diff --git a/content/Exam Preparation.md b/content/Exam Preparation.md new file mode 100644 index 00000000..ca520c59 --- /dev/null +++ b/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.) \ No newline at end of file diff --git a/content/Exercises/exer10.pdf b/content/Exercises/exer10.pdf new file mode 100644 index 00000000..6f55658e Binary files /dev/null and b/content/Exercises/exer10.pdf differ diff --git a/content/Exercises/exer11.pdf b/content/Exercises/exer11.pdf new file mode 100644 index 00000000..e22d26b6 Binary files /dev/null and b/content/Exercises/exer11.pdf differ diff --git a/content/Exercises/exer2.pdf b/content/Exercises/exer2.pdf new file mode 100644 index 00000000..5aa54c3f Binary files /dev/null and b/content/Exercises/exer2.pdf differ diff --git a/content/Exercises/exer3.pdf b/content/Exercises/exer3.pdf new file mode 100644 index 00000000..149438c3 Binary files /dev/null and b/content/Exercises/exer3.pdf differ diff --git a/content/Exercises/exer4.pdf b/content/Exercises/exer4.pdf new file mode 100644 index 00000000..3c0a3111 Binary files /dev/null and b/content/Exercises/exer4.pdf differ diff --git a/content/Exercises/exer5.pdf b/content/Exercises/exer5.pdf new file mode 100644 index 00000000..f444a8a6 Binary files /dev/null and b/content/Exercises/exer5.pdf differ diff --git a/content/Exercises/exer6.pdf b/content/Exercises/exer6.pdf new file mode 100644 index 00000000..dfaf9204 Binary files /dev/null and b/content/Exercises/exer6.pdf differ diff --git a/content/Exercises/exer7.pdf b/content/Exercises/exer7.pdf new file mode 100644 index 00000000..f79e9001 Binary files /dev/null and b/content/Exercises/exer7.pdf differ diff --git a/content/Exercises/exer8.pdf b/content/Exercises/exer8.pdf new file mode 100644 index 00000000..670d72ae Binary files /dev/null and b/content/Exercises/exer8.pdf differ diff --git a/content/Exercises/exer9.pdf b/content/Exercises/exer9.pdf new file mode 100644 index 00000000..ca9b532e Binary files /dev/null and b/content/Exercises/exer9.pdf differ diff --git a/content/Lectures/Lecture 14.md b/content/Lectures/Lecture 14.md index 829e5d78..4bc88421 100644 --- a/content/Lectures/Lecture 14.md +++ b/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 \ No newline at end of file diff --git a/content/Lectures/Lecture 16.md b/content/Lectures/Lecture 16.md index 2a7636a7..86894de5 100644 --- a/content/Lectures/Lecture 16.md +++ b/content/Lectures/Lecture 16.md @@ -5,8 +5,8 @@ date: 2025-03-10 # Recap Recap from the [[Lecture 15|previous]]. $L_{1}(\mu) = \left\{ \text{measure}\ f : \underbrace{X}_{\mu} \to \mathbb{C} \, | \, \underbrace{\int |f| \, d\mu}_{= \, \sup_{0 \leq s \leq |f|} \int s \, d\mu} \lt \infty \right\}$ -$s = \Sigma_{i = 1}^{n} a_{i} \times X_{A_{i}}$ -$\int s \, d\mu = \Sigma_{i=1}^{n} a_{i} \times \mu(A_{i})$ +$s = \sum_{i = 1}^{n} a_{i} \times X_{A_{i}}$ +$\int s \, d\mu = \sum_{i=1}^{n} a_{i} \times \mu(A_{i})$ $f = \mathrm{Re} f + i \mathrm{Im} f = (\mathrm{Re} f)^{+} - (\mathrm{Re} f)^{-} + i((\mathrm{Im} f)^{+} - (\mathrm{Im} f)^{-})$ $0 \leq \int (\mathrm{Re} f)^{\pm} \, d\mu \leq \int |f| \, d\mu$ @@ -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 $= \Sigma_{i j} \int (f + g) \, d\mu = \Sigma_{i j} \left( \int_{A_{i} \cap B_{j}} f \, d\mu + \int_{A_{i} \cap B_{j}} g \, d\mu\right)$ -$= \Sigma_{i j} \int_{A_{i} \cap B_{j}} f \, d\mu + \Sigma_{i j} \int_{A_{i} \cap B_{j}} g \, d\mu$ +LHS $= \sum_{i j} \int (f + g) \, d\mu = \sum_{i j} \left( \int_{A_{i} \cap B_{j}} f \, d\mu + \int_{A_{i} \cap B_{j}} g \, d\mu\right)$ +$= \sum_{i j} \int_{A_{i} \cap B_{j}} f \, d\mu + \sum_{i j} \int_{A_{i} \cap B_{j}} g \, d\mu$ $=^{L_{1}} \int_{X} f \, d\mu + \int_{X} g \, d\mu$ $L2$: diff --git a/content/Lectures/Lecture 17 - Lp Spaces.md b/content/Lectures/Lecture 17 - Lp Spaces.md index dd5bf97d..db3a8170 100644 --- a/content/Lectures/Lecture 17 - Lp Spaces.md +++ b/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 $(f|g) = \int f \times \bar{g} \, d\mu$. \ No newline at end of file +$L^{2}(\mu)$ 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 $\mu(A) \gt 0 \; \forall A \neq \emptyset$ +$V \setminus \sim = V$ diff --git a/content/Lectures/Lecture 4 - 1.2 Metric Spaces.md b/content/Lectures/Lecture 4 - 1.2 Metric Spaces.md index 9b9cfdc4..5c59a435 100644 --- a/content/Lectures/Lecture 4 - 1.2 Metric Spaces.md +++ b/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 $V$. diff --git a/content/Lectures/Lecture 5.md b/content/Lectures/Lecture 5.md index 2d1b83b3..7b4a3e36 100644 --- a/content/Lectures/Lecture 5.md +++ b/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 -> > $$\mathbb{C}^{n} (u | v) = \Sigma_{k = 1}^{n} u_{k} \overline{v_{k}}$$ -> > $$\mathbb{R}^{n} \| \, u \, \|_{2} = (\Sigma |u_{k}|^{2})^{\frac{1}{2}}$$ -> > $$\vec{u} \times \vec{v} = (\vec{u} | \vec{v})$$ +> > $$\mathbb{C}^{n} (u \mid v) = \sum_{k = 1}^{n} u_{k} \overline{v_{k}}$$ +> > $$\mathbb{R}^{n} \| \, u \, \|_{2} = \left( \sum \mid u_{k} \mid^{2} \right)^{\frac{1}{2}}$$ +> > $$\vec{u} \times \vec{v} = (\vec{u} \mid \vec{v})$$ diff --git a/content/Lectures/Lecture 7.md b/content/Lectures/Lecture 7.md index dbc24bdc..5eade645 100644 --- a/content/Lectures/Lecture 7.md +++ b/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; -$L(a(x_{1}, \dots, x_{n}) + b(y_{1}, \dots, y_{n})) = L((ax_{1} + by_{1}, \dots, ax_{n} + by_{n})) = \Sigma_{i=1}^n(ax_{i} + by_{i})v_{i} =$ -$\Sigma_{i=1}^n(ax_{i}v_{i} + by_{i}v_{i}) = a \times \Sigma_{i=1}^n x_{i}v_{i} + b \times \Sigma_{i=1}^ny_{i}v_{i} =$ +$n = \dim V$. 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; +$L(a(x_{1}, \dots, x_{n}) + b(y_{1}, \dots, y_{n})) = L((ax_{1} + by_{1}, \dots, ax_{n} + by_{n})) = \sum_{i=1}^n(ax_{i} + by_{i})v_{i} =$ +$\sum_{i=1}^n(ax_{i}v_{i} + by_{i}v_{i}) = a \times \sum_{i=1}^n x_{i}v_{i} + b \times \sum_{i=1}^ny_{i}v_{i} =$ $a \times L((x_{1},\dots,x_{n})) + b \times L((y_{1},\dots,y_{n}))$ **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