Linear Measurements, Transposes, and Topological Features
In earlier chapters, we usually treated vectors as the main objects:
\[ x=\begin{bmatrix}x_1\\ \vdots\\ x_n\end{bmatrix}. \]
But in many applications we do not observe a vector directly. We observe measurements of it.
A sensor records a number.
A linear constraint checks whether a vector satisfies an inequality.
A statistic extracts a score from data.
A cochain in topology assigns values to chains.
The common form is
\[ \ell(x)=a_1x_1+\cdots+a_nx_n. \]
This chapter develops the linear algebra of such measurements. The main idea is:
A vector space can be studied not only by its vectors, but also by all linear measurements on those vectors.
This viewpoint leads to dual spaces, dual bases, transpose maps, annihilators, Riesz representation, dual norms, and finally cohomology.
Let \(V\) be a vector space over a field \(\mathbb F\). The dual space of \(V\) is
\[ V^*=\operatorname{Hom}(V,\mathbb F), \]
the vector space of all linear maps from \(V\) to \(\mathbb F\).
An element \(\ell\in V^*\) is called a linear functional.
A functional \(\ell:V\to \mathbb F\) satisfies
\[ \ell(au+bv)=a\ell(u)+b\ell(v) \]
for all \(u,v\in V\) and all scalars \(a,b\in\mathbb F\).
Every row vector
\[ a^T=\begin{bmatrix}a_1&\cdots&a_n\end{bmatrix} \]
defines a linear functional on \(\mathbb R^n\) by
\[ \ell_a(x)=a^Tx=a_1x_1+\cdots+a_nx_n. \]
After choosing the standard basis, we identify
\[ (\mathbb R^n)^*\cong \mathbb R^{1\times n}. \]
This is why ordinary vectors are often columns, while dual vectors are often rows.
Let
\[ P_2=\{a+bt+ct^2:a,b,c\in\mathbb R\}. \]
The following are linear functionals on \(P_2\):
\[ \ell_1(p)=p(0),\qquad \ell_2(p)=p(1),\qquad \ell_3(p)=\int_0^1 p(t)\,dt. \]
Evaluation and integration preserve linear combinations, so they are linear measurements.
For \(V=\mathbb R^{m\times n}\) and fixed \(B\in\mathbb R^{m\times n}\), define
\[ \ell_B(A)=\operatorname{tr}(B^TA). \]
This is a linear functional on the matrix space. It is the matrix analogue of a dot product.
Let \(V\) be an \(n\)-dimensional vector space with basis
\[ \mathcal B=\{v_1,\ldots,v_n\}. \]
The dual basis of \(V^*\) is the collection
\[ \mathcal B^*=\{\varphi^1,\ldots,\varphi^n\} \]
defined by
\[ \varphi^i(v_j)=\delta_{ij} = \begin{cases} 1,&i=j,\\ 0,&i\ne j. \end{cases} \]
The dual basis extracts coordinates. If
\[ x=c_1v_1+\cdots+c_nv_n, \]
then
\[ \varphi^i(x)=c_i. \]
So coordinate functions are linear functionals.
If \(V\) is finite-dimensional and \(\mathcal B=\{v_1,\ldots,v_n\}\) is a basis of \(V\), then the dual basis
\[ \mathcal B^*=\{\varphi^1,\ldots,\varphi^n\} \]
is a basis of \(V^*\). In particular,
\[ \dim V^*=\dim V. \]
Let \(\ell\in V^*\). For any vector
\[ x=\sum_{j=1}^n c_jv_j, \]
linearity gives
\[ \ell(x)=\sum_{j=1}^n c_j\ell(v_j). \]
But
\[ \left(\sum_{j=1}^n \ell(v_j)\varphi^j\right)(x) = \sum_{j=1}^n \ell(v_j)c_j. \]
Therefore
\[ \ell=\sum_{j=1}^n \ell(v_j)\varphi^j, \]
so the dual basis spans \(V^*\).
Now suppose
\[ a_1\varphi^1+\cdots+a_n\varphi^n=0. \]
Evaluating at \(v_j\) gives \(a_j=0\). Therefore the dual basis is linearly independent. Hence it is a basis.
Let
\[ v_1=\begin{bmatrix}1\\1\end{bmatrix}, \qquad v_2=\begin{bmatrix}1\\2\end{bmatrix}. \]
Let
\[ P=[v_1\ v_2] = \begin{bmatrix} 1&1\\ 1&2 \end{bmatrix}. \]
The rows of \(P^{-1}\) form the dual basis:
\[ P^{-1} = \begin{bmatrix} 2&-1\\ -1&1 \end{bmatrix}. \]
Therefore
\[ \varphi^1(x)=\begin{bmatrix}2&-1\end{bmatrix}x, \qquad \varphi^2(x)=\begin{bmatrix}-1&1\end{bmatrix}x. \]
These satisfy
\[ \varphi^1(v_1)=1,\quad \varphi^1(v_2)=0,\quad \varphi^2(v_1)=0,\quad \varphi^2(v_2)=1. \]
import numpy as np
P = np.array([[1, 1],
[1, 2]], dtype=float)
Pinv = np.linalg.inv(P)
print("P =")
print(P)
print("\nP^{-1} =")
print(Pinv)
print("\nRows of P^{-1} are the dual basis functionals.")
print("\nP^{-1} P =")
print(Pinv @ P)P =
[[1. 1.]
[1. 2.]]
P^{-1} =
[[ 2. -1.]
[-1. 1.]]
Rows of P^{-1} are the dual basis functionals.
P^{-1} P =
[[1. 0.]
[0. 1.]]The identity matrix confirms that the \(i\)th row of \(P^{-1}\) evaluates to \(1\) on \(v_i\) and \(0\) on the other basis vector.
Let \(T:V\to W\) be a linear map. The dual map, or pullback, of \(T\) is
\[ T^*:W^*\to V^* \]
defined by
\[ T^*(\ell)=\ell\circ T. \]
Equivalently,
\[ (T^*\ell)(v)=\ell(Tv). \]
The direction reverses:
\[ V\xrightarrow{\ T\ }W, \qquad W^*\xrightarrow{\ T^*\ }V^*. \]
Suppose \(T:V\to W\) has matrix \(A\) with respect to chosen bases of \(V\) and \(W\). Then the dual map \(T^*:W^*\to V^*\) is represented by the transpose matrix \(A^T\) when functionals are represented by column coordinate vectors.
Write vectors as columns and functionals as row vectors. If \(T\) has matrix \(A\), then
\[ Tx=Ax. \]
Let \(\ell\in W^*\) be represented by a row vector \(y^T\). Then
\[ (T^*\ell)(x)=\ell(Tx)=y^T(Ax)=(y^TA)x. \]
Thus \(T^*\ell\) is represented by the row vector \(y^TA\). If functionals are represented as column coordinate vectors, this becomes multiplication by \(A^T\).
For complex vector spaces, distinguish:
\[ A^*=\overline A^T. \]
Let \(S\subseteq V\). The annihilator of \(S\) is
\[ S^\circ=\{\ell\in V^*:\ell(s)=0\text{ for all }s\in S\}. \]
If \(U\le V\) is a subspace, then \(U^\circ\) is a subspace of \(V^*\).
An annihilator is the space of all linear measurements that vanish on a given subspace.
Let \(V\) be finite-dimensional and let \(U\le V\). Then
\[ \dim U^\circ=\dim V-\dim U. \]
Choose a basis \(\{u_1,\ldots,u_k\}\) of \(U\) and extend it to a basis
\[ \{u_1,\ldots,u_k,v_{k+1},\ldots,v_n\} \]
of \(V\). Let
\[ \{\varphi^1,\ldots,\varphi^k,\varphi^{k+1},\ldots,\varphi^n\} \]
be the dual basis. A functional vanishes on \(U\) if and only if it has no components in \(\varphi^1,\ldots,\varphi^k\). Hence
\[ U^\circ=\operatorname{span}\{\varphi^{k+1},\ldots,\varphi^n\}. \]
Therefore
\[ \dim U^\circ=n-k=\dim V-\dim U. \]
Let \(T:V\to W\) be linear. Then
\[ \ker(T^*)=(\operatorname{im}T)^\circ. \]
Let \(\ell\in W^*\). Then
\[ \ell\in\ker(T^*) \Longleftrightarrow T^*\ell=0. \]
This means
\[ (T^*\ell)(v)=\ell(Tv)=0 \]
for all \(v\in V\). Equivalently, \(\ell\) vanishes on every vector in \(\operatorname{im}T\). Therefore
\[ \ell\in(\operatorname{im}T)^\circ. \]
For a matrix \(A\in\mathbb R^{m\times n}\),
\[ \operatorname{Null}(A^T)=(\operatorname{Col}(A))^\circ. \]
After identifying \((\mathbb R^m)^*\) with \(\mathbb R^m\) using the dot product, this becomes
\[ \operatorname{Null}(A^T)=\operatorname{Col}(A)^\perp. \]
import sympy as sp
A = sp.Matrix([
[1, 2],
[2, 4],
[1, 1]
])
print("A =")
sp.pretty_print(A)
print("\nColumn space basis:")
for v in A.columnspace():
sp.pretty_print(v)
print("\nAnnihilator of Col(A) = Null(A.T):")
for v in A.T.nullspace():
sp.pretty_print(v)
print("\nRank(A) =", A.rank())
print("dim Col(A) =", A.rank())
print("dim annihilator =", len(A.T.nullspace()))
print("ambient dimension =", A.rows)A =
⎡1 2⎤
⎢ ⎥
⎢2 4⎥
⎢ ⎥
⎣1 1⎦
Column space basis:
⎡1⎤
⎢ ⎥
⎢2⎥
⎢ ⎥
⎣1⎦
⎡2⎤
⎢ ⎥
⎢4⎥
⎢ ⎥
⎣1⎦
Annihilator of Col(A) = Null(A.T):
⎡-2⎤
⎢ ⎥
⎢1 ⎥
⎢ ⎥
⎣0 ⎦
Rank(A) = 2
dim Col(A) = 2
dim annihilator = 1
ambient dimension = 3The double dual of \(V\) is
\[ V^{**}=(V^*)^*. \]
There is a natural map
\[ \iota:V\to V^{**} \]
defined by
\[ \iota(v)(\ell)=\ell(v). \]
Thus \(\iota(v)\) is a linear functional on \(V^*\): it takes a measurement \(\ell\) and returns the measured value \(\ell(v)\).
If \(V\) is finite-dimensional, then the natural map
\[ \iota:V\to V^{**} \]
is an isomorphism.
The map \(\iota\) is linear. If \(\iota(v)=0\), then
\[ \ell(v)=0 \]
for every \(\ell\in V^*\). If \(v\ne 0\), extend \(v\) to a basis of \(V\) and choose the dual basis functional that sends \(v\) to \(1\). This contradicts \(\ell(v)=0\) for all \(\ell\).
Hence \(\iota\) is injective. Since
\[ \dim V=\dim V^*=\dim V^{**}, \]
an injective linear map from \(V\) to \(V^{**}\) is also surjective. Therefore \(\iota\) is an isomorphism.
In a finite-dimensional inner product space, each vector defines a linear functional by taking inner products.
Let \(V\) be a finite-dimensional real inner product space. For every \(\ell\in V^*\), there exists a unique vector \(a\in V\) such that
\[ \ell(x)=\langle x,a\rangle \]
for all \(x\in V\).
Let \(\{e_1,\ldots,e_n\}\) be an orthonormal basis of \(V\). Define
\[ a=\sum_{i=1}^n \ell(e_i)e_i. \]
If \(x=\sum_i x_ie_i\), then
\[ \langle x,a\rangle = \left\langle \sum_i x_ie_i,\sum_j\ell(e_j)e_j\right\rangle = \sum_i x_i\ell(e_i) = \ell(x). \]
For uniqueness, suppose
\[ \langle x,a\rangle=\langle x,b\rangle \]
for all \(x\). Then
\[ \langle x,a-b\rangle=0 \]
for all \(x\). Taking \(x=a-b\) gives
\[ \|a-b\|^2=0, \]
so \(a=b\).
Let \((V,\|\cdot\|)\) be a finite-dimensional normed vector space. The dual norm on \(V^*\) is
\[ \|\ell\|_*=\sup_{x\ne 0}\frac{|\ell(x)|}{\|x\|} = \sup_{\|x\|\le 1}|\ell(x)|. \]
For all \(\ell\in V^*\) and \(x\in V\),
\[ |\ell(x)|\le \|\ell\|_*\|x\|. \]
If \(x=0\), the result is immediate. If \(x\ne0\), then
\[ \left\|\frac{x}{\|x\|}\right\|=1. \]
By definition of the dual norm,
\[ \left|\ell\left(\frac{x}{\|x\|}\right)\right|\le \|\ell\|_*. \]
Multiplying by \(\|x\|\) gives
\[ |\ell(x)|\le \|\ell\|_*\|x\|. \]
For finite-dimensional \(\mathbb R^n\), if
\[ \ell_a(x)=a^Tx, \]
then the dual of the \(p\)-norm is the \(q\)-norm, where
\[ \frac1p+\frac1q=1. \]
Important examples:
\[ (\ell^1)^*=\ell^\infty,\qquad (\ell^2)^*=\ell^2,\qquad (\ell^\infty)^*=\ell^1. \]
Homology studies cycles modulo boundaries. Cohomology studies linear measurements on chains modulo trivial measurements.
A chain complex is a sequence
\[ \cdots \xrightarrow{\partial_{p+2}} C_{p+1} \xrightarrow{\partial_{p+1}} C_p \xrightarrow{\partial_p} C_{p-1} \xrightarrow{\partial_{p-1}}\cdots \]
such that
\[ \partial_p\partial_{p+1}=0. \]
The \(p\)th cochain space is the dual vector space
\[ C^p=C_p^*=\operatorname{Hom}(C_p,\mathbb F). \]
An element \(\alpha\in C^p\) is called a \(p\)-cochain.
A \(p\)-cochain assigns a scalar to each \(p\)-chain.
The \(p\)th coboundary map is
\[ \delta^p:C^p\to C^{p+1} \]
defined by
\[ (\delta^p\alpha)(c)=\alpha(\partial_{p+1}c). \]
Equivalently,
\[ \delta^p=\partial_{p+1}^*. \]
Thus, if \(D_p\) is the matrix of \(\partial_p\), then the matrix of \(\delta^{p-1}\) is
\[ D_p^T. \]
For every \(p\),
\[ \delta^{p+1}\delta^p=0. \]
Let \(\alpha\in C^p\) and let \(c\in C_{p+2}\). Then
\[ ((\delta^{p+1}\delta^p)\alpha)(c) = (\delta^p\alpha)(\partial_{p+2}c) = \alpha(\partial_{p+1}\partial_{p+2}c). \]
Since \(\partial_{p+1}\partial_{p+2}=0\), this equals \(0\). Hence
\[ \delta^{p+1}\delta^p=0. \]
The space of \(p\)-cocycles is
\[ Z^p=\ker\delta^p. \]
The space of \(p\)-coboundaries is
\[ B^p=\operatorname{im}\delta^{p-1}. \]
The \(p\)th cohomology group over \(\mathbb F\) is
\[ H^p=Z^p/B^p=\ker\delta^p/\operatorname{im}\delta^{p-1}. \]
| Homology | Cohomology |
|---|---|
| chains \(C_p\) | cochains \(C^p=C_p^*\) |
| boundary matrix \(D_p\) | coboundary matrix \(D_p^T\) |
| cycles \(\ker D_p\) | cocycles \(\ker D_{p+1}^T\) |
| boundaries \(\operatorname{im}D_{p+1}\) | coboundaries \(\operatorname{im}D_p^T\) |
| \(H_p=\ker D_p/\operatorname{im}D_{p+1}\) | \(H^p=\ker D_{p+1}^T/\operatorname{im}D_p^T\) |
If coefficients are taken in a field \(\mathbb F\), then
\[ \dim H^p=\dim H_p. \]
Let \(n_p=\dim C_p\). Since \(\delta^p\) has matrix \(D_{p+1}^T\),
\[ \dim\ker\delta^p = \dim\ker D_{p+1}^T = n_p-\operatorname{rank}(D_{p+1}^T) = n_p-\operatorname{rank}(D_{p+1}). \]
Also,
\[ \dim\operatorname{im}\delta^{p-1} = \operatorname{rank}(D_p^T) = \operatorname{rank}(D_p). \]
Therefore
\[ \dim H^p = \dim\ker\delta^p-\dim\operatorname{im}\delta^{p-1} = n_p-\operatorname{rank}(D_{p+1})-\operatorname{rank}(D_p). \]
For homology,
\[ \dim H_p = \dim\ker D_p-\dim\operatorname{im}D_{p+1} = (n_p-\operatorname{rank}D_p)-\operatorname{rank}D_{p+1}. \]
The two formulas are equal.
Consider vertices \(\{1,2,3\}\) and edges
\[ [1,2],\qquad [2,3],\qquad [1,3], \]
with no filled triangle.
Choose oriented edges
\[ e_1=[1,2],\qquad e_2=[2,3],\qquad e_3=[1,3]. \]
The boundary matrix \(\partial_1:C_1\to C_0\) is
\[ D_1= \begin{bmatrix} -1&0&-1\\ 1&-1&0\\ 0&1&1 \end{bmatrix}. \]
Since there are no \(2\)-simplices, \(D_2\) is an empty matrix. The coboundary map \(\delta^0:C^0\to C^1\) has matrix
\[ D_1^T= \begin{bmatrix} -1&1&0\\ 0&-1&1\\ -1&0&1 \end{bmatrix}. \]
Every \(1\)-cochain is a cocycle because \(\delta^1=0\). Since \(\operatorname{rank}D_1=2\),
\[ \dim H^1=3-2=1. \]
The one-dimensional cohomology detects the loop.
Now add the \(2\)-simplex \([1,2,3]\). Then
\[ \partial_2([1,2,3])=[2,3]-[1,3]+[1,2]. \]
In the edge basis \(([1,2],[2,3],[1,3])\),
\[ D_2= \begin{bmatrix} 1\\ 1\\ -1 \end{bmatrix}. \]
The coboundary map \(\delta^1:C^1\to C^2\) has matrix
\[ D_2^T= \begin{bmatrix} 1&1&-1 \end{bmatrix}. \]
A \(1\)-cochain \(\alpha=(a_1,a_2,a_3)\) is a cocycle if
\[ a_1+a_2-a_3=0. \]
Now
\[ \dim H^1=0. \]
The loop is now the boundary of a filled face, so it is no longer a hole.
import sympy as sp
def betti_numbers(D1, D2, n0, n1, n2):
r1 = D1.rank()
r2 = D2.rank()
beta0 = n0 - r1
beta1 = n1 - r1 - r2
beta2 = n2 - r2
return beta0, beta1, beta2, r1, r2
D1 = sp.Matrix([
[-1, 0, -1],
[ 1,-1, 0],
[ 0, 1, 1]
])
# Hollow triangle: no 2-simplex
D2_hollow = sp.zeros(3, 0)
print("Hollow triangle:")
print(betti_numbers(D1, D2_hollow, n0=3, n1=3, n2=0))
# Filled triangle: one 2-simplex
D2_filled = sp.Matrix([[1], [1], [-1]])
print("Filled triangle:")
print(betti_numbers(D1, D2_filled, n0=3, n1=3, n2=1))
print("\nCoboundary matrices for filled triangle:")
print("delta^0 = D1.T")
sp.pretty_print(D1.T)
print("delta^1 = D2.T")
sp.pretty_print(D2_filled.T)Hollow triangle:
(1, 1, 0, 2, 0)
Filled triangle:
(1, 0, 0, 2, 1)
Coboundary matrices for filled triangle:
delta^0 = D1.T
⎡-1 1 0⎤
⎢ ⎥
⎢0 -1 1⎥
⎢ ⎥
⎣-1 0 1⎦
delta^1 = D2.T
[1 1 -1]In topological data analysis, one often builds a filtration
\[ K_0\subseteq K_1\subseteq\cdots\subseteq K_m \]
from a point cloud.
Homology tracks cycles that appear and disappear. Cohomology gives a dual measurement-based view of the same information.
For example, if a data set has circular structure, a nonzero class in \(H^1\) can be used to construct a circular coordinate on the data.
Let
\[ v_1=\begin{bmatrix}1\\2\end{bmatrix}, \qquad v_2=\begin{bmatrix}3\\5\end{bmatrix}. \]
Find the dual basis functionals \(\varphi^1,\varphi^2\).
Let
\[ P=\begin{bmatrix}1&3\\2&5\end{bmatrix}. \]
Then
\[ P^{-1} = \begin{bmatrix} -5&3\\ 2&-1 \end{bmatrix}. \]
Therefore
\[ \varphi^1(x)=\begin{bmatrix}-5&3\end{bmatrix}x, \qquad \varphi^2(x)=\begin{bmatrix}2&-1\end{bmatrix}x. \]
Let \(A\in\mathbb R^{m\times n}\). Explain why
\[ \operatorname{Null}(A)=\operatorname{Row}(A)^\circ. \]
Each row of \(A\) is a linear functional on \(\mathbb R^n\). A vector \(x\) belongs to \(\operatorname{Null}(A)\) exactly when every row of \(A\) gives measurement zero on \(x\). Thus \(x\) is annihilated by all row measurements, so
\[ \operatorname{Null}(A)=\operatorname{Row}(A)^\circ. \]
Explain why the hollow triangle has \(\dim H^1=1\) but the filled triangle has \(\dim H^1=0\).
For the hollow triangle, there are three edge cochains and no \(2\)-simplex, so every \(1\)-cochain is a cocycle. Since \(\operatorname{rank}D_1=2\),
\[ \dim H^1=3-2=1. \]
For the filled triangle, the \(2\)-simplex introduces one independent condition
\[ a_1+a_2-a_3=0 \]
on \(1\)-cocycles. Then \(\dim Z^1=2\) and \(\dim B^1=2\), so
\[ \dim H^1=0. \]
Let \(V=\mathbb R^3\) and let \(U=\operatorname{span}\{(1,1,0),(0,1,1)\}\). Find \(\dim U^\circ\).
The two spanning vectors are linearly independent, so \(\dim U=2\). Since \(\dim V=3\),
\[ \dim U^\circ=3-2=1. \]
Find a nonzero linear functional that vanishes on
\[ U=\operatorname{span}\{(1,1,0),(0,1,1)\}. \]
Let \(\ell(x,y,z)=ax+by+cz\). We need
\[ a+b=0,\qquad b+c=0. \]
Choose \(b=1\). Then \(a=-1\) and \(c=-1\). Thus one answer is
\[ \ell(x,y,z)=-x+y-z. \]
Let
\[ A=\begin{bmatrix} 1&2\\ 2&4\\ 1&1 \end{bmatrix}. \]
Find a basis for \(\operatorname{Null}(A^T)\).
Solve
\[ A^Ty=0. \]
That is,
\[ \begin{bmatrix} 1&2&1\\ 2&4&1 \end{bmatrix} \begin{bmatrix}y_1\\y_2\\y_3\end{bmatrix}=0. \]
The equations are
\[ y_1+2y_2+y_3=0, \qquad 2y_1+4y_2+y_3=0. \]
Subtract twice the first equation from the second:
\[ -y_3=0, \]
so \(y_3=0\). Then \(y_1=-2y_2\). A basis is
\[ \begin{bmatrix}-2\\1\\0\end{bmatrix}. \]
For the filled triangle, compute \(\dim H^0\) and \(\dim H^1\).
The filled triangle is connected, so
\[ \dim H^0=1. \]
The loop is filled by a \(2\)-simplex, so
\[ \dim H^1=0. \]
Using matrices,
\[ \beta_0=n_0-\operatorname{rank}D_1=3-2=1, \]
and
\[ \beta_1=n_1-\operatorname{rank}D_1-\operatorname{rank}D_2 =3-2-1=0. \]
Use an AI assistant as a study partner, not as a replacement for your reasoning.
Duality turns vectors into measurements. The key ideas are:
From an applied linear algebra perspective, cohomology is not mysterious: it is a systematic way to measure topological features using linear functionals.