Part 2 · Tensors

2.5 Inverse and determinant

Source: Favata’s notes, appendix A.2, p. 15 (inverse, its characterisations and properties, determinant and its properties)

Objectives

Know when a tensor can be undone, in four equivalent ways.
Understand the determinant as the factor by which a tensor multiplies signed volumes.
Recipe: compute $det A$ from the columns; use $det (AB) = det A det B$ .
Pictures: the unit cube and its image; an orientation flip; a collapse.
Theorem: the Invertible Tensor Theorem, first version.
Vocabulary: invertible, inverse, kernel, orientation.
Essential vocabulary: determinant.

Inverse

Definition 2.5.1Inverse

$A$ is invertible if there is a tensor $A^{- 1}$ , the inverse, with $A A^{- 1} = A^{- 1} A = I .$

It is unique: if $B$ and $C$ are both inverses, $B = B (AC) = (BA) C = C$ .

The picture: $A$ can be undone when no information is lost, when no two vectors land on the same image. By linearity, $Au = Av$ iff $A (u - v) = 0$ , so it is enough to ask what is sent to $0$ .

Theorem 2.5.2The Invertible Tensor Theorem

For $A \in Lin$ the following are equivalent:

$A$ is invertible;
for every $u$ , $Au = 0 ⟹ u = 0$ : only $0$ is sent to $0$ ;
for every basis ${u, v, w}$ , ${Au, Av, Aw}$ is a basis;
for every $u, v$ , $u \times v \neq = 0 ⟹ Au \times Av \neq = 0$ : independent pairs stay independent;
$det A \neq = 0$ (below);
the columns of $[A]$ are a basis.

The theorem grows: §2.7 Eigenvalues and the spectral theorem and §3.3 Change in length and angle add statements to it, and the complete list is in the back matter.

Proof of 1–4 and 6 (5 after the determinant)beyond the notes

1 ⇒ 2: $u = A^{- 1} Au = A^{- 1} 0 = 0$ .
2 ⇒ 3: suppose $α Au + β Av + γ Aw = 0$ . Then $A (α u + β v + γ w) = 0$ , so by 2 $α u + β v + γ w = 0$ , so $α = β = γ = 0$ . The three images are independent, hence a basis of the 3D space $V$ .
3 ⇒ 1: ${A e_{i}}$ is a basis. Define $B$ by $B (A e_{i}) := e_{i}$ and extend linearly. Then $BA = I$ on the basis ${e_{i}}$ , and $AB = I$ on the basis ${A e_{i}}$ , so $B = A^{- 1}$ .
2 ⇔ 4: the argument of 2 ⇒ 3 with two vectors gives 2 ⇒ 4. Conversely, if $Au = 0$ with $u \neq = 0$ , pick $v$ not parallel to $u$ : then $u \times v \neq = 0$ but $Au \times Av = 0$ , against 4.
3 ⇔ 6: the columns are $A e_{i}$ . 3 with ${e_{i}}$ gives 6; and 6 is 3 ⇒ 1’s starting point.

Key observation 2.5.3beyond the notes

In finite dimension, “nothing nonzero is sent to zero” is already enough for an inverse: a linear map that loses no information can always be undone.

Theorem 2.5.4Rules for inverses

$(AB)^{- 1} = B^{- 1} A^{- 1}$ ;
$(A^{- 1})^{⊤} = (A^{⊤})^{- 1} =: A^{- ⊤}$ .

Proof (the notes call these immediately verifiable)beyond the notes

$(AB) (B^{- 1} A^{- 1}) = AI A^{- 1} = I$ , and the same in the other order.
$(A^{- 1})^{⊤} A^{⊤} = (A A^{- 1})^{⊤} = I^{⊤} = I$ , and the same in the other order.

Determinant

Definition 2.5.5DeterminantEssential

The determinant of $A \in Lin$ is the scalar $det A := \frac{Au \cdot ( Av \times Aw )}{u \cdot ( v \times w )} for any basis {u, v, w} .$

The denominator is the signed volume of the parallelepiped on $u, v, w$ (Key observation 1.3.7); the numerator, that of the parallelepiped on their images. So:

Key observation 2.5.6What the determinant is

$det A$ is the factor by which $A$ multiplies signed volumes.

$∣ det A ∣$ is the ratio of the volume of the image of any parallelepiped to its own.
${u, v, w}$ and ${Au, Av, Aw}$ have the same orientation iff $det A > 0$ .
$det A = 0$ iff the images are coplanar: the box is crushed flat.

Why the basis does not matterbeyond the notes

Numerator and denominator are both trilinear and alternating in $(u, v, w)$ (Scalar triple product). On a 3D space all such forms are multiples of one another: writing $u, v, w$ in the basis ${e_{i}}$ and expanding by trilinearity, an alternating form is fixed by its value on $(e_{1}, e_{2}, e_{3})$ . So their ratio is a constant that depends only on $A$ .

Recipe 2.5.7Determinant from the columnsbeyond the notes

Choose the basis ${e_{i}}$ : the denominator is $1$ , and $A e_{j}$ are the columns of $[A]$ : $det A = A e_{1} \cdot (A e_{2} \times A e_{3}) = ϵ_{ijk} A_{i 1} A_{j 2} A_{k 3} = det [A],$ the usual determinant of the matrix. In the plane, $det A = A_{11} A_{22} - A_{12} A_{21}$ , the signed area of the parallelogram on the two columns.

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Figure 2.5.1 The columns

A e_{1}, A e_{2}, A e_{3}

are draggable. The dashed unit cube goes to the shaded box, whose volume is

det A

. Push one column through the plane of the other two: it snaps there (the box is flat,

det A = 0

), and beyond it the box turns red: orientation reversed.

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Figure 2.5.2 The same in the plane, where you can see the letter. Drag the columns: the shaded square’s area is

∣ det A ∣

. Pass one column across the other and the F is mirrored. Line them up: the plane collapses onto a line, and the dashed kernel appears, the directions that

A

sends to

0

With condition 3 of Theorem 2.5.2: $det A \neq = 0$ iff ${Au, Av, Aw}$ is a basis, iff $A$ is invertible. That is statement 5 of the theorem.

Properties

Theorem 2.5.8Properties of the determinant

$det A = det A^{⊤}$ ;
$det (AB) = det (BA) = det A det B$ ;
$det (α A) = α^{3} det A$ ;
$det A^{- 1} = (det A)^{- 1}$ .

Proofbeyond the notes

In components it is the matrix fact $det [A]^{⊤} = det [A]$ : the expansion runs over the same permutations of rows as of columns.
If $B$ is invertible, ${Bu, Bv, Bw}$ is a basis, so $det (AB) = \frac{A ( Bu ) \cdot ( A ( Bv ) \times A ( Bw ))}{Bu \cdot ( Bv \times Bw )} \cdot \frac{Bu \cdot ( Bv \times Bw )}{u \cdot ( v \times w )} = det A det B .$ If $B$ is not invertible, $Bu = 0$ for some $u \neq = 0$ , so $ABu = 0$ too, and both sides are $0$ . Swapping the roles gives $det (BA)$ .
Each of the three factors in the numerator picks up $α$ . In dimension $n$ it is $α^{n}$ .
$det A det A^{- 1} = det (A A^{- 1}) = det I = 1$ .

Volume factors multiply when maps are applied one after the other: that is all of property 2.

Warning

$det (A + B) \neq = det A + det B$ , and $det (2 A) = 8 det A$ in 3D, not $2 det A$ : doubling every edge of a box multiplies its volume by 8.

Examples:

A diagonal tensor $λ_{1} e_{1} \otimes e_{1} + λ_{2} e_{2} \otimes e_{2} + λ_{3} e_{3} \otimes e_{3}$ turns the unit cube into a $λ_{1} \times λ_{2} \times λ_{3}$ box: $det = λ_{1} λ_{2} λ_{3}$ .
A dyad flattens everything onto a line: $det (a \otimes b) = 0$ (Exercise 2.5.1).

This is where the kinematics gets its key condition: $det F > 0$ , volumes stay positive and orientation is kept (Definition 3.2.4).

Exercises

Exercise 2.5.1beyond the notespractice (A.3)

Show that a dyad $a \otimes b$ is never invertible, in two ways.

Solution

Determinant: for any basis, $(a \otimes b) u$ , $(a \otimes b) v$ , $(a \otimes b) w$ are all multiples of $a$ . Two of them are parallel, their cross product is $0$ , and the numerator of $det$ vanishes.
Kernel: pick $u \neq = 0$ orthogonal to $b$ ; then $(a \otimes b) u = (b \cdot u) a = 0$ , failing condition 2.

Exercise 2.5.2beyond the notes

Find the inverse of the simple shear $F = I + γ e_{1} \otimes e_{2}$ without matrices, and its determinant.

Solution

Guess $I - γ e_{1} \otimes e_{2}$ : $(I + γ e_{1} \otimes e_{2}) (I - γ e_{1} \otimes e_{2}) = I - γ^{2} (e_{2} \cdot e_{1}) e_{1} \otimes e_{2} = I$ . The matrix is triangular with ones on the diagonal, so $det F = 1$ : a shear keeps volumes.

Exercise 2.5.3beyond the notes

Show that a skew tensor in 3D is never invertible, using the determinant.

Solution

$det W = det W^{⊤} = det (- W) = (- 1)^{3} det W = - det W$ , so $det W = 0$ . (Geometrically, the axis $w$ is sent to $0$ , Theorem 2.3.3.)