• Linear Algebra: #15 Why is the Determinant Important?

I am sure there are many points which could be advanced in answer to this question. But here I will concentrate on only two special points.

• The transformation formula for integrals in higher-dimensional spaces.
  This is a theorem which is usually dealt with in the Analysis III lecture. Let G ⊂ ℜⁿ be some open region, and let f : G → ℜ be a continuous function. Then the integral
  
  
  has some particular value (assuming, of course, that the integral converges). Now assume that we have a continuously differentiable injective mapping φ : G → ℜⁿ and a continuous function F : φ(G) → ℜ. Then we have the formula
  
  
  Here, D(φ(x)) is the Jacobi matrix of φ at the point x.
  
  This formula reflects the geometric idea that the determinant measures the change of the volume of n-dimensional space under the mapping φ.
  
  If φ is a linear mapping, then take Q ⊂ ℜⁿ to be the unit cube: Q = {(x₁, . . . , x_n) : 0 ≤ x_i ≤ 1, ∀i}. Then the volume of Q, which we can denote by vol(Q) is simply 1. On the other hand, we have vol(φ(Q)) = det(A), where A is the matrix representing φ with respect to the canonical coordinates for ℜⁿ. (A negative determinant — giving a negative volume — represents an orientation-reversing mapping.)


• The characteristic polynomial.
  Let f : V → V be a linear mapping, and let v be an eigenvector of f with f(v) = λv. That means that (f − λI_n)(v) = 0; therefore the mapping (f  − λI_n) : V → V is singular. Now consider the matrix A, representing f with respect to some particular basis of V. Since λI_n is the matrix representing the mapping λI_n, we must have that the difference A − λI_n is a singular matrix. In particular, we have det(A − λI_n) = 0.
  
  Another way of looking at this is to take a “variable” x, and then calculate (for example, using the Leibniz formula) the polynomial in x
  
  
  P(x) = det(A − xI_n). 
  
  This polynomial is called the characteristic polynomial for the matrix A. Therefore we have the theorem:
  
  
  Theorem 41
  The zeros of the characteristic polynomial of A are the eigenvalues of the linear mapping f : V → V which A represents.
  
  Obviously the degree of the polynomial is n for an n × n matrix A. So let us write the characteristic polynomial in the standard form
  
  
  P(x) = c_nxⁿ + c_n−1xⁿ⁻¹ + · · · + c₁x + c₀ . 
  
  The coefficients c₀, . . . , c_n are all elements of our field F.
  
  Now the matrix A represents the mapping f with respect to a particular choice of basis for the vector space V. With respect to some other basis, f is represented by some other matrix A', which is similar to A. That is, there exists some C ∈ GL(n, F) with A' = C⁻¹AC. But we have
  
  
  
  Therefore we have:
  
  
  Theorem 42
  The characteristic polynomial is invariant under a change of basis; that is, under a similarity transformation of the matrix.
  
  In particular, each of the coefficients c_i of the characteristic polynomial P(x) = c_nxⁿ + c_n−1xⁿ⁻¹ + · · · + c₁x + c₀ remains unchanged after a similarity transformation of the matrix A.
  
  What is the coefficient c_n? Looking at the Leibniz formula, we see that the term xⁿ can only occur in the product
  
  
  (a₁₁ − x)(a₂₂ − x) · · · (a_nn − x) = (−1)xⁿ − (a₁₁ + a₂₂ + · · · + a_nn)xⁿ⁻¹ + · · · . 
  
  Therefore c_n = 1 if n is even, and c_n = −1 if n is odd. This is not particularly interesting.
  
  So let us go one term lower and look at the coefficient c_n−1. Where does xⁿ⁻¹ occur in the Leibniz formula? Well, as we have just seen, there certainly is the term
  
  
  (−1)ⁿ⁻¹(a₁₁ + a₂₂ + · · · + a_nn)xⁿ⁻¹, 
  
  which comes from the product of the diagonal elements in the matrix A − xI_n. Do any other terms also involve the power xⁿ⁻¹? Let us look at Leibniz formula more carefully in this situation. We have
  
  
  
  
  Here, δ_ij = 1 if i = j. Otherwise, δ_ij = 0. Now if σ is a non-trivial permutation — not just the identity mapping — then obviously we must have two different numbers i₁ and i₂, with σ(i₁) ≠ i₁ and also σ(i₂) ≠ i₂. Therefore we see that these further terms in the sum can only contribute at most n − 2 powers of x. So we conclude that the (n − 1)-st coefficient is
  
  
  c_n−1 + = (−1)ⁿ⁻¹(a₁₁ + a₂₂ + · · · + a_nn).
  
  Definition
  
  
  
  Let A be an n × n matrix. The trace of A (in German, the spur of A) is the sum of the diagonal elements:
  
  
  tr(A) = a₁₁ + a₂₂ + · · · + a_nn. 
  
  
  Theorem 43
  tr(A) remains unchanged under a similarity transformation.

An example
Let f : ℜ² → ℜ² be a rotation through the angle θ. Then, with respect to the canonical basis of ℜ², the matrix of f is

That is to say, if λ ∈ ℜ is an eigenvalue of f, then λ must be a zero of the characteristic polynomial. That is,

 λ² − 2λ cos θ + 1 = 0. 

But, looking at the well-known formula for the roots of quadratic polynomials, we see that such a λ can only exist if |cos θ| = 1. That is, θ = 0 or π. This reflects the obvious geometric fact that a rotation through any angle other than 0 or π rotates any vector away from its original axis. In any case, the two possible values of θ give the two possible eigenvalues for f, namely +1 and −1.