The inner product goes away for the rest of the course. Two constructions replace it: the dual space, which collects all linear functionals, and the quotient, which collapses a subspace to zero. Both work over any field, and both are the standard tools for the canonical forms ahead.
Unit 3 used an inner product to turn functionals into vectors. Without one there is no such identification, and the dual becomes a genuinely separate space. Quotients let you factor out the part of a space you do not care about — the algebraic analogue of $W^\perp$, available over any field.
In economics a price vector is a functional on the space of bundles. In physics covectors are the dual objects to vectors, which is what upper and lower indices track. In optimization, duality theory pairs a linear program with a dual program living in the dual space.
The natural map $V\to V^{**}$ is an isomorphism in finite dimensions but not in general — spaces where it is are called reflexive, a central distinction in functional analysis. That failure in infinite dimensions is why the finite-dimensional theory here is so clean.
The dual space $V^*$ is the vector space of all linear functionals $f:V\to F$, with pointwise operations. Its elements are also called covectors.
If $\beta=\{v_1,\dots,v_n\}$ is a basis of $V$, define $f_i\in V^*$ by $f_i(v_j)=\delta_{ij}$, extended linearly. Then $\beta^*=\{f_1,\dots,f_n\}$ is a basis of $V^*$, so $\dim V^*=\dim V$.
On $\mathbb{R}^3$ with the standard basis, $f_i$ is the map picking off the $i$-th coordinate. A general functional is $f(x)=a_1x_1+a_2x_2+a_3x_3$, so $V^*$ looks like row vectors while $V$ is column vectors.
Changing even one $v_j$ changes every $f_i$, because the conditions $f_i(v_j)=0$ involve all the other vectors. There is no basis-free way to match up $V$ and $V^*$ — which is exactly what an inner product supplies, via Riesz.
For linear $T:V\to W$, the transpose (or dual map) $T^t:W^*\to V^*$ is defined by $T^t(g)=g\circ T$. Note the direction reverses.
If $A=[T]_\beta^\gamma$ then $[T^t]_{\gamma^*}^{\beta^*}=A^t$, the ordinary transpose — with no conjugation and no orthonormality needed.
This is the honest source of the transpose operation. The conjugate transpose of Unit 3 arises only when an inner product is used to identify $V$ with $V^*$, and the conjugate comes from that identification, not from the transpose itself.
The map $\Psi:V\to V^{**}$ given by $\Psi(v)(f)=f(v)$ is an isomorphism when $\dim V<\infty$, and it requires no choice of basis.
It is linear and injective: if $v\ne0$, extend $v$ to a basis and take the functional dual to it, which is nonzero on $v$. Since $\dim V^{**}=\dim V^*=\dim V$, injectivity gives surjectivity.
For a subspace $W\subseteq V$, the annihilator is $W^0=\{f\in V^*:f(w)=0\text{ for all }w\in W\}$, a subspace of $V^*$.
$\dim W+\dim W^0=\dim V$.
This is the exact algebraic analogue of $\dim W+\dim W^\perp=\dim V$ from Unit 2, with no inner product in sight.
In $\mathbb{R}^3$ let $W$ be the plane $z=0$, spanned by $e_1,e_2$. Then $W^0$ is spanned by $f_3$, the functional $x\mapsto x_3$ — one-dimensional, as $3-2=1$ requires. Every functional vanishing on the plane is a multiple of "read off the third coordinate".
For a subspace $W\subseteq V$, the coset of $v$ is $v+W=\{v+w:w\in W\}$. The quotient space $V/W$ is the set of cosets, with
These are well defined: if $u+W=u'+W$ and $v+W=v'+W$ then $u-u'$ and $v-v'$ lie in $W$, so their sum does too. The zero element is $0+W=W$ itself.
$\dim(V/W)=\dim V-\dim W$.
The quotient map $\pi:V\to V/W$, $\pi(v)=v+W$, is linear and surjective with null space exactly $W$. Rank–nullity from MAT240H1 gives $\dim V=\dim W+\dim(V/W)$.
For linear $T:V\to W$, the induced map $\overline{T}:V/N(T)\to R(T)$ given by $\overline{T}(v+N(T))=T(v)$ is a well-defined isomorphism.
In $\mathbb{R}^2$ let $W$ be the line $y=x$. The cosets of $W$ are the lines parallel to it, and $\mathbb{R}^2/W$ is one-dimensional: a coset is determined by which parallel line it is. Choosing the representative on the $x$-axis identifies $\mathbb{R}^2/W$ with $\mathbb{R}$ — but that choice is not canonical, which is the recurring theme of this unit.
• Treating $V^*$ as equal to $V$. They have the same dimension, hence
are isomorphic, but not canonically — the isomorphism depends on a basis
or an inner product.
• Getting the direction of $T^t$ wrong. If $T:V\to W$ then
$T^t:W^*\to V^*$.
• Treating elements of $V/W$ as vectors of $V$. They are sets; every
computation must be checked independent of the representative.
• Confusing $W^0\subseteq V^*$ with $W^\perp\subseteq V$. They live in
different spaces and only the inner product relates them.
Original problems written for this guide in the style of Friedberg, Insel & Spence exercises. Write a full solution before opening the answer.
Find the dual basis of $\{(1,0),(1,1)\}$ in $\mathbb{R}^2$.
Find $W^0$ for $W=\operatorname{span}\{(1,1,0)\}$ in $\mathbb{R}^3$.
What is $\dim(\mathbb{R}^4/W)$ for $W$ a plane?
For $T(x,y)=(x+y,\,2x)$ on $\mathbb{R}^2$, find the matrix of $T^t$ in the dual of the standard basis.
In $\mathbb{R}^2$ with $W$ the $x$-axis, describe the cosets and give a natural representative for each.
Prove $(W_1+W_2)^0=W_1^0\cap W_2^0$.
If $W$ is $T$-invariant, show $T$ induces a well-defined operator on $V/W$.
Explain why $\Psi:V\to V^{**}$ is called canonical while $V\cong V^*$ is not.
Course coverage. This unit is part of the complete MAT247H1 topic map. It should cover the following items before you move on:
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.
Dual basis gives $\dim V^*=\dim V$; annihilators mirror orthogonal complements without an inner product; quotients package rank–nullity as an isomorphism. Unit 9 uses the induced operator on a quotient throughout.
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.
What characterises the dual basis?
If $T:V\to W$, where does $T^t$ go?
State the dimension formula for $W^0$.
Why prefer $V/W$ to a complement of $W$?