-module structure on M*: indeed, $(f\cdot r_1)\cdot r_2$ takes m to $(f\cdot r_1)(m)r_2 = (f(m)r_1)r_2$ which is the image of $f\cdot (r_1 r_2)$ acting on m.

The module $M^*$ is called the dual because it’s a right module instead of a left one. Note that if N were a right-module, the resulting space Hom(N, R) of all right-module homomorphisms would give us a left module $N^*.$ It’s not true in general that $M^{**} \cong M$ but it holds for finite-dimensional vector spaces over a division ring.

Theorem. If V is a finite-dimensional vector space over division ring D, then $V^{**} \cong V.$

Proof.

Consider the map $V^* \times V\to D$ which takes (f, v) to f(v). Fixing f, we get a map $v\mapsto f(v)$ which is a left-module homomorphism. Fixing v, we get a right-module homomorphism $f\mapsto f(v)$ since (f·r) corresponds to the map $v\mapsto f(v)r$ by definition. This gives a left-module homomorphism $\phi:V\to V^{**}.$

Since V is finite dimensional, it suffices to show $\text{ker}\phi = 0$ . But if $v\in V-\{0\}$ , we can extend {v} to a basis of V. Define a linear map f : V → D which takes v to 1 and all other basis elements to 0. Then $(\phi(v))(f) = f(v) \ne 0$ so $\phi(v) \ne 0.$ This shows that $\phi$ is injective and thus an isomorphism. ♦

One way to visualise the duality is via this diagram:

Exercise

It’s tempting to define a module structure on Hom(M, R) via $(f\cdot r)(m) = f(rm).$ What’s wrong with this definition? [ Answer: the resulting f·r : M → R is not a left-module homomorphism. ]

Dual Basis

Suppose $\{ v_1, v_2, \ldots, v_n\}$ is a basis of V. Let $f_i : V\to D$ (i = 1, …, n) be linear maps defined as follows:

$f_i(v_j) = \begin{cases} 1, \quad &\text{ if } j = i, \\ 0, \quad &\text{ if } j\ne i.\end{cases}$

Each $f_i$ is well-defined by the universal property of the free module V. Using the Kronecker delta function, we can just write $f_i(v_j) = \delta_{ij}.$ This is called the dual basis for $\{v_1, \ldots, v_n\}.$

[ Why is this a basis, you might ask? We know that dim(V*) = dim(V) = n, so it suffices to check that $f_1, \ldots, f_n$ is linearly independent. For that, we write $\sum_i f_i\cdot r_i=0$ for some $r_1, \ldots, r_n \in D$ (recall that V* is a right module). Then for each j = 1, …, n, we have

$0 = \sum_i (f_i\cdot r_i)(v_j) = \sum_i f_i(v_j)r_i = \sum_i \delta_{ij}r_i = r_j$

and we’re done. ]

Now if $f\in V^*$ and $v\in V,$ we can write $f = \sum_{i=1}^n f_i c_i$ and $v = \sum_{j=1}^n d_j v_j$ for some $c_i, d_j \in D.$ Then

$\begin{aligned}f(v) &= \left(\sum_{i=1}^n f_i c_i\right)\left(\sum_{j=1}^n d_j v_j\right) = \sum_{i=1}^n f_i\left(\sum_{j=1}^n d_j v_j\right)c_i \\ &= \sum_{i=1}^n \sum_{j=1}^n d_j\delta_{ij}c_i = \sum_{i=1}^n d_i c_i\end{aligned}$

which is the product between a row vector & a column vector.

One thus gets a natural inner product between a vector space and its dual. Recall that in an Euclidean vector space $V = \mathbf{R}^3$ , there’s a natural inner product given by the usual dot product which is inherent in the geometry of the space. However, for generic vector spaces, it’s hard to find a natural inner product. E.g. what would one be for the space of all polynomials of degree at most 2? Thus, the dual space provides a “cheap” and natural way to get an inner product.

Example

Consider the space $V = \{a + bx + cx^2 : a, b, c\in \mathbf{R}\}$ over the reals R=R. Examples of elements of V* are:

$f\mapsto f(1)$ which takes $(a+bx+cx^2)\mapsto a+b+c$ ;
$f\mapsto \left.\frac {df}{dx}\right|_{x=-1}$ which takes $(a+bx+cx^2) \mapsto -b+2c$ ;
$f\mapsto \int_0^1 (a+bx+cx^2) dx$ which takes $(a+bx+cx^2) \mapsto a + \frac b 2 + \frac c 3$ .

It’s easy to check that these three elements of V* are linearly independent and hence form a basis. Note: in this case, the base ring is a field so right modules are also left, i.e. V* and V are isomorphic as abstract vector spaces! However, there’s no “natural” isomorphism between them since in order to establish an isomorphism, one needs to pick a basis of V, a basis of V* and map the corresponding elements to each other. On the other hand, the isomorphism between V** and V is completely natural.

Exercise.

[ All vector spaces in this exercise are of finite dimension. ]

Let $\{v_1, \ldots, v_n\}$ be a basis of V and $\{f_1, \ldots, f_n\}$ be its dual basis for V*. Denote the dual basis of $\{f_1, \ldots, f_n\}$ by $\{\alpha_1, \ldots, \alpha_n\}$ in V**. Prove that under the isomorphism $V\cong V^{**}$ , we have $v_i = \alpha_i.$

Let $\{v_i\}$ be a basis of V and $\{w_j\}$ be a basis of W. If T : V → W is a linear map, then the matrix representation of T with respect to bases $\{v_i\}, \{w_j\}$ is denoted M.

Prove that the map T* : W* → V* which takes g : W → D to the composition g º T : V → D is a linear map of right modules.
Let $\{f_i\}$ be the dual basis of $\{v_i\}$ for V* and $\{g_j\}$ be the dual basis of $\{w_j\}$ for W*. Prove that the matrix representation of T* with respect to bases $\{f_i\}, \{g_j\}$ is the transpose of M.

More on Duality

Let V be a finite-dimensional vector space over D and V* be its dual. We claim that there’s a 1-1 correspondence between subspaces of V and those of V*, which is inclusion-reversing. Let’s describe this:

if $W\subseteq V$ is a subspace, define $W^\perp := \{ f\in V^* : f(w) = 0 \text{ for all } w\in W\};$
if $X\subseteq V^*$ is a subspace, define $X^\perp := \{v\in V : f(v) = 0 \text{ for all } f\in X\}.$

The following preliminary results are easy to prove.

Proposition.

$W^\perp$ is a subspace of V*;

$X^\perp$ is a subspace of V;

if $W_1\subseteq W_2\subseteq V$ , then $W_1^\perp \supseteq W_2^\perp$ ;

if $X_1\subseteq X_2 \subseteq V^*$ , then $X_1^\perp \supseteq X_2^\perp$ ;

$W\subseteq W^{\perp\perp}$ and $X\subseteq X^{\perp\perp}$ .

We’ll skip the proof, though we’ll note that the above result in fact holds for any subsets $W\subseteq V$ and $X\subseteq V^*$ . This observation also helps us to remember the direction of inclusion for $W\subseteq W^{\perp\perp}$ since in this general case, $W^{\perp\perp}$ is the subspace of V generated by W.

The main thing we want to prove is the following:

Theorem. If $W\subseteq V$ is a subspace, then $W^{\perp\perp} = W$ . Likewise if $X\subseteq V^*$ is a subspace, then $X^{\perp\perp} = X.$

Proof.

Pick a basis $\{v_1, \ldots, v_k\}$ of W and extend it to a basis $\{v_1, \ldots, v_n\}$ of V, where dim(W) = k and dim(V) = n. Let $\{f_1, \ldots, f_n\} \subset V^*$ be the dual basis.

If $v\in V-W,$ write $v = \sum_{i=1}^n r_i v_i$ where each $r_i\in D.$ Since v is outside W, $r_j\ne 0$ for some j>k. This gives $f_j(v) = f_j(\sum_i r_i v_i) = r_j \ne 0$ and $f_j\in W^\perp$ since j>k. Hence $v\not\in W^{\perp\perp}$ and we have $W^{\perp\perp} \subseteq W.$

The case for X is obtained by replacing V with V* and identifying $V^{**} \cong V$ . ♦

Thus we get the following correspondence:

Furthermore, the dimensions “match”. E.g. suppose dim(V) = n, so dim(V*) = n. Then we claim that for any subspace W of V of dimension k,

$\dim(W^{\perp}) = n-k$ ;
$V^* / W^\perp \cong W^*$ naturally.

Since dim(W*) = dim(W) = k, the first statement follows from the second. From results above, the inclusion map W → V induces a map of the dual spaces V* → W*. The kernel of this map is precisely the set of all $f\in V^*$ such that f(w) = 0 for all w in W, which is exactly $W^\perp.$ This proves our claim. ♦