Foundations — Invertible matrix theorem — 12+ equivalent conditions

Before you can read the parent theorem, you must own every piece of notation it fires at you. We build each one from nothing — plain words, then a picture, then the reason the theorem needs it. Nothing is used before it is earned.

1. A vector — the arrow we push around

$x=\left(\begin{array}{c}{x}_1\\ x_2\end{array}\right)\text{means the arrow to the point }(x_1,x_2).$

• Plain words: a location, or equivalently a "push" of a certain length and direction.
• Picture: an arrow from the origin. Look at figure s01 — the black arrow is the vector $(3,2)$.
• Why the topic needs it: the whole theorem is about the equation $Ax=b$. Both $x$ and $b$ are vectors. If you can't see them as arrows, you can't see what the matrix does to them.

The symbol ${\mathbb{R}}^n$ (read "R-n") just names the whole space of all such arrows with $n$ numbers. ${\mathbb{R}}^2$ is the flat plane; ${\mathbb{R}}^3$ is 3D space.

The special arrow $0$ (bold zero) is the vector all of whose numbers are $0$ — an arrow of zero length, sitting exactly at the origin, pointing nowhere.

2. A matrix — the machine

$A=\left(\begin{array}{cc}2& 1\\ 1& 1\end{array}\right)$

• Plain words: a rulebook that takes an arrow in and spits a new arrow out.
• Picture: think of it as a transformation of the whole plane — it grabs every arrow and moves its tip somewhere new. See Linear Transformations — Injective and Surjective.
• Why the topic needs it: $A$ is the star of the show. "Is $A$ invertible?" is the entire question.

Square matters. A grid that is $3\times 5$ (3 rows, 5 columns) takes 5-number arrows in and gives 3-number arrows out — the input and output live in different spaces, so it can never "undo" cleanly. The theorem only applies when input space = output space, i.e. when $A$ is square.

3. Matrix times vector — how the machine runs

This is the single most important motion in the whole subject, so we picture it two ways.

• Plain words: the output is a weighted sum of the columns. The number $x_1$ says "how much of column 1," $x_2$ says "how much of column 2," and so on. See Linear Independence and Span.
• Why this view and not the dry row-by-row dot product? Because the theorem is about the columns — whether they're independent, whether they span. Writing $Ax$ as a column combination makes the equation $Ax=b$ literally say: "can I mix the columns to build $b$?" That single reframing powers half the theorem.
• Picture: figure s02 stretches column arrows $a_1,a_2$ by the weights and adds them tip-to-tail. The red arrow is the result $Ax$.

4. The identity $I$ and the inverse $A^{-1}$

$I_2=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)$

• Plain words: run $A$, then run $A^{-1}$, and you're back where you started — total information preserved.
• Why the topic needs it: condition 1 of the theorem is "$A^{-1}$ exists." Every other condition is a cheaper test for this exact thing.

5. The equation $Ax=0$ — the "does it squash?" test

This is the heartbeat of the theorem. Every arrow $x$ that the machine crushes down to the zero arrow is a witness that information was lost.

• Picture: figure s03 shows a good machine (only the origin maps to the origin) versus a bad one (a whole red line of arrows all collapse onto the origin).
• Why the topic needs it: if even one nonzero arrow gets crushed, two different inputs ($x$ and $0$) share the output $0$ — the machine is not undoable. That's condition 4, and it drives conditions 5, 6, 14, 17.

The set of all arrows that $A$ crushes to $0$ has a name: the null space, written $\mathrm{Nul}A$. See Rank and Nullity. A good (invertible) machine has $\mathrm{Nul}A=\{0\}$ — only the zero arrow gets crushed.

6. The determinant $\det A$ — the area-scaling number

• Plain words: feed the machine a unit square; $\det A$ is the (signed) area of the parallelogram that comes out.
• Picture: the unit square becomes a squashed parallelogram. If the machine flattens the square onto a line, that parallelogram has zero area — $\det A=0$. See Determinant.
• Why the topic needs it: $\det A=0$ exactly means the machine flattened space onto something lower-dimensional — it lost a whole dimension of information. That is condition 13, and it is often the cheapest test to run.

A flattened square = crushed arrows = non-trivial null space. Zero determinant and squashing are the same event seen from different windows — that's the theorem in miniature.

7. Rank, span, independence, basis, eigenvalue — the supporting cast

Quick plain-words definitions so the theorem's vocabulary is complete:

• Why the topic needs them: conditions 5, 8, 15, 16, 14 are each phrased in exactly this language. "Full rank $=n$" means no dimension was lost — the same good-machine idea one more time.

8. Elementary row operations — how we test cheaply

• Plain words: tidy the grid step by step without changing which arrows it crushes.
• Why the topic needs it: repeatedly tidying $A$ either arrives at the identity $I$ (good machine) or gets stuck with a row of zeros (crushed dimension). That's conditions 2 and 3.

How these feed the theorem

Every arrow into the box "Invertible Matrix Theorem" is one equivalent condition — and they all trace back to the same root idea: does the machine crush any arrow?

Equipment checklist

Test yourself — cover the right side and answer before revealing.