• Linear Algebra: #16 Complex Numbers

On the other hand, looking at the characteristic polynomial, namely x² − 2x cos θ +1 in the previous example, we see that in the case θ = ±π this reduces to x² + 1. And in the realm of the complex numbers ℂ, this equation does have zeros, namely ±i. Therefore we have the seemingly bizarre situation that a “complex” rotation through a quarter of a circle has vectors which are mapped back onto themselves (multiplied by plus or minus the “imaginary” number i). But there is no need for panic here! We need not follow the example of numerous famous physicists of the past, declaring the physical world to be “paradoxical”, “beyond human understanding”, etc. No. What we have here is a purely algebraic result using the abstract mathematical construction of the complex numbers which, in this form, has nothing to do with rotations of real physical space!

So let us forget physical intuition and simply enjoy thinking about the artificial mathematical game of extending the system of real numbers to the complex numbers. I assume that you all know that the set of complex numbers ℂ can be thought of as being the set of numbers of the form x + yi, where x and y are elements of the real numbers ℜ and i is an abstract symbol, introduced as a “solution” to the equation x² + 1 = 0. Thus i²  = −1. Furthermore, the set of numbers of the form x + 0 · i can be identified simply with x, and so we have an embedding ℜ ⊂ ℂ. The rules of addition and multiplication in ℂ are

Let z = x +yi be some complex number. Then the absolute value of z is defined to be the (non-negative) real number |z| = √(x² + y²). The complex conjugate of z is z = x − yi. Therefore |z| = √(zz).

It is a simple exercise to show that ℂ is a field. The main result — called (in German) the Hauptsatz der Algebra — is that ℂ is an algebraically closed field. That is, let ℂ[z] be the set of all polynomials with complex numbers as coefficients. Thus, for P(z) ∈ ℂ[z] we can write P(z) = c_nzⁿ + c_n−1zⁿ⁻¹ + · · · + c₁z + c₀, where c_j ∈ ℂ, for all j = 0, . . . , n. Then we have:


Theorem 44 (Hauptsatz der Algebra)
Let P(z) ∈ ℂ[z] be an arbitrary polynomial with complex coefficients. Then P has a zero in ℂ. That is, there exists some λ ∈ ℂ with P(λ) = 0.

The theory of complex numbers (Funktionentheorie in German) is an extremely interesting and pleasant subject. Complex analysis is quite different from the real analysis.


Theorem 45
Every complex polynomial can be completely factored into linear factors. That is, for each P(z) ∈ ℂ[z] of degree n, there exist n complex numbers (perhaps not all different) λ₁, . . . , λ_n, and a further complex number c, such that

P(z) = c(λ₁ − z) · · · (λ_n − z).

Proof
Given P(z), theorem 44 tells us that there exists some λ₁ ∈ ℂ, such that P(λ₁) = 0. Let us therefore divide the polynomial P(z) by the polynomial (λ₁− z). We obtain

P(z) =  (λ₁− z) · Q(z) + R(z), 

where both Q(z) and R(z) are polynomials in ℂ[z]. However, the degree of R(z) is less than the degree of the divisor, namely  (λ₁− z), which is 1. That is, R(z) must be a polynomial of degree zero, i.e. R(z) = r ∈ ℂ, a constant. But what is r? If we put λ₁ into our equation, we obtain

0 = P((λ₁) = (λ₁ − λ₁)Q(z) + r = 0 + r.

Therefore r = 0, and so

P(z) = (λ₁− z)Q(z), 

where Q(z) must be a polynomial of degree n−1. Therefore we apply our argument in turn to Q(z), again reducing the degree, and in the end, we obtain our factorization into linear factors.

So the consequence is: let V be a vector space over the field of complex numbers ℂ. Then every linear mapping f : V → V has at least one eigenvalue, and thus at least one eigenvector.

• 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.

• Linear Algebra: #14 Leibniz Formula

Definition
A permutation of the numbers {1, . . . , n} is a bijection

σ : {1, . . . , n} → {1, . . . , n}. 

The set of all permutations of the numbers {1, . . . , n} is denoted S_n. In fact, S_n is a group: the symmetric group of order n. Given a permutation σ ∈ S , we will say that a pair of numbers (i, j), with i, j ∈ {1, . . . , n} is a “reversed pair” if i < j, yet σ(i) > σ(j). Let s(σ) be the total number of reversed pairs in σ. Then the sign of sigma is defined to be the number

sign(σ) = (−1)^s(σ)

Theorem 37 (Leibniz)
Let the elements in the matrix A be a_ij , for i, j between 1 and n. Then we have

As a consequence of this formula, the following theorems can be proved:


Theorem 38
Let A be a diagonal matrix

Then det(A) = λ₁ λ₂ · · · λ_n .


Theorem 39
Let A be a triangular matrix

Then det(A) = a₁₁a₂₂ · · · a_nn .

Leibniz formula also gives:

Definition
Let A ∈ M(n × n, F). The transpose A^t of A is the matrix consisting of elements a_ij^t  such that for all i and j we have a_ij^t = a_ji, where a_ji are the elements of the original matrix A.


Theorem 40
det(A^t) = det(A).


14.1 Special rules for 2 × 2 and 3 × 3 matrices

For the 2 × 2 matrix, the Leibniz formula reduces to the simple formula

det(A) =  a₁₁a₂₂ - a₁₂a₂₁ 

For the 3 × 3 matrix, the formula is a little more complicated.

det(A) =  a₁₁a₂₂a₃₃ + a₁₂a₂₃a₃₃ + a₁₃a₂₁a₃₂ - a₁₁a₂₃a₃₂ - a₁₂a₂₁a₃₃ - a₁₁a₂₃a₃₂

14.2 A proof of Leibniz Formula 
Let the rows of the n × n identity matrix be ε₁ , . . . ,ε_n. Thus

ε₁ = (1 0 0 · · · 0), ε₂ = (0 1 0 · · · 0), . . . , ε_n = (0 0 0 · · · 1). 

Therefore, given that the i-th row in a matrix is

ξ_i = (a_i1a_i2· · · a_in),

 then we have

So let the matrix A be represented by its rows,

It was an exercise to show that the determinant function is additive. That is, if B and C are n × n matrices, then we have det(B + C) = det(B) + det(C). Therefore we can write

To begin with, observe that if ε_jk = ε_jl for some j_k ≠ j_l, then two rows are identical, and therefore the determinant is zero. Thus we need only the sum over all possible permutations (j₁, j₂, . . . , j_n) of the numbers (1, 2, . . . , n). Then, given such a permutation, we have the matrix

This can be transformed back into the identity matrix

by means of successively exchanging pairs of rows.

Each time this is done, the determinant changes sign (from +1 to -1, or from -1 to +1). Finally, of course, we know that the determinant of the identity matrix is 1.

Therefore we obtain the Leibniz formula

• Linear Algebra: #13 Determinant

Let M(n × n, F) be the set of all n × n matrices of elements of the field F.

Definition
A mapping det : M(n × n, F) → F is called a determinant function if it satisfies the following three conditions.

1. det(I_n) = 1, where I_n is the identity matrix. 


2. If A ∈ M(n×n, F) is changed to the matrix A' by multiplying all the elements in a single row with the scalar a ∈ F, then det(A') = a · det(A). (This is our row operation S_i(a).) 


3. If A' is obtained from A by adding one row to a different row, then det(A') = det(A). (This is our row operation S_ij(1).) 

Simple consequences of this definition
Let A ∈ M(n × n, F) be an arbitrary n × n matrix, and let us say that A is transformed into the new matrix A' by an elementary row operation. Then we have:

• If A' is obtained by multiplying row i by the scalar a ∈ F, then det(A') = a · det(A). This is completely obvious! It is just part of the definition of “determinants”.


• Therefore, if A' is obtained from A by multiplying a row with −1 then we have det(A' ) = −det(A).


• Also, it follows that a matrix containing a row consisting of zeros must have zero as its determinant. 


• If A has two identical rows, then its determinant must also be zero. For can we multiply one of these rows with −1, then add it to the other row, obtaining a matrix with a zero row. 


• If A ′ is obtained by exchanging rows i and j, then det(A') = −det(A). This is a bit more difficult to see. Let us say that A = (u₁, . . . , u_i, . . . , u_j, . . . , u_n), where u_k is the k-th row of the matrix, for each k. Then we can write
  
  
  (This is the elementary row operation S_ij.)


• If A' is obtained from A by an elementary row operation of the form S_ij(c), then det(A') = det(A). For we have: 
  
  

Therefore we see that each elementary row operation has a well-defined effect on the determinant of the matrix. This gives us the following algorithm for calculating the determinant of an arbitrary matrix in M(n × n, F).

How to find the determinant of a matrix
Given: An arbitrary matrix A ∈ M(n × n, F).
Find: det(A).

Method:

1. Using elementary row operations, transform A into a matrix in step form, keeping track of the changes in the determinant at each stage. 


2. If the bottom line of the matrix we obtain only consists of zeros, then the determinant is zero, and thus the determinant of the original matrix was zero. 


3. Otherwise, the matrix has been transformed into an upper triangular matrix, all of whose diagonal elements are 1. But now we can transform this matrix into the identity matrix I_n( by elementary row operations of the type S_ij(c). Since we know that det(I_n) must be 1, we then find a unique value for the determinant of the original matrix A. In particular, in this case det(A) ≠ 0.

Note that in both this algorithm, as well as in the algorithm for finding the inverse of a regular matrix, the method of Gaussian elimination was used. Thus we can combine both ideas into a single algorithm, suitable for practical calculations in a computer, which yields both the matrix inverse (if it exists), and the determinant. This algorithm also proves the following theorem.


Theorem 34
There is only one determinant function and it is uniquely given by our algorithm. Furthermore, a matrix A ∈ M(n × n, F) is regular if and only if det(A) ≠ 0.

In particular, using these methods it is easy to see that the following theorem is true.


Theorem 35
Let A, B ∈ M(n×n, F). Then we have det(A· B) = det(A) · det(B).

Proof
If either A or B is singular, then A · B is singular. This can be seen by thinking about the linear mappings V → V which A and B represent. At least one of these mappings is singular. Thus the dimension of the image is less than n, so the dimension of the image of the composition of the two mappings must also be less than n. Therefore A · B must be singular. That means, on the one hand, that det(A · B) = 0. And on the other hand, that either det(A) = 0 or else det(B) = 0. Either way, the theorem is true in this case.

If both A and B are regular, then they are both in GL(n, F). Therefore, as we have seen, they can be written as products of elementary matrices. It suffices then to prove that det(S₁)det(S₂) = det(S₁S₂), where S₁ and S₂ are elementary matrices. But our arguments above show that this is, indeed, true.

Remembering that A is regular if and only if A ∈ GL(n, F), we have:

Corollary

If A ∈ GL(n, F) then det(A⁻¹) = (det(A))⁻¹. 

In particular, if det(A) = 1 then we also have det(A⁻¹) = 1. The set of all such matrices must then form a group.

Another simple corollary is the following.

Corollary
Assume that the matrix A is in block form, so that the linear mapping which it represents splits into a direct sum of invariant subspaces (see theorem 29). Then det(A) is the product of the determinants of the blocks.

Proof
If

That is, for the matrix A_i^*, all the blocks except the i-th block are replaced with identity-matrix blocks. Then A = A₁^* · · · A_p^*, and it is easy to see that det(A_i^*) = det(A_i) for each i.

Definition
The special linear group of order n is defined to be the set

 SL(n, F) = {A ∈ GL(n, F) : det(A) = 1}. 

Theorem 36
Let A' = C⁻¹AC. Then det(A ′ ) = det(A).

Proof
This follows, since det(C⁻¹) = (det(C))⁻¹.

• Linear Algebra: #12 Elementary Matrices

These are n × n matrices which we denote by S_ij, S_i(a), and S_ij(c). They are such that when any n × n matrix A is multiplied on the right by such an S, then the given elementary column operation is performed on the matrix A. Furthermore, if the matrix A is multiplied on the left by such an elementary matrix, then the given row operation on the matrix is performed. It is a simple matter to verify that the following matrices are the ones we are looking for.

Here, everything is zero except for the two elements at the positions ij and ji, which have the value 1. Also the diagonal elements are all 1 except for the elements at ii and jj, which are zero.

Then we have

That is, S_i(a) is a diagonal matrix, all of whose diagonal elements are 1 except for the single element at the position ii, which has the value a.

Finally we have

So this is again just the n × n identity matrix, but this time we have replaced the zero in the ij-th position with the scalar c. It is an elementary exercise to see that:


Theorem 32
Each of the n × n elementary matrices are regular. And thus we can prove that these elementary matrices generate the group GL(n, F). Furthermore, for every elementary matrix, the inverse matrix is again elementary.


Theorem 33
Every matrix in GL(n, F) can be represented as a product of elementary matrices.

Proof

Let A be some arbitrary regular matrix. We have already seen that A can be transformed into a matrix in step form by means of elementary row operations. That is, there is some sequence of elementary matrices: S₁, . . . , S_p, such that the product

 A^∗ = S_p · · · S₁A 

is an n ×n matrix in step form. However, since A was a regular matrix, the number of steps must be equal to n. That is, A^∗ must be a triangular matrix whose diagonal elements are all equal to 1.

But now it is obvious that the elements above the diagonal can all be reduced to zero by elementary row operations of type S_ij(c). These row operations can again be realized by multiplication of A on the right by some further set of elementary matrices: S_p+1, · · · , S_q. This gives us the matrix equation

 S_q· · · S_p+1S_p · · · S₁A = I_n

or

A =  S₁⁻¹ . . . S_p⁻¹S_p+1⁻¹· · · S_q⁻¹

Since the inverse of each elementary matrix is itself elementary, we have thus expressed A as a product of elementary matrices.

This proof also shows how we can go about programming a computer to calculate the inverse of an invertible matrix. Namely, through the process of Gauss elimination, we convert the given matrix into the identity matrix I_n. During this process, we keep multiplying together the elementary matrices which represent the respective row operations. In the end, we obtain the inverse matrix

A⁻¹ = S_q· · · S_p+1S_p · · · S₁. 

We also note that this is the method which can be used to obtain the value of the determinant function for the matrix. But first we must find out what the definition of determinants of matrices is!

• Linear Algebra: #11 Eigenvalues, Eigenspaces, Matrices which can be Diagonalized

Definition
Let f : V → V be a linear mapping of an n-dimensional vector space into itself. A subspace U ⊂ V is called invariant with respect to f if f(U) ⊂ U. That is, f(u) ∈ U for all u ∈ U.


Theorem 28
Assume that the r dimensional subspace U ⊂ V is invariant with respect to f : V → V. Let A be the matrix representing f with respect to a given basis {v₁, . . . , v_n} of V. Then A is similar to a matrix A' which has the following form

Proof
Let {u₁, . . . , u_r} be a basis for the subspace U. Then extend this to a basis {u₁, . . . , u_r, u_r+1, . . . , u_n} of V. The matrix of f with respect to this new basis has the desired form.

Definition
Let U₁, . . . , U_p ⊂ V be subspaces. We say that V is the direct sum of these subspaces if V = U₁+ · · · +U_p, and furthermore if v = u₁+ · · · +u_p such that u_i ∈ U_i, for each i, then this expression for v is unique. In other words, if v = u₁+ · · · +u_p = u₁'+ · · · +u_p' with u_i' ∈ U_i for each i, then u_i = u_i', for each i. In this case, one writes V = U₁ ⊕ · · · ⊕ U_p

This immediately gives the following result:


Theorem 29
Let f : V → V be such that there exist subspaces U_i ⊂ V, for i = 1, . . . , p, such that V = U₁ ⊕ · · · ⊕ U_p and also f is invariant with respect to each U_i. Then there exists a basis of V such that the matrix of f with respect to this basis has the following block form.

where each block A_i is a square matrix, representing the restriction of f to the subspace U_i.

Proof
Choose the basis to be a union of bases for each of the U_i.

A special case is when the invariant subspace is an eigenspace.

Definition
Assume that λ ∈ F is an eigenvalue of the mapping f : V → V. The set {v ∈ V : f(v) = λv} is called the eigenspace of λ with respect to the mapping f. That is, the eigenspace is the set of all eigenvectors (and with the zero vector 0 included) with eigenvalue λ.


Theorem 30
Each eigenspace is a subspace of V.

Proof
Let u, w ∈ V be in the eigenspace of λ. Let a, b ∈ F be arbitrary scalars. Then we have

f(au + bw) = af(u) + bf(w) = aλu + bλw = λ(au + bw). 

Obviously if λ₁ and λ₂ are two different (λ₁ ≠ λ₂ ) eigenvalues, then the only common element of the eigenspaces is the zero vector 0. Thus if every vector in V is an eigenvector, then we have the situation of theorem 29. One very particular case is that we have n different eigenvalues, where n is the dimension of V.


Theorem 31
Let λ₁, . . . , λ_n, be eigenvalues of the linear mapping f : V → V, where  λ_i ≠ λ_j for i ≠ j. Let v₁, . . . , v_n be eigenvectors to these eigenvalues. That is, v_i ≠ 0 and f(v_i) = λ_iv_i, for each i = 1, . . . , n. Then the set {v₁, . . . , v_n} is linearly independent.

Proof
Assume to the contrary that there exist a₁, . . . , a_n, not all zero, with

a₁v₁ + · · · + a_nv_n = 0. 

Assume further that as few of the a_i as possible are non-zero. Let a_p be the first non-zero scalar. That is, a_i = 0 for i < p, and a_p≠ 0. Obviously some other a_k is non-zero, for some k ≠ p, for otherwise we would have the equation 0 = a_pv_p , which would imply that v_p = 0, contrary to the assumption that v_p is an eigenvector. Therefore we have

But, remembering that  λ_i ≠ λ_j for i ≠ j, we see that the scalar term for v_p is zero, yet all other non-zero scalar terms remain non-zero. Thus we have found a new sum with fewer non-zero scalars than in the original sum with the a_is. This is a contradiction.

Therefore, in this particular case, the given set of eigenvectors {v₁, . . . , v_n} form a basis for V. With respect to this basis, the matrix of the mapping is diagonal, with the diagonal elements being the eigenvalues.

• Linear Algebra: #10 Similar Matrices; Changing Bases

Definition
Let A and A' be n ×n matrices. If a matrix C ∈ GL(n, F) exists, such that A' = C⁻¹AC then we say that the matrices A and A' are similar.


Theorem 27
Let f : V → V be a linear mapping and let {u₁, . . . , u_n}, {v₁, . . . , v_n} be two bases for V. Assume that A is the matrix for f with respect to the basis {v₁, . . . , v_n} and furthermore A' is the matrix for f with respect to the basis {u₁, . . . , u_n}.

 Then we have A' = C⁻¹AC.

Proof
From the definition of A', we have

for all i = 1, . . . , n. On the other hand we have

Therefore we have A' = C⁻¹AC.

Note that we have written here v_k (in terms of the summation from l = 1, ...., n), and then we have said that the resulting matrix (which we call C^*) is, in fact, C⁻¹. To see that this is true, we begin with the definition of C itself. We have

That is, CC^* = I_n, and therefore C^* = C⁻¹.

Which mapping does the matrix C represent? From the equations

we see that it represents a mapping g : V → V such that g(v_i) =  u_i for all i, expressed in terms of the original basis {v₁, . . . , v_n}. So we see that a similarity transformation, taking a square matrix A to a similar matrix A' = C⁻¹AC is always associated with a change of basis for the vector space V.

Much of the theory of linear algebra is concerned with finding a simple basis (with respect to a given linear mapping of the vector space into itself), such that the matrix of the mapping with respect to this simpler basis is itself simple — for example diagonal, or at least trigonal.