Vector spaces

Lecture 11

Author

Affiliation

Minjae Park

Auburn University
MATH 2660 - Spring 2026

Published

February 4, 2026

Recap

Transition matrix as redistribution

We can interpret the transition matrix
\(P = [\vec{p}_1 \ \cdots \ \vec{p}_N]\) of a Markov chain as a redistribution rule.
Suppose there are \(N\) types of an object, and let
\(\vec{\lambda}_0 = \langle \lambda_1, \dots, \lambda_N \rangle\) represent the number of objects of each type.
After one unit of time, each type-\(m\) object changes type according to the proportions \(\vec{p}_m = \langle p_{1m}, \dots, p_{Nm} \rangle\).
Since \(p_{1m} + \cdots + p_{Nm} = 1\), the total number of objects is preserved. In particular, if there are \(\lambda_m\) objects of type \(m\), then \(p_{1m}\lambda_m\) become type \(1\), \(p_{2m}\lambda_m\) become type \(2\), and so on.
Thus, the transition matrix redistributes the total amount \(\lambda_m\) of type \(m\) among all types.

Transition matrix as redistribution

Recall that \(P = [\vec{p}_1 \ \cdots \ \vec{p}_N]\), viewed as a linear transformation, sends \(\vec{e}_m\) (one object of type \(m\)) to \(\vec{p}_m\). This exactly matches our redistribution model.
After one unit of time, the numbers of each type are given by \[ \vec{\lambda}_1 = P \vec{\lambda}_0. \]
After \(k\) time units, the numbers of each type are given by \[ \vec{\lambda}_k = P^k \vec{\lambda}_0. \] The total amount remains unchanged.

Example

Suppose there are three chemicals: \(100\) grams of A, \(100\) grams of B, and \(0\) grams of C.
Every minute:
- \(20\%\) of A transforms into B, \(30\%\) transforms into C, and the rest remains A;
- \(50\%\) of B transforms into A, and the rest remains B;
- \(30\%\) of C transforms into A, and \(70\%\) transforms into B.
Find the column transition matrix.
Write a formula using matrix multiplication to compute the amounts of A, B, and C after \(10\) minutes.

Answer

Since we are using the column convention, the \(m\)th column lists the fractions that go from type \(m\) to types \(A,B,C\) (in that row order).
From A: \(50\%\) stays A, \(20\%\) goes to B, \(30\%\) goes to C, so \[ \vec p_A=\begin{pmatrix}0.5\\0.2\\0.3\end{pmatrix}. \]
From B: \(50\%\) goes to A, \(50\%\) stays B, \(0\%\) goes to C, so \[ \vec p_B=\begin{pmatrix}0.5\\0.5\\0\end{pmatrix}. \]
From C: \(30\%\) goes to A, \(70\%\) goes to B, \(0\%\) stays C, so \[ \vec p_C=\begin{pmatrix}0.3\\0.7\\0\end{pmatrix}. \]
Therefore the column transition matrix is \[ P=\begin{pmatrix} 0.5 & 0.5 & 0.3\\ 0.2 & 0.5 & 0.7\\ 0.3 & 0 & 0 \end{pmatrix}. \]
Let \[ \vec{\lambda}_0=\begin{pmatrix}100\\100\\0\end{pmatrix} \] be the initial amounts (in grams). Then after \(10\) minutes, \[ \vec{\lambda}_{10}=P^{10}\vec{\lambda}_0 \approx \langle 94.34, 77.36, 28.30 \rangle. \]
Check: \(94.34+77.36+28.30=200.00\), so total mass is preserved.

Markov chain and probabilities

In a Markov chain, there is a single object whose type changes randomly according to given transition probabilities.
You can think of the object’s total “presence” as \(100\%\) (or \(1\)), which is redistributed among the states at each step.
This leads to a probability vector, whose entries always sum to \(1\), since total probability is conserved.
As before, the probability vector evolves according to \[ \vec{\pi}_k = P^k\,\vec{\pi}_0. \]
The \(n\)th entry of \(\vec{\pi}_k\) can be interpreted as
“the probability that the object is of type \(n\) after \(k\) time steps, given the initial probability distribution \(\vec{\pi}_0\).”

Abstract Vector Spaces

Motivation

So far, we have discussed vectors in \(\mathbb{R}^n\), which can be represented using coordinates.
However, we often wrote vectors abstractly as \(\vec v\), without referring to coordinates, and discussed concepts such as linear combinations \(\vec w = c_1\vec v_1 + c_2\vec v_2\) or linear transformations \(\vec w = A\vec v\).
In fact, linear combinations can be defined for many objects that do not live in Euclidean space.
This allows us to study linear algebra on more abstract spaces, even though each “vector” may not look like an arrow in \(\mathbb{R}^n\).

Examples of Linear Combinations

Let \(c_1, c_2\) be real numbers.

If \(A, B\) are \(3\times 3\) matrices, then \[C = c_1 A + c_2 B\] is also a \(3\times 3\) matrix.
If \(x, y\) are complex numbers, then \[z = c_1 x + c_2 y\] is also a complex number.

If \(A(x), B(x)\) are polynomials, then \[C(x) = c_1 A(x) + c_2 B(x)\] is also a polynomial.
If \(A(x), B(x)\) are continuous functions, then \[C(x) = c_1 A(x) + c_2 B(x)\] is also a continuous function.
If \(A(x), B(x)\) are twice-differentiable functions, then \[C(x) = c_1 A(x) + c_2 B(x)\] is also a twice-differentiable function.

Vector Space

A vector space is a “safe ground” where linear combinations are always allowed and stay inside the space.
To even talk about linear combinations, we must be able to add vectors and multiply them by scalars.
Moreover, the space must be closed under these operations: performing them should never take us outside the space.

Definition of a Vector Space

Let \(V\) be a set equipped with two operations: vector addition and scalar multiplication. If the following axioms hold for all \(\vec u,\vec v,\vec w \in V\) and all scalars \(c,d \in \mathbb{R}\), then \(V\) is called a vector space.

(Addition axioms)

Closure under addition. \[\vec u + \vec v \in V.\]
Commutative property. \[\vec u + \vec v = \vec v + \vec u.\]
Associative property. \[\vec u + (\vec v + \vec w) = (\vec u + \vec v) + \vec w.\]
Additive identity. There exists a vector \(\vec 0 \in V\) such that \[\vec u + \vec 0 = \vec u.\]
Additive inverse. For every \(\vec u \in V\), there exists \(-\vec u \in V\) such that \[\vec u + (-\vec u) = \vec 0.\]

(Scalar multiplication axioms)

Closure under scalar multiplication. \[c\vec u \in V.\]
Distributive property (over vector addition). \[c(\vec u + \vec v) = c\vec u + c\vec v.\]
Distributive property (over scalar addition). \[(c + d)\vec u = c\vec u + d\vec u.\]
Associative property. \[c(d\vec u) = (cd)\vec u.\]
Scalar identity. \[1\vec u = \vec u.\]

Example

Let \[ V = \{a_3x^3 + a_2x^2 + a_1x + a_0 \mid a_0,a_1,a_2,a_3 \in \mathbb{R}\}, \] the set of all polynomials of degree at most \(3\). We use the usual polynomial addition and scalar multiplication.

We check that \(V\) is a vector space:

Closure under addition: The sum of two degree-\(\le 3\) polynomials is still degree \(\le 3\).
Commutativity: Polynomial addition is commutative.
Associativity: Polynomial addition is associative.
Additive identity: The zero polynomial \(0(x)=0\) satisfies \(p(x)+0(x)=p(x)\).
Additive inverse: For any \(p(x)\in V\), the polynomial \(-p(x)\) is also in \(V\) and \(p(x)+(-p(x))=0\).
Closure under scalar multiplication: Multiplying a degree-\(\le 3\) polynomial by a scalar does not increase its degree.
Distributivity over vector addition: \[c(p(x)+q(x)) = cp(x)+cq(x).\]
Distributivity over scalar addition: \[(c+d)p(x) = cp(x)+dp(x).\]
Associativity of scalar multiplication: \[c(dp(x)) = (cd)p(x).\]
Scalar identity: \[1\cdot p(x) = p(x).\]

Therefore, \(V\) is a vector space.

Non-examples

The set of fruits is not a vector space, since there is no meaningful way to define vector addition or scalar multiplication.
The set of integers \(\mathbb{Z}\) is not a vector space because it is not closed under scalar multiplication. For example, \(1 \in \mathbb{Z}\) but \(0.5 \cdot 1 \notin \mathbb{Z}\).
Let \(V=\{(x,y)\in\mathbb{R}^2 : x \ge 0,\ y \ge 0\}\). This set is not a vector space because it is not closed under scalar multiplication (e.g. multiplying by \(-1\) leaves the set).
Let \(V=\{(x,y)\in\mathbb{R}^2 : xy = 0\}\). This set is not a vector space because it is not closed under vector addition: \[(1,0)\in V,\ (0,1)\in V,\quad \text{but } (1,0)+(0,1)=(1,1)\notin V.\]

Definition of a Subspace

Let \(V\) be a vector space. A subspace \(W\) is a vector space contained inside \(V\).
Since vector addition and scalar multiplication are already defined in \(V\), to check whether \(W\) is a subspace we only need to verify closure.
A subset \(W \subset V\) is a subspace if for all \(\vec u,\vec v \in W\) and all \(c\in\mathbb{R}\), \[\vec u + \vec v \in W, \qquad c\vec u \in W.\]

Examples

Let \(a,b,c \in \mathbb{R}\) be fixed.

The zero set \(W=\left\{\langle 0,0 \rangle\right\}\) is a subspace of \(V=\mathbb{R}^2\).
The set of \(4\times 4\) upper triangular matrices is a subspace of the space of all \(4\times 4\) matrices \(V=M_{4\times 4}\).
A line through the origin \(W=\{t\langle a,b \rangle : t\in\mathbb{R}\}\) is a subspace of \(V=\mathbb{R}^2\), since it is closed under vector addition and scalar multiplication.
The plane \(W=\{(x,y,z)\in\mathbb{R}^3 : ax+by+cz=0\}\) is a subspace of \(V=\mathbb{R}^3\); we check this on the next slide.

Checking the plane example

Let \(\vec u=(x_1,y_1,z_1)\) and \(\vec v=(x_2,y_2,z_2)\) be in \(W\). Then \[ax_1+by_1+cz_1=0,\qquad ax_2+by_2+cz_2=0.\]
Adding, \[a(x_1+x_2)+b(y_1+y_2)+c(z_1+z_2)=0,\] so \(\vec u+\vec v\in W\).
For any \(d\in\mathbb{R}\), \[a(dx_1)+b(dy_1)+c(dz_1)=d(ax_1+by_1+cz_1)=0,\] so \(d\vec u\in W\).
Therefore, \(W\) is a subspace of \(\mathbb{R}^3\).