The Invertible Matrix Theorem

Main idea

For a square matrix, a whole list of properties you've met separately turn out to be the same fact in different clothes: either the matrix keeps every dimension intact, or it flattens something. So checking any one of them, like a nonzero determinant, tells you all the rest.

A \text{ invertible} \iff A\vec x = \vec 0 \text{ has only } \vec x = \vec 0 \iff \text{columns independent} \iff \text{columns span } \mathbb R^n \iff \det A \ne 0

Explanation

The Invertible Matrix Theorem is simply a promise that, for a square matrix, a long list of yes-or-no questions all have the same answer. It's called a "theorem" because it's been proved once and for all, and it's good for saving work: answer the easiest question on the list and you get every other answer for free. The chain of $\iff$ signs looks intimidating, but each piece is something you already know: "invertible" (it can be undone), " $A\vec x = \vec 0$ has only $\vec x = \vec 0$ " (nothing nonzero gets crushed to zero), "columns independent" (no column is redundant), "columns span" (every output can be reached) and " $\det A \ne 0$ " (area doesn't collapse). Underneath, they're all saying one thing: $A$ doesn't flatten space.

Think about why they have to go together. If $A$ squashes even one direction down to zero, then some nonzero $\vec x$ goes to $\vec 0$ , the columns are dependent, the outputs fill less than all of $\mathbb R^n$ , the area collapses, and there's no way to undo it. If it squashes nothing, every one of those flips the other way.

For an $n\times n$ matrix $A$ , these are all equivalent (all true or all false):

$A$ is invertible.
$A$ row reduces to $I$ ; it has $n$ pivots.
$A\vec x = \vec 0$ has only the trivial solution.
The columns are linearly independent.
The columns span $\mathbb R^n$ ; $A\vec x = \vec b$ is solvable for every $\vec b$ .
The transformation $\vec x \mapsto A\vec x$ is one-to-one and onto.
$\det A \ne 0$ , and 0 is not an eigenvalue.
$\operatorname{rank}A = n$ .

So checking any one of them settles all the others.

A common mistake. It's tempting to use the theorem on any matrix, since rank, spanning and independence make sense for every shape. It only applies to square matrices. A $3\times2$ matrix can have independent columns without spanning $\mathbb R^3$ , because two columns can never reach all three dimensions.

See it

The parallelogram spanned by the columns of A = [[1, k], [2, 3]] as k slides from 0 to 3, with det A = 3 − 2k going from 3 down to −3. Notice what happens at k = 1.5: the two columns line up, the parallelogram collapses to a line, and the determinant hits 0. At that one moment every statement of the theorem fails together: no inverse, dependent columns, and outputs that reach only a line.

Worked examples

Example 1

A $3\times3$ matrix has columns $\vec a_1, \vec a_2$ and $\vec a_3 = \vec a_1 + \vec a_2$ . Is it invertible?

1
The third column is built from the other two, so the columns are dependent. That's one statement on the list failing, which means all of them fail.

No, so also $\det A = 0$ and some $A\vec x = \vec b$ has no solution.

Practice

Type your answer and check it. Solve them all to complete the lesson.

Problem 1+5 XP

A $3\times3$ matrix has 2 pivots. Is it invertible?

Problem 2+5 XP

For a square $A$ , $A\vec x = \vec 0$ has only the trivial solution. Is $A$ invertible?

Problem 3+5 XP

If $\det A = 0$ , can $A\vec x = \vec b$ have exactly one solution?

Problem 4+5 XP

The columns of a $4\times4$ matrix span $\mathbb R^4$ . Are they independent?

Problem 5+5 XP

A $5\times5$ matrix has a column of zeros. Is it invertible?

Problem 6+5 XP

A $4\times4$ sensor-calibration matrix has one column equal to the sum of two others. Will the calibration equation $A\vec x = \vec b$ have exactly one solution for every reading $\vec b$ ?

Challenge problem+25 XP

The hardest one here. Get it, and you've got this lesson.

For which values of $k$ is $\begin{bmatrix}1 & 2 & 3\\ 0 & k & 4\\ 0 & 0 & k - 2\end{bmatrix}$ not invertible? List all of them.

Where it's used

Numerical analysts check invertibility (via pivots or the condition number) before trusting a solution: a nearly singular matrix magnifies tiny errors enormously.

When a Transform Flattens Your SpriteLin Alg · Game Development

The Inverse of a Matrix LU Factorization