path: root/mathematics/linear-algebra.md

---
layout: algebra
title: mathematics/linear algebra
---

# Linear Algebra

$$ℝ^n$$

## Fields and Vector Spaces

A **field** $F$ is a *set* with two *binary operation* $+$ and $×$ satisfying the following axioms:
- $(F, +)$ is a *commutative group*:
  - associativity: $∀a,b,c : a + (b + c) = (a + b) + c$
  - additive identity: $∃0, ∀a : 0 + a = a + 0 = a$
  - additive inverse: $∀a, ∃-a : a + (-a) = 0$
  - commutativity: $∀a,b : a + b = b + a$
- $(F, ×)$ is a *commutative group*
  - associativity: $∀a,b,c : a(bc) = (ab)c$
  - multiplicative identity: $∃1, ∀a : 1a = a1 = a$
  - multiplicative inverse: $∀a ≠ 0, ∃\frac{1}{a} : a\frac{1}{a} = 1$
  - commutativity: $∀a,b : ab = ba$
- $×$ is distributive with respect to $+$
  - distributivity: $∀a,b,c : a(b + c) = (ab) + (ac)$

Intuitively, a field is a set on which addition, subtraction, multiplication and division are defined and behave as they do on $ℝ$. We often omit the multiplication sign, and write $a × a$ as simply $aa$. It can also be thought of as a *commutative ring* with a multiplicative inverse (sans 0).

A **vector space** $V$ *over* a field $F$ is a set with a binary operation $+$ and a binary function satisfying the following axioms:
- $(V, +)$ is a *commutative group*:
  - associativity: $∀u,v,w : u + (v + w) = (u + v) + w$
  - additive identity: $∃0, ∀v : 0 + v = v + 0 = v$
  - additive inverse: $∀v, ∃-v : v + (-v) = 0$
  - commutativity: $∀u,v : u + v = v + u$
- $(V,)$ is a *scalar operation*
  - scalar identity: $∃1 ∈ F, ∀v : 1v = v1 = v$
  - commutativity: $∀a,b ∈ F, ∀v : (ab)v = a(bv)$
- The *distributive laws* hold:
  - $∀a ∈ F, ∀u,v ∈ V : a(u + v) = au + av$
  - $∀a,b ∈ F, ∀v ∈ V : (a + b)v = av + bv$

Our definition of our vector space leads us to some facts:
- The zero vector is *unique* and always present.
  - Proof: Suppose there were two zero vectors: $0$ and $0'$. Then $0' = 0 + 0' = 0' + 0 = 0$.
- Vector spaces are *non-empty*.
  - Proof: By definition, the zero vector exists.
- The additive inverse for some $x$ is *unique*.
  - Proof: Suppose there were two inverses: $-x$ and $-x'$. Then $-x + x = 0$ and $-x + x + -x' = -x'$, and so as $x + -x' = 0$ $-x = -x'$.
- If $V$ is a vector space over $F$ and $V ≠ \\{0\\}$, then $V$ is an *infinite set* over $F$.
  - Proof: you can just keep adding things

<details markdown="block">
<summary>Examples</summary>

Let $S = \\{(a_1, a_2) | a_1, a_2 ∈ ℝ\\}$.
For $(a_1, a_2), (b_1, b_2) ∈ S$ and $c ∈ ℝ$, we define:
- $(a_1, a_2) + (b_1, b_2) = (a_1 + b_1, a_2 - b_2)$
- $c(a_1, a_2) = (ca_1, ca_2)$.

This fails commutativity! It is thus not a vector space.

Let $S = \\{(a_1, a_2) | a_1, a_2 ∈ ℝ\\}$. We define:
- $(a_1, a_2) + (b_1, b_2) = (a_1 + b_1)$
- $c(a_1, a_2) = (ca_1, 0)$

This fails the existence of zero! It is thus not a vector space.
</details>

## Subspaces

A subset $W$ of a vector space $V$ over a field 𝔽 is called a **subspace** of $V$ if $W$ is a *vector space* over 𝔽 with the operations of addition and scalar multiplication from $V$.
- .
- .

A subset of $V$ is a **subspace** of V iff:
- the subset is non-empty
- the subset contains the zero vector
- it is closed under addition and multiplication

Let $V$ be a vector space over $F$ and $S$ a nonempty subset of $V$. A vector $v \in V$ is a **linear combination** of vectors $s,t ∈ V$ if there exists a *finite* number of vectors $u_1, u_2, ..., u_n ∈ S$ and scalars $a_1, a_2, ..., a_n ∈ F$ such that $v = a_1 u_1 + a_2 u_2 + ... a_n u_n$.

We call $a_1 ... a_n$ the *coefficients* of the linear combination.



---


## Introduction: The Complex Numbers

A **complex number** is of the form $a + b\text{i}$, where $a, b$ are real numbers and $i$ represents the imaginary base.

We denote the set of complex numbers as $ℂ$.
The complex numbers form a *vector space*.
Every element in the vector space can be expressed as a *linear combination* of $a + bi$.

- ℂ: the set of complex numbers
- ℝ: the set of real numbers
- ℚ: the set of rational numbers

Elementary operations on the complex numbers as are follows:
- $(a + bi) ± (c + di) = a ± c ± (b ± d)$
- $(a + bi)(c + di) = (ac - bd) + (ad + bc)$
- $i^2 = -1$

The **conjugate** of $z = a + bi$ is ... defined as $\bar{z} = a - bi$.
As long as $a^2 + b^2 ≠ 0$, the inverse of $z = a + bi$ is given by $z^{-1} = \frac{a - b}{a^2 + b^2} = \frac{\bar{z}}{a^2 + b^2}$.

**Theorem**: Let $z, w ∈ ℂ$. Then:
- $\bar{\bar{z}} = z$ the conjugate of the conjugate
- $\bar{z ± w} = \bar{z} ± \bar{w}$
- $\bar{zw} = \bar{z}\bar{w}$
- $\bar{\frac{z}{w}} = \frac{\bar{z}}{\bar{w}}$ (if $w ≠ 0$)
  - (where $\frac{z}{w} = z ⋅ w^{-1} = z \frac{1}{w}$)
- $z$ is a *real number* iff $\bar{z} = z$

Let $z = a + bi$, where $a, b ∈ ℝ$. The **absolute value** of $z$ is defined as the real number $\sqrt{a^2 + b^2}$.

This absolute value shares many properties. For $z, w ∈ ℂ$:
- $z \bar{z} = a^2 + b^2 = |z|^2$, where $|z| = \sqrt{a^2 + b^2}$
- $\frac{z}{w} = \frac{|z|}{|w|}$ where $w ≠ 0$
- $|z + w| ≤ |z| + |w|$
- $|z| - |w| ≤ |z + w|$

**The Fundamental Theorem of Algebra**:
Consider the polynomial $p(z) = a_n z^n + a_{n-1}z^{n-1} + ... + a_1 z + a_0$ where all $a$ are complex numbers.
If $n ≥ 1$, then $p(z)$ has a zero (*root*), i.e. there exists a complex number $c$ such tbar $p(c) = 0$.

If $c$ is a root, $\bar{c}$ is also a root!

Let 𝔽 be one of the following sets: ℚ, ℝ, ℂ

https://math.stackexchange.com/questions/3492590/linear-combination-span-independence-and-bases-for-infinite-dimensional-vector

Let $S$ be a nonempty subset of a vector space $V$. The **span** of $S$, denoted $span(S)$, is the set consisting of all linear combination of vectors in S. For convenience, we define $span(∅) = \\{0\\}$.

The span of any subset $S$ of a vector space $V$ is a subspace of $V$.

---

A subspace $S$ over a vector space $V$ is **linearly dependent** if there exists a finite number of distinct vectors $u_1, u_2, ..., u_n ∈ S$ and scalars $a_1, a_2, ..., a_n ∈ F$ such that $a_1, a_2, ..., a_n$ are not all zero and $a_1 u_1 + a_2 u_2 + ... a_n u_n = 0$.

A subset $S$ of a vector space that is not linearly dependent is **linearly independent**.


Example: Consider the following set: $S = \\{(1, 0, 0, -1), (0, 1, 0, -1), (0, 0, 1, -1), (0, 0, 0, 1)\\}$
Assume that $a v_1 + a_2 v_2 + a_3 v_3 + a_4 v_4 = 0$. then...

as the determinant is nonzero, S is linearly independent.


Let $V$ be a vector space, and let $S_1 ⊆ S_2 ⊆ V$. If $S_1$ is linearly dependent, then $S_2$ is linearly dependent. If $S_2$ is linearly independent, then $S_1$ is also linearly independent.

Let $S$ be a linearly independent subset of a vector space $V$, and let $v ∈ V : v ∉ S$. Then $S ∪ \\{v\\}$ is linearly independent iff $v ∈ span(S)$.

A basis $B$ for a vector space $V$ is a *linearly independent* subset of $V$ that *spans* $V$. If $B$ is a basis for $V$, we also say that the vectors of $B$ form a basis for $V$.

Let $V$ be  a vector space and $β = \\{v_1, ..., v_n\\}$ be a subset of V. Then β is a basis for V iff every $v ∈ V$ can be **uniquely expressed** as a linear combination of vectors of β. that is, V can be written in the form $v = a_1 u_1 + a_2 u_2 ... a_n u_n$ for unique scalars a.

If a vector space V is spanned by a finite set S, then some subset of S is a basis of V. So, V has a finite basis.
Proof: If $S = ∅$, then $V = span{S} = span{∅} = \span{𝕆}$ in which case ∅ is a basis for $V$.
If S ≠ ∅, then ∃ u_1 ∈ S : u_1 ≠ 𝕆. and we have two cases: span(u_1) = V we are done...

(Replacement theorem) Let $V$ be a vector space that is spanned by a set G containing exactly n vectors, and let L be a linearly independent subset of V containing exactly m vectors. Then m ≤ n. Moreover, you can find a subset H of G

Let V be a vector space with dimension n.
- any finite spanning set for V contains at least n vectors, and a spanning set for V that contains exactly n vectors is a basis for V.

Theorem 1.4: Let $W$ be a subspace of a finite-dimensional vector space $V$. Then $W$ is also finite-dimensional (and dim W ≤ dim V). Moreover if dim W = dim V, then V = W.

---

## Linear Transformations

Let $V$ and $W$ be vector spaces (over a field $F$).

A function $T: V → W$ is a **linear transformation** from $V$ into $W$ if $∀x,y ∈ V, c ∈ F$ we have $T(cx + y) = cT(x) + T(y)$. Subsequently:
- $T(x + y) = T(x) + T(y)$
- $T(cx) = cT(x)$
- $T(0) = 0$
- $T(\sum_{i=1}^n a_i x_i) = \sum_{i=1}^n a_i T(x_i)$

Let $T: V → W$ be a linear transformation.

The **kernel** (or null space) $N(T)$ of $T$ is the set of all vectors in $V$ such that $T(x) = 0$: that is, $N(T) = \\{ x ∈ V : T(x) = 0 \\}$.

The **image** (or range) $R(T)$ of $T$ is the subset of $W$ consisting of all images (under $T$) of elements of $V$: that is, $R(T) = \\{ T(x) : x ∈ V \\}$.

Theorem: The kernel $N(T)$ and image $R(T)$ are subspaces of $V$ and $W$, respectively.
<details markdown="block">
<summary>Proof</summary>
We shall denote the zero vector of $V$ and $W$ as $0_v$ and $0_w$, respectively.

Let $x,y ∈ N(T)$ and $c ∈ F$. As $T(0_v) = 0_w$, $0_v ∈ N(T)$. Then $T(cx + y) = cT(x) + T(y) = 0_w + 0_w = 0_w$, as $x$ and $y$ are in the null space. Hence any linear combination of $x$ and $y$ in the null space is in the null space. So as $N(T) ⊆ V$ by definition, $N(T)$ is a subspace of $V$. ∎

Let $x,y ∈ R(T)$ and $c ∈ F$. As $T(0_v) = 0_w$, $0_w ∈ R(T)$. Then there exist $v,w ∈ V$ such that $T(v) = x$ and $T(w) = y$. So $T(v + cw) = T(v) + cT(w) = x + cy$. Hence any linear combination of $x$ and $y$ is in the image. So as $R(T) ⊆ W$ by definition, $R(T)$ is a subspace of $W$. ∎

</details>

Theorem: If $β = \\{ v_1, v_2, ... v_n \\}$ is a basis for $V$, then $R(T) = span(\\{ T(v_1), T(v_2), ..., T(v_n) \\})$.
<details markdown="block">
<summary>Proof</summary>

Clearly, $T(v_i) ∈ R(T)$ for all $i$. As $R(T)$ is a subspace of $W$ by the previous theorem, $R(T)$ *contains* $span(\\{ v_1, v_2, ... v_n \\}) = span(T(β))$.

Now consider an arbitrary $w ∈ R(T)$. By definition, there exists a $v ∈ V$ such that $w = T(v)$. As $β$ is a basis for $V$, we can consider $v$ as a linear combination of some basis vectors in $β$. As $T$ is linear, it thus must be the case that $T(v) ∈ span(T(β))$. So $R(T) = span(T(β))$. ∎

</details>

For a finite-dimensional $N(T)$ and $R(T)$: the **nullity** and **rank** of $T$ are the dimensions of $N(T)$ and $R(T)$, respectively.

**Rank-Nullity Theorem**: If $V$ is *finite-dimensional*, then $dim(V) = nullity(T) + rank(T)$.
<details markdown="block">
<summary>Proof</summary>

Let $dim(V) = n$, let $nullity(T) = k$, and let $β_N$ be a basis for $N(T)$. As $N(T) ⊆ V$, we may extend $β_N$ to a basis $β$ for $V$.

We assert that $S = \\{T(v_{k+1}), T(v_{k+2}), ..., T(v_n) \\}$. We must prove this.

...

So $S$ is linearly independent. As it contains $n-k$ linearly independent vectors in $R(T)$, it is a basis for $R(T)$.

</details>

Recall that a *function* definitionally maps *each* element of its domain to *exactly* one element of its codomain.
- A function is **injective** (or **one-to-one**) iff each element of its domain maps to a *distinct* element of its codomain, that is, $f(x) = f(y) → x = y$.
- A function is **surjective** (or **onto**) iff each element of the codomain is mapped to by *at least* one element in the domain, that is, $R(T) = W$.
- A function is **bijective** iff it is surjective and injective. Necessarily, a bijective function is invertible, which will be formally stated & proven later.

Theorem: $T$ is injective iff $N(T) = \\{0\\}$.
<details markdown="block">
<summary>Proof</summary>

Suppose that $T$ is injective. Consider some $x ∈ N(T)$. Then $T(x) = 0 = T(0)$. As $T$ is injective, $x = 0$, and so $N(T) = \\{0\\}$. ∎

Now suppose that $N(T) = \\{0\\}$. Consider some $T(x) = T(y)$. Then $T(x) - T(y) = T(x-y) = 0$. So $x-y ∈ N(T) = \\{0\\}$, and so $x-y = 0$ and $x = y$. So $T$ is injective. ∎

</details>

Theorem: For $V$ and $W$ of equal (and finite) dimension: $T$ is injective iff it is surjective.
<details markdown="block">
<summary>Proof</summary>

We have that $T$ is injective iff $N(T) = \\{0\\}$ and thus iff $nullity(T) = 0$. So $T$ is injective iff $rank(T) = dim(R(T)) = dim(W)$, from the Rank-Nullity Theorem. $dim(R(T)) = dim(W)$ is equivalent to $R(T) = W$, which is the definition of surjectivity. So $T$ is injective iff it is surjective. ∎

</details>

Theorem: Suppose that $V$ has a finite basis $\\{ v_1, v_2, ..., v_n \\}$. For any vectors $w_1, w_2, ... w_n$ in $W$, there exists *exactly* one linear transformation such that $T(v_i) = w_i$ for $i = 1, 2, ..., n$.
<details markdown="block">
<summary>Proof</summary>
...
</details>

Theorem: Suppose that $V$ has a finite basis $\\{ v_1, v_2, ..., v_n \\}$.  If $U, T : V → W$ are linear and $U(v_i) = T(v_i)$ for all $i$, then $U = T$.
<details markdown="block">
<summary>Proof</summary>
...
</details>

## Composition of Linear Transformations

Let $V$, $W$, and $Z$ be vector spaces.

Theorem: Let $T, U : V → W$ be linear. Then their sum and scalar products are linear.
<details markdown="block">
<summary>Proof</summary>
...
</details>

Theorem: The set of all linear transformations (via our definitions of addition and scalar multiplication above) $V → W$ forms a vector space over $F$. We denote this as $\mathcal{L}(V, W)$.
- If $V = W$, we write $\mathcal{L}(V)$.
<details markdown="block">
<summary>Proof</summary>
...
</details>

Let $T, U : V → W$ be arbitrary functions. We define **addition** $T + U : V → W$ as $∀x ∈ V : (T + U)(x) = T(x) + U(x)$, and **scalar multiplication** $aT : V → W$ as $∀x ∈ V : (aT)(x) = aT(x)$ for all $a ∈ F$.

Theorem: Let $T : V → W$ and $U : W → Z$ be linear. Then their composition $UT : V → Z$ is linear.
<details markdown="block">
<summary>Proof</summary>
Let $x,y ∈ V$ and $c ∈ F$. Then:

$$UT(cx + y)$$

$$= U(T(cx + y)) = U(cT(x) + T(y))$$

$$= cU(T(x)) + U(T(y)) = c(UT)(x) + UT(y)$$

</details>

Theorem: Let $T, U_1, U_2 ∈ \mathcal{L}(V)$, and $a ∈ F$. Then:
- $T(U_1 + U_2) = TU_1 + TU_2$
- $(U_1 + U_2)T = U_1 T + U_2 T$
- $T(U_1 U_2) = (TU_1) U_2$
- $TI = IT = T$
- $a(U_1 U_2) = (aU_1) U_2 = U_1 (aU_2)$
<details markdown="block">
<summary>Proof</summary>
...
<!-- A more general result holds for linear transformations with domains unequal to their codomains, exercise 7 -->
</details>

## Invertibility and Isomorphism

- Let $V$ and $W$ be vector spaces.
- Let $T: V → W$ be a linear transformation.
- Let $I_V: V → V$ and $I_W: W → W$ denote the identity transformations within $V$ and $W$, respectively.

A function $U: W → V$ is an **inverse** of $T$ if $TU = I_W$ and $UT = I_V$.<br>
If $T$ has an inverse, then $T$ is **invertible**.

Theorem: Consider a linear function $T: V → W$.
- If $T$ is invertible, it has a *unique* inverse $T^{-1}$.
- If $T$ is invertible, $T^{-1}$ is invertible with the inverse $T$.
- A function is invertible if and only iff it is bijective.
<details markdown="block">
<summary>Proof</summary>
...
</details>

Theorem: If $T$ is linear and invertible, $T^{-1}$ is linear and invertible.
<details markdown="block">
<summary>Proof</summary>

Let $y_1, y_2 ∈ W$ and $c ∈ F$. As $T$ is bijective, there exist unique vectors $x_1$ and $x_2$ such that $T(x_1) = y_1$ and $T(x_2) = y_2$. So $x_1 = T^{-1}(y_1)$ and $x_2 = T^{-1}(y_2)$. So:

$$T^{-1}(y_1 + cy_2) = T^{-1}[T(x_1) + cT(x_2)] = T^{-1}[T(x_1 + cx_2)] = x_1 + cx_2 = T^{-1}(y_1) + cT^{-1}(y_2)$$

Thus $T^{-1}$ is linear. We may see it is invertible by definition. ∎

</details>

$V$ is **isomorphic** to $W$ if there exists an *invertible* linear $T : V → W$ (an **isomorphism**).

Lemma: For finite-dimensional $V$ and $W$: If $T: V → W$ is invertible, then $dim(V) = dim(W)$.
<details markdown="block">
<summary>Proof</summary>

As $T$ is invertible, it is bijective. So $nullity(T) = 0$, $rank(T) = dim(R(T)) = dim(W)$. So by the Rank-Nullity Theorem $dim(V) = dim(W)$. ∎

</details>

Theorem: $V$ is isomorphic to $W$ iff $dim(V) = dim(W)$. Subsequently:
- $V$ is isomorphic to $F^n$ iff $dim(V) = n$.
<details markdown="block">
<summary>Proof</summary>

Suppose that $V$ is isomorphic to $W$ and that $T : V → W$ is an isomorphism. As $T$ is an isomorphism, it is invertible, and so by our earlier lemma $dim(V) = dim(W)$.

Now suppose that $dim(V) = dim(W)$. Let $β = \\{v_1, v_2, ..., v_n\\}$ and $γ = \\{w_1, w_2, ..., w_n\\}$ be bases for $V$ and $W$, respectively. Then as $V$ and $W$ are of equal dimension there must exist $T : V → W$ such that $T$ is linear and $T(v_i) = w_i$ for all $i$. Thus $R(T) = span(T(β)) = span(γ) = W$, and $T$ is surjective. As $V$ and $W$ are of equal dimension, $T$ is also injective. So $T$ is bijective, and so $T$ is an isomorphism. ∎

</details>

## Linear Transformations as Matrices

- Let $V, W$ be finite-dimensional vector spaces (over a field $F$).
- Let $T, U : V → W$ be linear transformations from $V$ to $W$.
- Let $β$ and $γ$ be ordered bases of $V$ and $W$, respectively.
- Let $a ∈ F$ be a scalar.

An **ordered basis** of a finite-dimensional vector space $V$ is, well, an ordered basis of $V$. We represent this with exactly the same notation as a standard unordered basis, but will call attention to it whenever necessary.
- For the vector space $F^n$ we call $\\{ e_1, e_2, ..., e_n \\}$ the **standard ordered basis** for $F^n$.
- For the vector space $P_n(F)$ we call $\\{ 1, x, ..., x^n \\}$ the **standard ordered basis** for $P_n(F)$.

Let $a_1, a_2, ... a_n$ be the unique scalars such that $x = Σ_{i=1}^n a_i u_i$ for all $x ∈ V$. The **coordinate vector** of $x$ relative to $β$ is $\begin{pmatrix} a_1 \\ a_2 \\ ... \\ a_n \end{pmatrix}$ (vert) and denoted $[x]_β$.

The $m × n$ matrix $A$ defined by $A_{ij} = a_{ij}$ is called the **matrix representation of $T$ in the ordered bases $β$ and $γ$**, and denoted as $A = [T]_β^γ$. If $V = W$ and $β = γ$, we write $A = [T]_β$.

Theorem: $[T + U]_β^γ = [T]_β^γ + [U]_β^γ$ and $[aT]_β^γ = a[T]_β^γ$.
<details markdown="block">
<summary>Proof</summary>
...
</details>

---

Let $A$ be a $n × n$ matrix. Then $A$ is **invertible** iff there exists an $n × n$ matrix $B$ such that $AB = BA = I$.

Theorem: If $A$ is invertible, the matrix $B$ is unique, and denoted $A^{-1}$.
<details markdown="block">
<summary>Proof</summary>
Suppose there existed another inverse matrix $C$. Then $C = CI = C(AB) = (CA)B = IB = B$.
</details>

Theorem: For finite-dimensional $V$ and $W$: $T$ is invertible iff $[T]_β^γ$ is invertible, and $[T^{-1}]_γ^β = ([T]_β^γ)^{-1}$. Subsequently:
- Let $U : V → V$ be linear. $U$ is invertible iff $[U]_β$ is invertible, and $[U^{-1}]_β = ([T]_β)^{-1}$.
- Let $A$ be an $n × n$ matrix. $A$ is invertible iff $[L_A]$ is invertible, and $(L_A)^{-1} = L_{A-1}$.
<details markdown="block">
<summary>Proof</summary>
...
</details>

Theorem: Let $V$ and $W$ be of finite dimensions $n$ and $m$ with ordered finite bases $β$ and $γ$, respectively. The function $Φ : \mathcal{L}(V,W) → M_{m×n}(F)$ where $Φ(T) = [T]_β^γ$ for all $T ∈ \mathcal{L}(V,W)$ is an isomorphism.
That is, the set (really *vector space*) of all linear transformations between two vector spaces $V$ and $W$ itself is isomorphic to the vector space of all $m × n$ matrices. Subsequently:
- For $V$ and $W$ of finite dimensions $n$ and $m$, $\mathcal{L}(V,W)$ is of finite dimension $mn$.
<details markdown="block">
<summary>Proof</summary>
...
</details>

The **standard representation** of an $n$-dimensional $V$ with respect to its ordered basis $β$ is the function $ϕ_β : V → F^n$ where $ϕ_β(x) = [x]_β$ for all $x ∈ V$.

Theorem: For any $V$ with ordered basis $β$, $ϕ_β$ is an isomorphism.
<details markdown="block">
<summary>Proof</summary>
...
</details>

## The Change of Coordinate Matrix

- Let $V$ be a finite-dimensional vector space (over a field $F$).
- Let $B$ and $β'$ be two ordered bases for $V$.
- Let $T$ be a linear operator on $V$.

Theorem: Let $Q = [I_V]_{β'}^β$. Then $Q$ is invertible, and for any $v ∈ V$, $[v]_β = Q[v]_{β'}$.
<details markdown="block">
<summary>Proof</summary>
...
</details>

We call such a matrix $Q = [I_V]_{β'}^β$ a **change of coordinate matrix** and say that $Q$ **changes β'-coordinates into β-coordinates**.

Let $Q = [I_V]_{β'}^β$.

Theorem: $Q^{-1}$ changes β-coordinates into β'-coordinates.
<details markdown="block">
<summary>Proof</summary>
...
</details>

Theorem: $[T]_{β'} = Q^{-1} [T]_β Q$
<details markdown="block">
<summary>Proof</summary>s
...
</details>

## Dual Spaces

The **dual space** of a vector space $V$ is the vector space $\mathcal{L}(V,F)$ and is denoted by $V^*$.

- Let $V, W$ be finite-dimensional vector spaces (over a field $F$).
- Let $T, U : V → W$ be linear transformations from $V$ to $W$.
- Let $β$ and $γ$ be ordered bases of $V$ and $W$, respectively.

Theorem: For a finite-dimensional $V$: $dim(V^*) = dim(\mathcal{L}(V,F)) = dim(V) ∙ dim(F) = dim(V)$.

Corollary: $V$ and $V^*$ are isomorphic.
<details markdown="block">
<summary>Proof</summary>
$dim(V) = dim(V^*)$ and so they are isomorphic.
</details>

The **dual basis** of a basis $β$ (of a vector space $V$) is the ordered basis $β^* = \\{ f_1, f_2, ..., f_n \\}$ of $V^*$ that satisfies the $i$th coordinate function $f_i(x_j) = δ_{ij} (1 ≤ i, j ≤ n)$. Recall that $δ_{ij} = 1$ if $i=j$, and $0$ otherwise.

Theorem: Let $β = \\{ x_1, x_2, ..., x_n \\}$. Let $f_i(1 ≤ i ≤ n)$ be the $i$th coordinate function wrt. $β$, and let $β^* = \\{ f_1, f_2, ..., f_n \\}$. Then $

...

Theorem: For any linear transformation $T : V → W$, the mapping $T^t : W^* → V^*$ defined by $T^t(g) = gT$ for all $g ∈ W^*$ is a linear transformation, and $[T^t]_{γ^*}^{β^*} = ([T]_β^γ)^t$
<details markdown="block">
<summary>Proof</summary>
...
</details>

<!-- ## Homogeneous Linear Differential Equations -->

## Systems of Linear Equations

This section is mostly review.

<!--

## Determinants

...

Let $Ax = b$ be the matrix form of a system of $n$ linear equations in $n$ unknowns (where $x = (x_1, x_2, ..., x_n)^t$).

Cramer's Rule: If $det(A) ≠ 0$, then the system $Ax = b$ has a *unique* solution, and for each $k$ from $1$ to $n$, $x_k = [det(A)]^{-1} ∙ det(M_k)$, where $M_k$ is the $n × n$ matrix obtained from $A$ by replacing column $k$ of $A$ by $b$.
<details markdown="block">
<summary>Proof</summary>
...
</details>

...

-->

---

## Determinants, summarized

Determinants are important for future sections. We state facts here without proof.

Let $A$ be a matrix containing entries from a field $F$.

The **determinant** of an $n × n$ (square) matrix $A$ is a scalar in $F$, and denoted $|A|$. The determinant is calculated as follows:
- For a $1 × 1$ matrix $A = [a]$, $|A| = a$
- For a $2 × 2$ matrix $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$, $|A| = ac - bd$
- For an $n × n$ matrix $A = \begin{bmatrix} a_{1,1} & a_{1,2} & ... & a_{1,n} \\ a_{2,1} & a_{2,2} & ... & a_{2,n} \\ ⋮ & ⋮ & ⋱ & ⋮ \\ a_{n,1} & a_{n,2} & ... & a_{n,n} \end{bmatrix}$, $|A| = Σ^n_{i=1} (-1)^{i+j} A_{i,j} |A^~_{i,j}|$.

The last one deserves some additional exposition: todo

The determinant has a number of nice properties that make it of fair interest.
1. $|B| = -|A|$ if $B$ is a matrix obtained by exchanging any two rows or columns of $A$
2. $|B| = k|A|$ if $B$ is a matrix obtained by multiplying $A$ by some scalar $k$
3. $|B| = |A|$ if $B$ is a matrix obtained by adding a multiple of a column or row to a *different* column or row
4. $|A| = 1$ if $A$ is the identity matrix
5. $|A| = 0$ if either the rows or columns of $A$ are not linearly independent
6. $|AB| = |A||B|$ if $A$ and $B$ are both $n × n$ matrices
7. $|A| ≠ 0$ iff $A$ is invertible
8. $|A| = |A^t|$
9. $|A| = |B|$ if $A$ and $B$ are *similar*

Thus, we can say that the determinant *characterizes* square matrices (and thus linear operations), somewhat. It is a scalar value with a deep relation to the core identity of the matrix, and changes regularly as the matrix changes.

## Eigenvalues and Eigenvectors

- Let $V$ be a finite-dimensional vector space over a field $F$.
- Let $T: V → V$ be a linear operator on $V$.
- Let $β$ be an ordered basis on $V$.
- Let $A$ be in $M_{n×n}(F)$ (a square $n×n$ matrix with entries in $F$).

$T$ is **diagonalizable** if there exists an ordered basis $β$ for $V$ such that $[T]_β$ is a *diagonal matrix*.
$A$ is **diagonalizable** if $L_A$ is diagonalizable.

A nonzero vector $v ∈ V$ is an **eigenvector** if $∃λ ∈ F$: $T(v) = λv$.
The corresponding scalar $λ$ is the **eigenvalue** corresponding to the eigenvector $v$.
A nonzero vector $v ∈ F^n$ is an **eigenvector** of $A$ if $v$ is an eigenvector of $L_A$ (that is, $∃λ ∈ F$: $Av = λv$)
The corresponding scalar $λ$ is the **eigenvalue** of $A$ corresponding to the eigenvector $v$.

The terms *characteristic vector* and *proper vector* are also used in place of *eigenvector*.
The terms *characteristic value* and *proper value* are also used in place of *eigenvalue*.

Theorem: $T: V → V$ is diagonalizable if and only if $V$ has an ordered basis $β$ consisting of eigenvectors of $T$.
<details markdown="block">
<summary>Proof</summary>
...
</details>

Corollary: If $T$ is diagonalizable, and $β = \{v_1, v_2, ..., v_n\}$ is an ordered basis of eigenvectors of $T$, and $D = [T]_β$, then $D$ is a diagonal matrix with $D_{i,i}$ being the eigenvalue corresponding to $v_n$ for any $i ≤ n$.
<details markdown="block">
<summary>Proof</summary>
...
</details>

To *diagonalize* a matrix (or a linear operator) is to find a basis of eigenvectors and the corresponding eigenvalues.

Theorem: A scalar $λ$ is an eigenvalue of $A$ if and only if $|A - λI_n| = 0$, that is, the eigenvalues of a matrix are exactly the zeros of its characteristic polynomial.
<details markdown="block">
<summary>Proof</summary>
A scalar $λ$ is an eigenvalue of $A$ iff $∃v ≠ 0 ∈ F^n$: $Av = λv$, that is, $(A - λI_n)(v) = 0$.
... todo
</details>

The **characteristic polynomial** of $A$ is the polynomial $f(t) = |A - tI_n|$.
The **characteristic polynomial** of $T$ is the characteristic polynomial of $[T]_β$, often denoted $f(t) = |T - tI|$.

Theorem: The characteristic polynomial of $A$ is a polynomial of degree $n$ with leading coefficient $(-1)^n$.
<details markdown="block">
<summary>Proof</summary>
...
</details>

Corollary: $A$ has at most $n$ distinct eigenvalues.
<details markdown="block">
<summary>Proof</summary>
...
</details>

Theorem: A vector $v ∈ V$ is an eigenvector of $T$ corresponding to an eigenvalue $λ$ iff $v ≠ 0 ∈ N(T - λI)$.
<details markdown="block">
<summary>Proof</summary>
...
</details>

<!-- look at examples 6 and 7 in chapter 5 -->

## Diagonalizability

- Let $V$ be a finite-dimensional vector space over a field $F$.
- Let $T: V → V$ be a linear operator on $V$.
- Let $A$ be in $M_{n×n}(F)$ (a square $n×n$ matrix with entries in $F$).
- Let $λ$ be an eigenvalue of $T$.
- Let $λ_1, λ_2, ..., λ_k$ be distinct eigenvalues of $T$.

Theorem: Let $λ_1, λ_2, ..., λ_k$ be distinct eigenvalues of $T$. If $v_1, v_2, ..., v_k$ are eigenvectors of $T$ such that $λ_i$ corresponds to $v_i$ (for all $i ≤ k$), then $\{v_1, v_2, ..., v_k\}$ is linearly independent. In fewer words, eigenvectors with distinct eigenvalues are all linearly independent from one another.
<details markdown="block">
<summary>Proof</summary>
...
</details>


Corollary: If $T$ has $n$ distinct eigenvalues, where $n = dim(V)$, then $T$ is diagonalizable.
<details markdown="block">
<summary>Proof</summary>
...
</details>

A polynomial $f(t) ∈ P(F)$ **splits over** $F$ if there are scalars $c, a_1, a_2, ..., a_n ∈ F$: $f(t) = c(a_1 - t)(a_2 - t)...(a_n - t)$.

Theorem: The characteristic polynomial of any diagonalizable linear operator splits.
<details markdown="block">
<summary>Proof</summary>
...
</details>

Let $λ$ be an eigenvalue of a linear operator or matrix with characteristic polynomial $f(t)$. The **(algebraic) multiplicity** of $λ$ is the largest positive integer $k$ for which $(t-λ)^k$ is a factor of $f(t)$.

Let $λ$ be an eigenvalue of $T$. The set $E_λ = \{x ∈ V: T(x) = λx\} = N(T - λI_V)$ is called the **eigenspace** of $T$ with respect to the eigenvalue $λ$. Similarly, the **eigenspace** of $A$ is the eigenspace of $L_A$.

Theorem: If $T$ has multiplicity $m$, $1 ≤ dim(E_λ) ≤ m$.
<details markdown="block">
<summary>Proof</summary>
...
</details>

Lemma: Let $S_i$ be a finite linearly independent subset of the eigenspace $E_{λ_i}$ (for all $i ≤ k$). Then $S = S_1 ∪ S_2 ∪ ... ∪ S_k$ is a linearly independent subset of $V$.
<details markdown="block">
<summary>Proof</summary>
...
</details>

Theorem: If the characteristic polynomial of $T$ splits, then $T$ is diagonalizable iff the multiplicity of $λ_i$ is equal to $dim(E_{λ_i})$ (for all $i ≤ k$).
<details markdown="block">
<summary>Proof</summary>
...
</details>

Corollary: If the characteristic polynomial of $T$ splits, $T$ is diagonalizable, $β_i$ is an ordered basis for $E_{λ_i}$ (for all $i ≤ k$), then $β = β_1 ∪ β_2 ∪ ... ∪ β_k$ is an ordered basis for $V$ consisting of eigenvectors of $T$.
<details markdown="block">
<summary>Proof</summary>
...
</details>

## Direct Sums

- Let $V$ be a finite-dimensional vector space over a field $F$.
- Let $T: V → V$ be a linear operator on $V$.
- Let $W_1, W_2, ..., W_k$ be subspaces of $V$.

The **sum** of some subspaces $W_i$ (for $1 ≤ i ≤ k$) is the set $\{v_1 + v_2 + ... + v_k : v_i ∈ W_i \}$, denoted $W_1 + W_2 + ... + W_k$ or $Σ^k_{i=1} W_i$.

The subspaces $W_i$ (for $1 ≤ i ≤ k$) form a **direct sum** of $V$, denoted $W_1 ⊕ W_2 ⊕ ... ⊕ W_k$, if $V = Σ^k_{i=1} W_i$ and $W_j ∩ Σ_{i≠j} W_i = \{0\}$ for all $j ≤ k$.

Theorem: The following conditions are equivalent:
1. $V = W_1 ⊕ W_2 ⊕ ... ⊕ W_k$.
2. $V = Σ^k_{i=1} W_i$ and ??? todo
3. Every vector $v ∈ V$ can be uniquely written as $v = v_1 + v_2 + ... + v_k$ where $v_i ∈ W_i$.
4. If $γ_i$ is an ordered basis for $W_i$ (for $1 ≤ i ≤ k$), then $γ_1 ∪ γ_2 ∪ ... ∪ γ_k$ is an ordered basis for $V$.
5. There exists an ordered basis $γ_i$ for $W_i$ for every $1 ≤ i ≤ k$ such that $γ_i ∪ γ_2 ∪ ... γ_k$ is an ordered basis for $V$.
<details markdown="block">
<summary>Proof</summary>
...
</details>

Theorem: $T: V → V$ is diagonalizable if and only if $V$ is the direct sum of the eigenspaces of $T$.
<details markdown="block">
<summary>Proof</summary>
...
</details>