3.5 Inverses and the Invertible Matrix Theorem

The previous section built compositions; this one asks for a specific composition — the one that gets you back. A matrix $A$ is invertible when there is a matrix $A^{- 1}$ with $A^{- 1} A = I$ , the identity, so that applying $A$ and then $A^{- 1}$ returns every vector to where it started. Section 3.3 already tells you when to expect one. If $T$ confuses two inputs, no return trip can know which of them to go back to; if $T$ misses a target, no return trip can start from there. Invertible should mean one-to-one and onto — and for a square matrix those were the same condition.

Work with a family, so the interesting case can be approached rather than merely described:

$A sub d equals open parenthesis 2, 1; 1, d close parenthesis$

Chapter 4 will recompute its determinant at length; for now take $det A_{d} = 2 d - 1$ , which section 3.2's area argument already gives you as the cross-product of the columns $open parenthesis 2, 1 close parenthesis$ and $open parenthesis 1, d close parenthesis$ .

Seen face-on in a single vertical plane: a faint gray unit square, a larger colored parallelogram which is that square stretched and sheared by a matrix, and a third quadrilateral drawn on top of the gray square, which is the parallelogram carried back by the inverse matrix. However the slider is dragged, the third shape sits exactly on the gray square - the round trip always returns what it started with. As the slider approaches its lowest setting the middle parallelogram grows thin and nearly collapses to a line, and the numbers in the inverse that undoes it grow without bound. — Seen face-on: the faint gray unit square, its image under $A_{d}$ — the parallelogram with red edge $open parenthesis 2, 1 close parenthesis$ and blue edge $open parenthesis 1, d close parenthesis$ — and, drawn over the gray square, the result of carrying that parallelogram back through $A_{d}^{- 1}$ . The slider $d$ runs from $0.55$ up to $2.5$ , stopping just short of the value where the determinant would vanish.

Explore the figure

At the default $d equals 1$ the middle parallelogram is a healthy shape and the returned quadrilateral sits exactly on the gray square. That coincidence is the whole definition: $A^{- 1}$ undoes $A$ .
Drag $d$ up to $2.5$ . The parallelogram grows — its area is $2 (2.5) - 1 = 4$ — and the return trip still lands precisely on the square. Nothing about being large makes a transformation hard to reverse.
Now drag $d$ down towards $0.55$ . The parallelogram thins alarmingly; at $d equals 0.55$ its area is $2 (0.55) - 1 = 0.1$ , a tenth of the square it came from. The return trip still works exactly — but look at what it has to do, stretching that sliver back into a full square.
Read the slider's lower limit as a warning. It stops at $0.55$ rather than $0.5$ because at $d equals 0.5$ the columns are $open parenthesis 2, 1 close parenthesis$ and $open parenthesis 1, 0.5 close parenthesis$ , the second exactly half the first: the parallelogram has collapsed to a segment, and there is nothing left to carry back.

The formula, and where it breaks

For a $2 times 2$ matrix the inverse can be written down outright.

[\begin{matrix} d \cdot {(- 1 + 2 \cdot d)}^{- 1} & - {(- 1 + 2 \cdot d)}^{- 1} \\ - {(- 1 + 2 \cdot d)}^{- 1} & 2 \cdot {(- 1 + 2 \cdot d)}^{- 1} \end{matrix}]

The inverse, for every d at once ✓ Computed · mojocas 0.1.0 ✓ Agrees with the text

The inverse of $A_{d}$ computed symbolically. Every entry carries the determinant $2 d minus 1$ in its denominator — swap the diagonal entries, negate the off-diagonal ones, divide by the determinant. The formula makes the failure case unmissable: at $d equals 0.5$ the denominator is zero and there is no matrix to write down. That is not a defect of the formula but a report about the geometry, since a collapsed parallelogram cannot be un-collapsed.

Substituting a particular value makes the shape concrete.

[\begin{matrix} 1 & - 1 \\ - 1 & 2 \end{matrix}]

The inverse at the default slider ✓ Computed · mojocas 0.1.0 ✓ Agrees with the text

At $d equals 1$ the determinant is $2 (1) - 1 = 1$ , so the division does nothing and the inverse comes out in whole numbers. This is the matrix the figure is applying when you first open it, and you can check the round trip by hand: it sends the red column $open parenthesis 2, 1 close parenthesis$ back to $open parenthesis 1, 0 close parenthesis$ and the blue column $open parenthesis 1, 1 close parenthesis$ back to $open parenthesis 0, 1 close parenthesis$ — the two basis vectors the square was built from.

Why must a zero determinant rule out an inverse? Because determinants multiply. If $A^{- 1}$ existed then $det (A^{- 1}) \cdot det (A) = det (I) = 1$ , and no number times zero is $1$ . The same fact told geometrically: $| det A |$ is the factor by which areas scale, so a determinant of zero means every region is flattened to zero area, and no transformation can restore area from nothing.

Everything, all at once

Collect what this chapter and the last have established about a square $n \times n$ matrix $A$ . Each of the following says the same thing; if any one holds, all of them do.

$A$ is invertible.
$det A not equal to 0$ .
The columns of $A$ are linearly independent.
The columns of $A$ span $ℝ^{n}$ .
$rank A = n$ .
The null space of $A$ is $open brace 0 close brace$ , i.e. $nullity A equals 0$ .
$T (𝐱) = A 𝐱$ is one-to-one.
$T (𝐱) = A 𝐱$ is onto.
$A 𝐱 = 𝐛$ has exactly one solution for every $𝐛$ .
The reduced row echelon form of $A$ is the identity matrix.

This is the Invertible Matrix Theorem, and its content is not that these are ten separate facts about invertibility but that they are one fact wearing ten costumes. You have watched most of them already: independence was a vector failing to escape a plane, rank and nullity were a budget that had to add up, onto and one-to-one were a missed point and a collapsed line, and the determinant was an area running through zero. The theorem's practical use is that you may check whichever condition is cheapest and conclude all the others — and its practical warning is that when one fails, they all fail together, so a matrix is never nearly invertible in some respects and fine in others.

Example 1 A matrix that fails every clause at once

Show that the matrix with columns $open parenthesis 2, 1 close parenthesis$ and $open parenthesis 1, 0.5 close parenthesis$ fails at least four conditions of the theorem, naming them explicitly.

This is $A_{d}$ at $d equals 0.5$ , the value the slider stops short of.

Determinant. $det = 2 (0.5) - 1 (1) = 0$ .

Independence. The second column is exactly half the first: $open parenthesis 1, 0.5 close parenthesis equals 0.5 open parenthesis 2, 1 close parenthesis$ , which is a nontrivial dependency relation, so the columns are dependent.

Span and rank. Both columns lie on the single line through $open parenthesis 2, 1 close parenthesis$ , so their span is that line rather than $ℝ^{2}$ , and the rank is $1$ , not $2$ .

Null space. By the rank theorem, nullity $= 2 - 1 = 1$ , so the null space is a line, not $open brace 0 close brace$ . Concretely, $A (1, - 2) = 1 (2, 1) - 2 (1, 0.5) = (2, 1) - (2, 1) = (0, 0)$ , so $open parenthesis 1, minus 2 close parenthesis$ is a nonzero vector sent to zero.

And therefore. The map is not one-to-one, since $open parenthesis 0, 0 close parenthesis$ and $open parenthesis 1, minus 2 close parenthesis$ share an image; it is not onto, since its outputs fill only a line; and $A 𝐱 = 𝐛$ has no solution for any $𝐛$ off that line and infinitely many for any $𝐛$ on it. Every clause fell at once, which is what the theorem promised.

An original work of XYZ Homework, built around interactive XYZ 3D figures. Its chapter sequence is aligned to Interactive Linear Algebra (Margalit & Rabinoff, Georgia Tech, GNU FDL); this work is original, copies nothing from it, and is not affiliated with or endorsed by its authors. License: CC-BY-NC-SA-4.0.