Linear Mappings and Their Matrices
3.9 Geometry of the Determinant: Orientation
3.8.3. Describe the geometric effect of multiplying by the 3-by-3 elementary matrices R2;3,1, R3;1,2, and S2,−3.
3.8.4. (a) Express the matrix 0−1
1 0
as a product of recombine and scale matrices (you may not need both types).
(b) Use part (a) to describe counterclockwise rotation of the plane through angle π/2 as a composition of shears and scales.
3.8.5. Describe counterclockwise rotation of the plane through angle θ (where cos θ6= 0 and sin θ 6= 0) as a composition of shears and scales.
3.8.6. In R3, describe the linear mapping that takes e1 to e2, e2to e3, and e3
to e1 as a composition of shears, scales, and transpositions.
3.8.7. Let P be the parallelogram in R2 spanned by (a, c) and (b, d). Cal-culate directly that | deta b
c d
| = area P. (Hint: area = base × height
= |(a, c)| |(b, d)| | sin θ(a,c),(b,d)|. It may be cleaner to find the square of the area.)
3.8.8. This exercise shows directly that| det | = volume in R3. LetP be the parallelepiped in R3 spanned by v1, v2, v3, let P′ be spanned by the vectors v1′, v2′, v3′ obtained from performing the Gram–Schmidt process on the vj’s, let A∈ M3(R) have rows v1, v2, v3 and let A′∈ M3(R) have rows v′1, v′2, v′3.
(a) Explain why det A′= det A.
(b) Give a plausible geometric argument that volP′ = volP.
(c) Show that
A′A′t=
|v1′|2 0 0 0 |v′2|2 0 0 0 |v3′|2
.
Explain why therefore| det A′| = vol P′. It follows from parts (a) and (b) that that | det A| = vol P.
3.9 Geometry of the Determinant: Orientation
Recall from section 2.1 that a basis of Rnis a set of vectors{f1,· · · , fp} such that any vector in Rn is a unique linear combination of the {fj}. Though strictly speaking a basis is only a set, we adopt here the convention that the basis vectors are given in the specified order indicated. Given such a basis, view the vectors as columns and let F denote the matrix in Mn,p(R) with columns f1,· · · , fp. Thus the order of the basis vectors is now relevant. For a standard basis vector ej of Rp, the matrix-by-vector product F ej gives the jth column fj of F . Therefore, for any vector x = (x1,· · · , xp)∈ Rp (viewed as a column),
F x = F·
Thus, multiplying all column vectors x∈ Rp by the matrix F gives precisely the linear combinations of f1,· · · , fp, and so we have the equivalences
{f1,· · · , fp} is a basis of Rn
⇐⇒ each y∈ Rn is uniquely expressible as a linear combination of the {fj}
!
Theorem 3.9.1 (Characterization of Bases). Any basis of Rn has n el-ements. The vectors {f1,· · · , fn} form a basis exactly when the matrix F having them as its columns has nonzero determinant.
Let{f1,· · · , fn} be a basis of Rn, and let F be the matrix formed by their columns. Abuse terminology and call det F the determinant of the basis, written det{f1,· · · , fn}. Again, this depends on the order of the {fj}. There are then two kinds of basis of Rn, positive and negative bases, according to the sign of their determinants. The standard basis{e1,· · · , en} forms the columns of the identity matrix I and is therefore positive.
The multilinear function det F is continuous in the n2entries of f1,· · · , fn (see exercise 3.6.1). If a basis{f1,· · · , fn} can be smoothly deformed via other bases to the standard basis then the corresponding determinants must change continuously to 1 without passing through 0. Such a basis must therefore be positive. Similarly, a negative basis can not be smoothly deformed via other bases to the standard basis. It is also true but less clear (and not proved here) that every positive basis deforms smoothly to the standard basis.
The plane R2is by convention drawn with{e1, e2} forming a counterclock-wise angle of π/2. Two vectors{f1, f2} form a basis if they are not collinear.
Therefore the basis {f1, f2} can be deformed via bases to {e1, e2} exactly when the angle θf1,f2 goes counterclockwise from f1to f2. (Recall from equa-tion (2.2) that the angle between two nonzero vectors is between 0 and π.) That is, in R2, the basis{f1, f2} is positive exactly when the angle from f1 to f2is counterclockwise. (See figure 3.20.)
Three-space R3 is by convention drawn with {e1, e2, e3} forming a right-handed triple, meaning that when the fingers of your right hand curl from e1 to e2 your thumb forms an acute angle with e3. Three vectors{f1, f2, f3}
3.9 Geometry of the Determinant: Orientation 125 f1
f1 f2
f2
Figure 3.20. Positive and negative bases of R2
form a basis if they are not coplanar. In other words they must form a right-or left-handed triple. Only right-handed triples defright-orm via other nonplanar triples to{e1, e2, e3}. Therefore in R3, the basis{f1, f2, f3} is positive exactly when it forms a right-handed triple. (See figure 3.21.)
f1
f1
f2
f2
f3
f3
Figure 3.21.Positive and negative bases of R3
The geometric generalization to Rnof a counterclockwise angle in the plane and a right-handed triple in space is not so clear, but the algebraic notion of positive basis is the same for all n.
Consider any invertible mapping T : Rn−→ Rn with matrix A∈ Mn(R), and any basis{f1,· · · , fn} of Rn. If F again denotes the matrix with columns f1,· · · , fnthen AF has columns{Af1,· · · , Afn} = {T (f1),· · · , T (fn)}. These form a new basis of Rn with determinant
det{T (f1),· · · , T (fn)} = det AF = det A det F = det A det{f1,· · · , fn}.
The calculation lets us interpret the sign of det A geometrically: If det A > 0 then T preserves the orientation of bases, and if det A < 0 then T reverses orientation. For example, the mapping with matrix
0 0 0 1 1 0 0 0 0 1 0 0 0 0 1 0
reverses orientation in R4.
To summarize: Let A be an n-by-n matrix. Whether det A is nonzero says whether A is invertible, the magnitude of det A is the factor by which A magnifies volume, and (assuming that det A6= 0) the sign of det A determines how A affects orientation.
Exercises
3.9.1. Any invertible mapping T : Rn −→ Rn is a composition of scales, shears and transpositions. Give conditions on such a composition to make the mapping orientation-preserving, orientation-reversing.
3.9.2. Does the linear mapping T : Rn −→ Rn that takes e1 to e2, e2 to e3,
· · · , en to e1 preserve or reverse orientation? (The answer depends on n.) More generally, if π is a permutation in Sn, does the linear mapping taking e1
to eπ(1),· · · , en to eπ(n)preserve or reverse orientation? (This depends on π.) 3.9.3. Argue geometrically in R2 that any basis can be smoothly deformed via other bases to the standard basis or to {e1,−e2}. Do the same for R3 and{e1, e2,−e3}.