MAT247H1/Unit 1
Unit 1 · Algebra II

Inner Product Spaces

MAT240H1 had no way to say how long a vector is or what angle two vectors make. An inner product supplies both from a single operation. Everything geometric in this course — orthogonality, projection, the spectral theorem — is built on the axioms below.

By the end you can
Why it matters & where it comes from
Why it exists

A vector space alone cannot express "closest", "perpendicular", or "best approximation". Those are the questions applications actually ask, and an inner product is the minimum extra structure that answers them.

In the real world

Least squares regression minimises a distance defined by an inner product. Fourier series are orthogonal expansions in a function space. Quantum states live in a complex inner product space, where $\langle u,v\rangle$ is a probability amplitude — which is why the complex case matters.

Story & open questions

Cauchy proved the finite sum version in 1821; Bunyakovsky and Schwarz found the integral version decades later, which is why the inequality carries up to three names. Extending these axioms to infinite dimensions and demanding completeness gives Hilbert space, the subject of MAT357H1.

The axioms

Definition

Let $V$ be a vector space over $F=\mathbb{R}$ or $\mathbb{C}$. An inner product is a map $\langle\cdot,\cdot\rangle:V\times V\to F$ with, for all $u,v,w\in V$ and $c\in F$:

Linear in the first slot: $\langle u+w,v\rangle=\langle u,v\rangle+\langle w,v\rangle$ and $\langle cu,v\rangle=c\langle u,v\rangle$.
Conjugate symmetry: $\langle v,u\rangle=\overline{\langle u,v\rangle}$. Over $\mathbb{R}$ the bar does nothing and this is plain symmetry.
Positive definite: $\langle u,u\rangle>0$ for $u\ne0$. Conjugate symmetry forces $\langle u,u\rangle$ to equal its own conjugate, so it is real and the inequality makes sense.

The norm is $\|u\|=\sqrt{\langle u,u\rangle}$.

Theorem The second slot is conjugate-linear

$\langle u,cv\rangle=\overline{c}\,\langle u,v\rangle$ and $\langle u,v+w\rangle=\langle u,v\rangle+\langle u,w\rangle$.

This is forced, not chosen: conjugate symmetry plus linearity in the first slot gives $\langle u,cv\rangle=\overline{\langle cv,u\rangle} =\overline{c\langle v,u\rangle}=\overline{c}\,\overline{\langle v,u\rangle} =\overline{c}\,\langle u,v\rangle$.

Example Why the conjugate is unavoidable
Suppose we tried plain symmetry on $\mathbb{C}$ with $\langle u,v\rangle=\sum u_iv_i$. Take $u=(1,i)$ in $\mathbb{C}^2$. Then $\langle u,u\rangle=1^2+i^2=1-1=0$ with $u\ne0$ — positive definiteness dies.
With the conjugate, $\langle u,v\rangle=\sum u_i \overline{v_i}$ gives $\langle u,u\rangle=|1|^2+|i|^2=2>0$. In general $\langle u,u\rangle=\sum|u_i|^2$, a sum of non-negative reals.

The standard examples

Example Four to keep in mind
Euclidean: on $\mathbb{R}^n$, $\langle u,v\rangle=\sum_{i=1}^n u_iv_i$. Then $\|u\|$ is ordinary length.
Standard complex: on $\mathbb{C}^n$, $\langle u,v\rangle=\sum_{i=1}^n u_i\overline{v_i}$.
Functions: on $C[0,1]$, $\langle f,g\rangle=\int_0^1 f(t)\overline{g(t)}\,dt$. Positive definiteness uses continuity: a continuous non-negative function with integral $0$ is identically $0$.
Matrices: on $M_n(\mathbb{C})$, the Frobenius inner product $\langle A,B\rangle=\operatorname{tr}(B^*A)$, which is just the standard inner product on $\mathbb{C}^{n^2}$ in disguise.
Not every norm comes from an inner product. The norm $\|u\|_1=\sum|u_i|$ is a perfectly good norm, but no inner product produces it. The test is the parallelogram law below — and this is why "inner product space" is a strictly stronger assumption than "normed space".

Cauchy–Schwarz and its consequences

Theorem Cauchy–Schwarz

$|\langle u,v\rangle|\le\|u\|\,\|v\|$, with equality exactly when $u$ and $v$ are linearly dependent.

Proof The one-line trick
If $v=0$ both sides are $0$. Otherwise set $c=\dfrac{\langle u,v\rangle}{\|v\|^2}$ and consider the vector $u-cv$, which is $u$ with its $v$-component removed.
Positive definiteness gives $0\le\|u-cv\|^2 =\langle u-cv,u-cv\rangle$. Expanding with linearity in the first slot and conjugate-linearity in the second, the cross terms cancel and this equals $\|u\|^2-\dfrac{|\langle u,v\rangle|^2}{\|v\|^2}$.
Rearranging gives $|\langle u,v\rangle|^2\le\|u\|^2\|v\|^2$. Equality means $\|u-cv\|=0$, so $u=cv$. ∎
Theorem Triangle inequality

$\|u+v\|\le\|u\|+\|v\|$.

Expand $\|u+v\|^2=\|u\|^2+2\operatorname{Re}\langle u,v\rangle+\|v\|^2$. Since $\operatorname{Re}\langle u,v\rangle\le|\langle u,v\rangle| \le\|u\|\|v\|$ by Cauchy–Schwarz, the right side is at most $(\|u\|+\|v\|)^2$.

Theorem Parallelogram law

$\|u+v\|^2+\|u-v\|^2=2\|u\|^2+2\|v\|^2$. Expanding both squares makes the cross terms cancel. A norm satisfies this identity if and only if it comes from an inner product, which is how you rule out $\|\cdot\|_1$.

Common pitfalls

• Pulling a scalar out of the second slot without conjugating. Over $\mathbb{C}$, $\langle u,cv\rangle=\overline{c}\langle u,v\rangle$.
• Assuming $\langle u,v\rangle$ is real. It is a complex number; only $\langle u,u\rangle$ is guaranteed real.
• Defining the angle between complex vectors. The formula $\cos\theta=\langle u,v\rangle/(\|u\|\|v\|)$ needs a real inner product; over $\mathbb{C}$ only orthogonality survives.
• Forgetting that the inner product is extra data. The same vector space carries many different inner products, and "orthogonal" means nothing until you name one.

Practice

Original problems written for this guide in the style of Friedberg, Insel & Spence exercises. Write a full solution before opening the answer.

Practice 1 · Axiom check

On $\mathbb{R}^2$, is $\langle u,v\rangle=u_1v_1-u_2v_2$ an inner product?

Answer
No. It is bilinear and symmetric, but $u=(0,1)$ gives $\langle u,u\rangle=-1<0$, so positive definiteness fails. This is the Minkowski form from relativity, an indefinite form.
Practice 2 · Weighted product

For which $a,b\in\mathbb{R}$ is $\langle u,v\rangle=au_1v_1+bu_2v_2$ an inner product on $\mathbb{R}^2$?

Answer
Exactly when $a>0$ and $b>0$. Symmetry and bilinearity hold for any $a,b$; taking $u=(1,0)$ and $u=(0,1)$ forces $a>0$ and $b>0$, and those conditions make $\langle u,u\rangle=au_1^2+bu_2^2>0$ for $u\ne0$.
Practice 3 · Complex computation

In $\mathbb{C}^2$ with the standard inner product, compute $\langle u,v\rangle$ and $\langle v,u\rangle$ for $u=(1,i)$, $v=(2,1-i)$.

Solution
$\langle u,v\rangle =1\cdot\overline{2}+i\cdot\overline{(1-i)}=2+i(1+i)=2+i-1=1+i$. And $\langle v,u\rangle=2\cdot\overline{1}+(1-i)\overline{i} =2+(1-i)(-i)=2-i-1=1-i$, which is indeed the conjugate.
Practice 4 · Function space

In $C[0,1]$ with $\langle f,g\rangle=\int_0^1 fg$, compute $\|f\|$ for $f(t)=t$.

Solution
$\|f\|^2=\int_0^1 t^2\,dt=\tfrac13$, so $\|f\|=1/\sqrt3$.
Practice 5 · Polarization

In a real inner product space, prove $\langle u,v\rangle=\tfrac14\left(\|u+v\|^2-\|u-v\|^2\right)$.

Solution
Expand both norms: $\|u\pm v\|^2=\|u\|^2\pm2\langle u,v\rangle+\|v\|^2$. Subtracting kills the $\|u\|^2$ and $\|v\|^2$ terms and leaves $4\langle u,v\rangle$. The inner product is recoverable from the norm.
Practice 6 · Ruling out a norm

Show $\|u\|_1=|u_1|+|u_2|$ on $\mathbb{R}^2$ comes from no inner product.

Solution
Test the parallelogram law with $u=(1,0)$, $v=(0,1)$. Then $\|u+v\|_1=\|u-v\|_1=2$, so the left side is $8$; but $2\|u\|_1^2+2\|v\|_1^2=4$. The law fails, so no inner product induces this norm.
Practice 7 · Equality case

When is $\|u+v\|=\|u\|+\|v\|$ in a real inner product space?

Solution
Tracing the proof, equality needs $\langle u,v\rangle=\|u\|\|v\|$: Cauchy–Schwarz equality with a non-negative value. So $v=cu$ with $c\ge0$ (or one of them is $0$) — the vectors point the same way, not merely parallel.
Practice 8 · Trace form

Verify $\langle A,B\rangle=\operatorname{tr}(B^*A)$ is positive definite on $M_n(\mathbb{C})$.

Solution
The $(i,i)$ entry of $A^*A$ is $\sum_j|a_{ji}|^2$, so $\operatorname{tr}(A^*A) =\sum_{i,j}|a_{ji}|^2\ge0$, vanishing only when every entry is $0$. It is the standard inner product on $\mathbb{C}^{n^2}$.

Expanded topic checklist

Course coverage. This unit is part of the complete MAT247H1 topic map. It should cover the following items before you move on:

Textbook/source alignment

Use the current UofT MAT247H1 calendar topics as the coverage spine, and pair lecture notes with a Friedberg-Insel-Spence-style linear algebra text for extra reading. The exact assigned textbook should still be checked against the instructor syllabus.

Core theorem map

The spine here is short: the axioms give Cauchy–Schwarz, which gives the triangle inequality, which makes $\|\cdot\|$ a genuine norm and $V$ a metric space.

Study routine. Whenever a proof stalls, expand $\|u\pm cv\|^2$ for a well-chosen scalar $c$. That single move proves most inequalities in this unit.

Deep practice extensions

Use these after the main practice set. They are designed to force definition-level reasoning, theorem selection, and mixed application rather than routine calculation only.

Extension 1 · Slot convention

If $\langle cu,v\rangle=c\langle u,v\rangle$, what is $\langle u,cv\rangle$?

Guide
$\overline{c}\langle u,v\rangle$. Conjugate symmetry converts one slot into the other, and conjugation of the scalar comes along.
Extension 2 · Definiteness

Which axiom fails for $\langle u,v\rangle=u_1v_1$ on $\mathbb{R}^2$?

Guide
Positive definiteness: $u=(0,1)$ is nonzero but $\langle u,u\rangle=0$. This is positive semi-definite — a degenerate form.
Extension 3 · Equality characterisation

State when Cauchy–Schwarz is an equality.

Guide
Exactly when $u$ and $v$ are linearly dependent, that is one is a scalar multiple of the other.
Extension 4 · Norm test

How do you decide whether a given norm arises from an inner product?

Guide
Test the parallelogram law $\|u+v\|^2+\|u-v\|^2=2\|u\|^2+2\|v\|^2$. If it holds for all $u,v$ the polarization identity reconstructs the inner product; if it fails anywhere, no inner product induces the norm.
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