Quartz sync: Mar 13, 2025, 2:03 PM

content/Exam Preparation.md

@@ -0,0 +1,57 @@
+Taken directly from Canvas:
+# Exam topics
+
+The following list is meant to provide a starting point for the type of questions (relating to the complex function theory part of the course) you will get at the exam (but this is not an exhaustive list). You are not expected to know the details of the proofs, but you need to show that you understand the concepts and know how to apply them.
+
+- Similarities and differences between C and R^2.
+- Holomorphic functions and their properties.
+- Complex exponential and logarithm.
+- Integration in the complex plane.
+- Cauchy's theorem and integral formula.
+- Taylor and Laurent series.
+- Singularities and their classification.
+- Residues and the residue theorem.
+- Applications to the computation of integrals.
+
+# Relevant Questions
+**Here are 20 relevant exam questions in the topology and measure theory part of the course:**
+
+What distinguishes the real numbers from the rational ones?
+
+What is an equivalence relation?
+
+What is a topological space? Examples?
+
+What is the ball topology on a metric space?
+
+What is the topology on a Banach space?
+
+What is a compact set?
+
+State the Heine-Borel theorem. Proof?
+
+What is a continuous function?
+
+Why does a real valued continuous function obtain its maximum on a compact set?
+
+What is a net? Given an example of an upward filtered ordered set.
+
+What is the initial topology?
+
+What is the product topology?
+
+What is a measure? Easy examples?
+
+Define the Lebesgue integral of a extended non-negative measurable function.
+
+State Lebesgue's monotone convergence theorem.
+
+Define L^p-spaces, and point out their crucial property.
+
+(Not relevant: State the Riesz representation theorem.
+
+What is the Lebesgue measure on R^n ?
+
+What is a complex measure?
+
+State the Lebesgue-Radon-Nikodym theorem.)

content/Lectures/Lecture 14.md

@@ -13,7 +13,7 @@ Going over [[Lecture 13 - Measure Theory]]
 Measure $\mu$ on $X$ with $M$:
 $\mu : M \to [0, \infty]$ such that
 $\mu(0) = 0$
-$\mu(\cup_{n=1}^\infty A_{n}) = \Sigma_{n=1}^\infty \mu(A_{n})$
+$\mu(\cup_{n=1}^\infty A_{n}) = \sum_{n=1}^\infty \mu(A_{n})$
 For $A_{n} \in M$ **pairwise disjoint**
 
 Example $M = \wp(X)$ define measure:
@@ -21,7 +21,7 @@ $\mu(A) = \begin{cases}\#A & \text{When}\ A\ \text{is finite}\\ \infty & \text{W
 
 ---
 # Simple Function
-A **simple function** on $X$ is a function $s : X \to \mathbb{R}$ of the form $s = \Sigma_{i=1}^{n} a_{i} \times X_{a_{i}}$ for pairwise disjoint $A_{i} \subset X$ and distinct real numbers $a_{i}$
+A **simple function** on $X$ is a function $s : X \to \mathbb{R}$ of the form $s = \sum_{i=1}^{n} a_{i} \times X_{a_{i}}$ for pairwise disjoint $A_{i} \subset X$ and distinct real numbers $a_{i}$
 
 $X_{A}(x) = \begin{cases}1 & \text{If}\ x \in A\\ 0 & \text{If}\ x \notin A\end{cases}$
 
@@ -30,7 +30,7 @@ $X_{A}(x) = \begin{cases}1 & \text{If}\ x \in A\\ 0 & \text{If}\ x \notin A\end{
 > 
 > $A_{i} = s^{-1}(\{ a_{i} \}) = s^{-1}(\mathbb{R} \setminus \{ a_{i} \})^{\complement}$
 
-If we have a [[Measure|measure]] $\mu$ on $X$, define $\int_{A} s \, d\mu \equiv \Sigma_{i=1}^{n} a_{i} \times \mu (A \cap A_{i})$ for $A \in M$  (they are all $\in [0, \infty])$.
+If we have a [[Measure|measure]] $\mu$ on $X$, define $\int_{A} s \, d\mu \equiv \sum_{i=1}^{n} a_{i} \times \mu (A \cap A_{i})$ for $A \in M$  (they are all $\in [0, \infty])$.
 # Lebesgue Integral
 Defines [[Lebesgue Integral]] in the lecture
 
@@ -53,13 +53,13 @@ Then $\mu(A_{n}) \to \mu(\cup_{m = 1}^{\infty} A_{m})$ as $n \to \infty$.
 ## Proof
 Consider $B_{n} = A_{n} \setminus A_{n-1}$ with $A_{0} \equiv \emptyset$.
 So $A_{n} = B_{1} \cup \dots \cup B_{n}$, and $\cup B_{n} = \cup A_{n}$, and $B_{n} \cap B_{m} = \emptyset$ when $n \neq m$.
-Hence $\mu(\cup A_{m}) = \mu(\cup B_{m}) = \Sigma_{m = 1}^{\infty} \mu (B_{m}) = \lim_{ N \to \infty } \Sigma_{m=1}^{N} \mu (B_{m})$.
+Hence $\mu(\cup A_{m}) = \mu(\cup B_{m}) = \sum_{m = 1}^{\infty} \mu (B_{m}) = \lim_{ N \to \infty } \sum_{m=1}^{N} \mu (B_{m})$.
 # Lemma
 $X$, $\mu$, $s : X \to [0, \infty]$ [[Measurable|measurable]] [[Simple Function|simple function]].
 Then $A \mapsto \int_{A} s \, d\mu$ defines a [[Measure|measure]] on $X$.
 ## Proof
 $A = \emptyset \implies$ measure is $0$. Given [[Continuous|continuous]] disjoint union $\cup B_{n}$ with $B_{n}$ [[Measure|measure]], then
 $$
-\int_{\cup B_{n}} s \, d\mu = \Sigma_{i} a_{i} \mu(\underbrace{A_{i} \cap (\cup B_{n})}_{\cup(A_{i} \cap B_{n})}) = \Sigma_{i} a_{i} \Sigma_{n}^{\infty} \mu (A_{i} \cap B_{n}) = \Sigma_{n}^{\infty} \Sigma_{i} a_{i} \mu(A_{i} \cap B_{n}) = \Sigma_{n}^{\infty} \int_{B_{n}} s \, d\mu 
+\int_{\cup B_{n}} s \, d\mu = \sum_{i} a_{i} \mu(\underbrace{A_{i} \cap (\cup B_{n})}_{\cup(A_{i} \cap B_{n})}) = \sum_{i} a_{i} \sum_{n}^{\infty} \mu (A_{i} \cap B_{n}) = \sum_{n}^{\infty} \sum_{i} a_{i} \mu(A_{i} \cap B_{n}) = \sum_{n}^{\infty} \int_{B_{n}} s \, d\mu 
 $$
 QED

content/Lectures/Lecture 16.md

@@ -5,8 +5,8 @@ date: 2025-03-10
 # Recap
 Recap from the [[Lecture 15|previous]].
 $L_{1}(\mu) = \left\{  \text{measure}\ f : \underbrace{X}_{\mu} \to \mathbb{C} \, | \, \underbrace{\int |f| \, d\mu}_{= \, \sup_{0 \leq s \leq |f|} \int s \, d\mu} \lt \infty \right\}$
-$s = \Sigma_{i = 1}^{n} a_{i} \times X_{A_{i}}$
+$s = \sum_{i = 1}^{n} a_{i} \times X_{A_{i}}$
-$\int s \, d\mu = \Sigma_{i=1}^{n} a_{i} \times \mu(A_{i})$
+$\int s \, d\mu = \sum_{i=1}^{n} a_{i} \times \mu(A_{i})$
 
 $f = \mathrm{Re} f + i \mathrm{Im} f = (\mathrm{Re} f)^{+} - (\mathrm{Re} f)^{-} + i((\mathrm{Im} f)^{+} - (\mathrm{Im} f)^{-})$
 $0 \leq \int (\mathrm{Re} f)^{\pm} \, d\mu \leq \int |f| \, d\mu$
@@ -26,8 +26,8 @@ $= \int_{A_{i} + B_{j}} f \, d\mu + \int_{A_{i} + B_{j}} g \, d\mu$
 > [[Measure]] $L1$.
 
 $\implies \int_{X} (f + g) \, d\mu = \int f \, d\mu + \int g \, d\mu$
-LHS $= \Sigma_{i j} \int (f + g) \, d\mu = \Sigma_{i j} \left( \int_{A_{i} \cap B_{j}} f \, d\mu + \int_{A_{i} \cap B_{j}} g \, d\mu\right)$
+LHS $= \sum_{i j} \int (f + g) \, d\mu = \sum_{i j} \left( \int_{A_{i} \cap B_{j}} f \, d\mu + \int_{A_{i} \cap B_{j}} g \, d\mu\right)$
-$= \Sigma_{i j} \int_{A_{i} \cap B_{j}} f \, d\mu + \Sigma_{i j} \int_{A_{i} \cap B_{j}} g \, d\mu$
+$= \sum_{i j} \int_{A_{i} \cap B_{j}} f \, d\mu + \sum_{i j} \int_{A_{i} \cap B_{j}} g \, d\mu$
 $=^{L_{1}} \int_{X} f \, d\mu + \int_{X} g \, d\mu$
 
 $L2$:

content/Lectures/Lecture 17 - Lp Spaces.md

@@ -15,7 +15,7 @@ Let $f, g : X \to [0, \infty]$ be measurable on $(X, \mu)$. Then $\| f \times g
 To show [[Hölder's Inequality]], we may assume that $\| f \|_{p}$, $\| g \|_{q}$ are positive real numbers
 
 > [!note] Need:
-> $\Pi_{i=1}^{n} y_{i}^{a_{i}} \leq^{(*)} \Sigma_{i=1}^{n} a_{i} \times y_{i}$ when $a_{i}, y_{i} \gt 0$ and $\Sigma_{i = 1}^{n} a_{i} = 1$.
+> $\Pi_{i=1}^{n} y_{i}^{a_{i}} \leq^{(*)} \sum_{i=1}^{n} a_{i} \times y_{i}$ when $a_{i}, y_{i} \gt 0$ and $\sum_{i = 1}^{n} a_{i} = 1$.
 
 Use (* from the note above) with $a_{1} = \frac{1}{p}$, $a_{2} = \frac{1}{q}$, $y_{1} = \left(\frac{\mid f(x) \mid}{\| f \|_{p}}\right)^{p}$, $y_{2} = \left(\frac{\mid g(x) \mid}{\| g \|_{q}} \right)^{q}$ for $x$ such that $y_{i} \gt 0$.
 
@@ -57,14 +57,64 @@ In fact $f = 0$ **almost everywhere**, in that $\exists$[[Measure|measure]] $A \
 
 > [!example]
 > $f \restriction_{A} = f$ on $A$.
-> $g(x) = \begin{cases} f(x) & \forall x \in A\\ 0 & \forall x \in A^{\complement}, A = f^{-1}(\mathbb{C} \setminus \{ 0 \})\end{cases}$
+> $g(x) = \begin{cases} f(x), & \forall x \in A\\ 0, & \forall x \in A^{\complement}, A = f^{-1}(\mathbb{C} \setminus \{ 0 \})\end{cases}$
 
 > [!example]
 > $\int g \, d\mu = \int_{A} g \, d\mu + \int_{A^{\complement}} g \, d\mu = \int_{A} g \, d\mu = \int f \, d\mu$
 
 Define equivalence relation on $V$ by $f \sim g$ if $f - g = 0$ almost everywhere.
-Then $V \setminus \sim$ will be a [[Vector Space|vector space]] and with [[Norm|norm]] $\| [ f ] \|_{p} \equiv \| f \|_{p}$. Then $\| [ f ] \|_{p} = 0 \implies f = 0$ almost everywhere, so $[f] = [0] = 0$.
+Then $V \setminus \sim$ will be a [[Vector Space|vector space]] and with [[Norm|norm]] $\| \, [ \, f \, ] \, \|_{p} \equiv \| \, f \, \|_{p}$. Then $\| \, [ \, f \, ] \, \|_{p} = 0 \implies f = 0$ almost everywhere, so $[\, f \, ] = [\, 0 \,] = 0$.
 
 # Theorem
 $L^{p}(\mu) = V \setminus \sim$ is a [[Banach Space]].
-$L^{2}(\mu)$ is a [[Hilbert space]] with $(f|g) = \int f \times \bar{g} \, d\mu$.
+$L^{2}(\mu)$ is a [[Hilbert space]] with $(f \mid g) = \int f \times \bar{g} \, d\mu$.
+
+---
+# Exercises
+This is the exercise part of the lecture.
+# Exercise 3.3.6 Question 2
+$(X, \mu)$, $f : X \to [0, \infty \rangle$ [[Measurable|measurable]] such that $\int f \, d\mu = 0$.
+Show that $f = 0$ almost everywhere.
+**Hint:** $A \equiv \{ x \mid f(x) \gt 0 \} = \cup_{n = 1}^{\infty} A_{n}$, where $A_{n} = \underbrace{\left\{  x \mid f(x) \gt \frac{1}{n}  \right\}}_{f^{-1} \left( \langle \frac{1}{n}, \infty \rangle \right)} \in \sigma$.
+## Proof
+If $\mu(A) \gt 0$, then $0 \lt \mu(A) \lt \sum_{n=1}^{\infty} \mu(A_{n})$, so $\mu(A_{m}) \gt 0$ for some $m$.
+
+Then $\int f \, d\mu \geq \int_{A} f \, d\mu \geq \int_{A_{m}} f \, d\mu \geq \int_{A_{m}} \frac{1}{m} \, d\mu = \frac{1}{m} \times \mu(A_{m}) \gt 0$, which is a contradiction.
+
+So $\mu(A) = 0$ and $f = 0$ almost everywhere.
+
+QED.
+# Exercise 3.4.5 Question 3
+What is $L^{p}(\mu)$ for $p \in [1, \infty \rangle$ when $\mu$ is the [[Measure|counting measure]]?
+> [!note] On "counting measure"
+> I thing "counting measure" is the same way that has been said in short hand with "measure" for all the previous lectures.
+
+## Proof
+$f : X \to \mathbb{C}$ all functions. $\sigma = \wp(X)$.
+
+> [!note] What is $L^{p}(\mu)?$
+> $\| f \|_{p} = (\int \mid f \mid^{p} \, d\mu)^{\frac{1}{p}}$
+> $V = \left\{  f : X \to \mathbb{C} \mid \int \mid f \mid^{p} \, d\mu \lt 0  \right\}$
+> $L^{p}(\mu) = V \setminus \sim$
+
+Say $g : X \to [0, \infty \rangle$.
+Then (define: $s = \sum_{i = 1}^{n} a_{i} X_{A_{i}}$) $\int g \, d\mu = \sup_{0 \leq s \leq g} \int s \, d\mu = \sup_{0\leq s\leq g} \sum_{i=1}^{n} \overbrace{a_{i}}^{\neq 0} \times \mu(A_{i}) = \infty$ if any $A_{i}$ is infinite.
+
+For any finite subset $F$ of $X$, define
+$$S_{F}(x) = \begin{cases}
+g(x), & x\in F \\
+0, & x \not\in F
+\end{cases}$$
+Then $S_{F}$ is a [[Simple Function|simple function]], and $0\leq S_{F} \leq g$. Have $\int S_{F} \, d\mu = \int_{F} g \, d\mu = \sum_{x \in F} g(x)$
+
+$S_{F} = \sum_{x \in F} g(x) X_{\{ x \}}$ since $\int S_{F} \, d\mu = \sum_{x \in F} g(x) \mu \underbrace{(\{ x \})}_{1}$
+$\sum_{x \in F} g(x) \times \underbrace{X_{\{ x \}} (y)}_{\delta x, y} = g(y) = S_{F}(y)$
+
+$X = \mathbb{N}$
+$F = \{ 1, \dots, n \}$
+
+Then $\int g \, d\mu = \sup_{0 \leq s \leq g} \int s \, d\mu \geq \sup_{\text{F finite}} \sum_{x \in F} g(x) \overbrace{=}^{x =\mathbb{N}} \sum_{n=1}^{\infty} g(n)$
+($\mathbb{N}$, $\mu$ [[Continuous|continuous]] [[Measure|measure]]) $\implies l^{p}(\mathbb{N}) = L^{p}(\mu) = \left\{  \{ x_{n} \} \mid \sum_{n=1}^{\infty} |x_{n}|^{p} \lt \infty  \right\}$
+
+$\| f \|_{p} = 0 \implies f = 0$ almost everywhere $\iff f = 0$ since $\mu(A) \gt 0 \; \forall A \neq \emptyset$
+$V \setminus \sim = V$

content/Lectures/Lecture 4 - 1.2 Metric Spaces.md

@@ -3,7 +3,7 @@ lecture: 4
 date: 2025-01-16
 ---
 # The Inverse Image
-The Inverse Image uses a [[Inverse Function]] $f^{-1} (z)$ of $Z \subset Y$ written $f: x \to y$ is $f^{-1}(z) \equiv \{ x \in X | f(x) \in Z \}$.
+The Inverse Image uses a [[Inverse Function]] $f^{-1} (z)$ of $Z \subset Y$ written $f: x \to y$ is $f^{-1}(z) \equiv \{ x \in X \mid f(x) \in Z \}$.
 # Complex Numbers
 In [[Complex Numbers]], $\mathbb{C} = \mathbb{R} \times \mathbb{R}$ with usual addition of vectors
 ## Multiplying Vectors 
@@ -39,7 +39,7 @@ $$v_{1} = (1, \, 0, \, 0, \, 0, \, \dots)$$
 $$v_{2} = (0, \, 1, \, 0, \, 0, \, \dots)$$
 $$v_{3} = (0, \, 0, \, 1, \, 0, \, \dots)$$
 and the way of writing this would be:
-$$(x_{1}, \, \dots, \, x_{n}) = \Sigma^{n}_{i=1} x_{i} \times v_{i} = 0$$
+$$(x_{1}, \, \dots, \, x_{n}) = \sum^{n}_{i=1} x_{i} \times v_{i} = 0$$
 $$\implies x_{i} = 0$$
 ## Proposition
 Any [[Vector Space]] $V$ has a [[Linear Basis]], and every basis has the same cardinality referred to as the $dim(V)$ (dimension of $V$) of $V$.

content/Lectures/Lecture 5.md

@@ -6,7 +6,7 @@ date: 2025-01-20
 > A normed space is a [[Banach Space]] when the corresponding metric space is complete. Any normed space can be completed to a [[Banach Space]], see the real numbers from $\mathbb{Q}$.
 
 > [!example] Example: $\mathbb{C}^n$ vector space, then
-> 1. $\| \, v \, \|_{1} \equiv \Sigma_{k=1}^{n} |v_{k}|, \; v=(v_{1}, \dots, v_{n})$
+> 1. $\| \, v \, \|_{1} \equiv \sum_{k=1}^{n} |v_{k}|, \; v=(v_{1}, \dots, v_{n})$
 > 2. $\| \, v \, \|_{\infty} \equiv \text{sub}_{k=1,\dots,n}\{ |v_{k} | \}$
 > 
 > Which are norms on $\mathbb{C}^n$
@@ -25,14 +25,14 @@ Has linear basis $\{ \delta_{x} \}$, $x \in X$
 ### Proof
 Given that $f \in C_{c}(X)$ then
 
-$f = \Sigma_{x \in X} f(x) \times \delta_{x}$ is a finite sum, and the only possibility.
+$f = \sum_{x \in X} f(x) \times \delta_{x}$ is a finite sum, and the only possibility.
 
 [[QED]]
 
 So $\text{dim} C_{c} (X) = |X|$.
 
 ---
-Definite norms on $C_{c}(X)$ by $\| \, f \, \|_{1} = \Sigma_{x \in X} | f(x) |$ and $\| \, f \, \|_{\infty} = \text{sup}_{x \in X} | f(x) |$ when $|X| = n$ we recover $\mathbb{C}^n$.
+Definite norms on $C_{c}(X)$ by $\| \, f \, \|_{1} = \sum_{x \in X} | f(x) |$ and $\| \, f \, \|_{\infty} = \sup_{x \in X} | f(x) |$ when $|X| = n$ we recover $\mathbb{C}^n$.
 
 ---
 We write $V \simeq W$and say that $V$ and $W$ are [[Isomorphic]]
@@ -45,11 +45,11 @@ $f$ should preserve all relevant structure.
 Follows from the [[Cauchy-Schwarz Inequality]]
 
 > [!example] Example on $C_{c}(X)$ then
-> $$(f | g) = \Sigma_{x \in X} f(x) \times \overline{g(x)}$$
+> $$(f \mid g) = \sum_{x \in X} f(x) \times \overline{g(x)}$$
 > This defines an [[Inner Product|inner product]], which can be completed to a [[Hilbert Spaces|Hilbert space]].
-> $$\| \, f \, \|_{2} = (\Sigma_{x \in X} | f(x) |^2)^{\frac{1}{2}}$$
+> $$\| \, f \, \|_{2} = \left( \sum_{x \in X} \mid f(x) \mid^2 \right)^{\frac{1}{2}}$$
 > > [!note] The [[Inner Product|inner product]] here
-> > $$\mathbb{C}^{n} (u | v) = \Sigma_{k = 1}^{n} u_{k} \overline{v_{k}}$$
+> > $$\mathbb{C}^{n} (u \mid v) = \sum_{k = 1}^{n} u_{k} \overline{v_{k}}$$
-> > $$\mathbb{R}^{n} \| \, u \, \|_{2} = (\Sigma |u_{k}|^{2})^{\frac{1}{2}}$$
+> > $$\mathbb{R}^{n} \| \, u \, \|_{2} = \left( \sum \mid u_{k} \mid^{2} \right)^{\frac{1}{2}}$$
-> > $$\vec{u} \times \vec{v} = (\vec{u} | \vec{v})$$
+> > $$\vec{u} \times \vec{v} = (\vec{u} \mid \vec{v})$$
 

content/Lectures/Lecture 7.md

@@ -66,18 +66,18 @@ $$(x, y) + (z, w) \equiv (x+z, y+w)$$
 $$a \times (x, y) \equiv (a x, a y)$$
 Linear map:
 ## Proof
-$n = \dim V$. Say $\{ v_{i} \}_{i=1}^n$ is a linear basis for $V$. Define [[Bijective|bijective]] [[Linear Map|linear map]] $L: \mathbb{C}^n \to V$ by $L((x_{1}, \dots, x_{n})) = \Sigma_{i=1}^n x_{i} \times v_{i}$. It is linear;
+$n = \dim V$. Say $\{ v_{i} \}_{i=1}^n$ is a linear basis for $V$. Define [[Bijective|bijective]] [[Linear Map|linear map]] $L: \mathbb{C}^n \to V$ by $L((x_{1}, \dots, x_{n})) = \sum_{i=1}^n x_{i} \times v_{i}$. It is linear;
-$L(a(x_{1}, \dots, x_{n}) + b(y_{1}, \dots, y_{n})) = L((ax_{1} + by_{1}, \dots, ax_{n} + by_{n})) = \Sigma_{i=1}^n(ax_{i} + by_{i})v_{i} =$
+$L(a(x_{1}, \dots, x_{n}) + b(y_{1}, \dots, y_{n})) = L((ax_{1} + by_{1}, \dots, ax_{n} + by_{n})) = \sum_{i=1}^n(ax_{i} + by_{i})v_{i} =$
-$\Sigma_{i=1}^n(ax_{i}v_{i} + by_{i}v_{i}) = a \times \Sigma_{i=1}^n x_{i}v_{i} + b \times \Sigma_{i=1}^ny_{i}v_{i} =$
+$\sum_{i=1}^n(ax_{i}v_{i} + by_{i}v_{i}) = a \times \sum_{i=1}^n x_{i}v_{i} + b \times \sum_{i=1}^ny_{i}v_{i} =$
 $a \times L((x_{1},\dots,x_{n})) + b \times L((y_{1},\dots,y_{n}))$
 
 **Surs.**
-$v \in V$. Basis $\implies \exists \underbrace{x_{i}}_{\in \mathbb{C}}$ such that $v = \Sigma_{i=1}^n x_{i} v_{i} = L((x_{1}, \dots, x_{n}))$.
+$v \in V$. Basis $\implies \exists \underbrace{x_{i}}_{\in \mathbb{C}}$ such that $v = \sum_{i=1}^n x_{i} v_{i} = L((x_{1}, \dots, x_{n}))$.
 So $v \in \text{lm} L$ 
 
 **Injective**
 Say $L((x_{1}, \dots, x_{n})) = L((y_{1}, \dots, y_{n}))$ then
-$\Sigma_{i=1}^n x_{i} v_{i} = \Sigma_{i=1}^n y_{i} v_{i} \xrightarrow{basis} x_{i} = y_{i}, \; \forall i$, so $x = y$
+$\sum_{i=1}^n x_{i} v_{i} = \sum_{i=1}^n y_{i} v_{i} \xrightarrow{basis} x_{i} = y_{i}, \; \forall i$, so $x = y$
 Injective for [[Linear Map|linear map]] $L \iff \ker L = \{ 0 \} \equiv \{ x \in \mathbb{C} | L(x) = 0 \}$.
 
 $\ker L$ is a vector space