Oct 27, 2026
I assume you know about vector space scalar multiplication, vector addition, and the laws they satisfy. The scalars must be a field such as the real numbers \boldsymbol{R} or complex numbers \boldsymbol{C}. Another example is \boldsymbol{Z}_p = \boldsymbol{Z}/p\boldsymbol{Z} of integers modulo a prime p. We will use \boldsymbol{F} to denote a generic field of scalars.
A linear operator is a function T\colon V\to W between two vector spaces that preserves the vector space structure: T(ax + y) = a(Tx) + Ty for scalar a\in\boldsymbol{F} and vectors x,y\in V. Note the scalar multiplication and vector addition on the left-hand side are those for V and on the right-hand side are those for W.
We write \boldsymbol{0}_V for the zero of vector addition in V: \boldsymbol{0}_V + v = v = v + \boldsymbol{0}_V.
Exercise: Show v + v = v implies v = \boldsymbol{0}_V.
Hint: Add -v to both sides.
Exercise: Show for any linear operator T\colon V\to W that T\boldsymbol{0}_V = \boldsymbol{0}_W.
Hint: T(\boldsymbol{0}+ \boldsymbol{0}) = T\boldsymbol{0}.
Exercise: Show T(x + y) = Tx + Ty.
Hint: Use 1x = x.
Exercise: Show T(ax) = a(Tx).
Hint: Use T(\boldsymbol{0}_V) = \boldsymbol{0}_W.
A subspace of a vector space V is a subset U\subseteq V that is closed under scalar multiplication and vector addition. We can write this as {\boldsymbol{F}U \subseteq U} and {U + U\subseteq U} where {\boldsymbol{F}U = \{au\mid a\in\boldsymbol{F}, u\in U\}} and {U + U = \{u + u'\mid u,u'\in U\}}.
Exercise: If U and U' are subspaces show U\cap U' and U + U' are subspaces.
Hint. Show they are closed under scalar multiplication and vector addition.
The subspaces of a vector space are a lattice. They are a partial order where U\subseteq U, $UU’ and U'\subseteq U implies U = U' and $U
Exercise (Extra Credit): If U, U' and ,U'' are subspaces then {U\cap(U' + U'') = (U\cap U') + (U\cap U'')}.
The subspaces of a vector space are a distributive lattice.
The quotient of a vector space V by a subspace U is {V/U = \{v + U\mid v\in V\}}.
Exercise: Show v + U = v' + U if and only if v - v'\in U.
Hint: Note w\in v + U if and only if w = v + u for some u\in U and w\in v' + U if and only if w = v' + u' for some u'\in U. This shows \boldsymbol{0}= w - w = v - v' + (u - u').
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The quotient space is also a vector space with scalar multiplication a(v + U) = av + U for a\in\boldsymbol{R}, v\in V, and vector addition (v + U) + (v' + U) = (v + v') + U for v, v'\in V.
Exercise. Show these are well-defined.
Hint: Show v + U = v' + U implies av + U = av' + U and
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The kernel of a linear operator T\colon V\to W is {\ker T = \{v\in V\mid Tv = \boldsymbol{0}\}\subseteq V} and the range is {\operatorname{ran}T = \{Tv\mid v\in V\}\subseteq W}.
Exercise Show \ker T is a subspace of V and \operatorname{ran}T is a subspace of W.
The first fundamental fact about the structure of a linear operator {T\colon V\to W} is there are linear operators {V \to V/\ker T \to \operatorname{ran}T \to W}. The first linear operator is onto/surjective, the second linear operator is one-to-one and onto/bijective and the third linear operater is one-to-one/injective.
The second operator demystifies the various pseudo inverse notions people still stuck in the Gilbert Strang row and column vector tediousness suffer through. Every linear operator from {V/\ker T\to\operatorname{ran}T} is invertible.
Exercise Show V\to V/\ker V where v\mapsto v + \ker T is onto.
Exercise Show V/\ker T\to\operatorname{ran}T where v + \ker T\mapsto Tv is well-defined, onto, and one-to-one.
Hint. It is well-defined if v + \ker T = v' + \ker T implies Tv = Tv'. It is clearly onto. Show if Tv = Tv' then v + \ker T = v' + \ker T using the previous exercises.
Exercise. Show \operatorname{ran}T\to W where Tv\to Tv is one-to-one.
Hint. Duh!
If S\colon U\to V and T\colon V\to W are linear operators we say they are exact if \operatorname{ran}S = \ker T, S is injective, and T is surjective.