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 vi‘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 fV → 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 fV → 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 fM → 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 fM → N is expressed as a matrix as follows:


Example 1

Take RR, 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 fM → 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 = {ab√2 : ab integers} which is a Z-module. Take fM → 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 fV → W depends on our choice of bases for V and W. If VW, 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(MR) 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(NR) 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.


Consider the map V^* \times V\to D which takes (fv) 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 fV → 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:



It’s tempting to define a module structure on Hom(MR) via (f\cdot r)(m) = f(rm). What’s wrong with this definition? [ Answer: the resulting f·r : MR 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.


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.


[ 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 TV → 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 W → D to the composition 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.


  • 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.


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. ♦

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