FINAL EXAM: SOLUTIONS

MATH 25

1. Simultaneous Diagonalisation

Let f and g be two endomorphisms of a d-dimensional vector space V over a field K. We say that f and g are simultaneously diagonalisable if there exists a basis e_{1}, ⋯, e_{d} of V and elements {λ_{i}}_{i=1···d} and {μ_{i}}_{i=1···d} of K such that f(e_{i}) = λ_{i}e_{i} and g(e_{i}) = μ_{i}e_{i} for i = 1 ⋯ d.

Observe that if f and g are simultaneously diagonalisable, then fg = gf. In the rest of this problem, we will assume that f and g are diagonalisable and that fg = gf. The goal of the problem is to prove that this implies that f and g are simultaneously diagonalisable.

1.1. Prove that if W is a sub-vector space of V, stable under f, then the restriction of f to W is diagonalisable.

Since f is diagonalisable, there exists a polynomial P ∈ K[X] with simple roots such that P (f) = 0 on V. Since W is a subspace of V, P (f) = 0 a fortiori on W and this shows that the restriction of f to W is diagonalisable.

1.2. Prove that if μ ∈ K, then the vector space ker(g − μ Id) is stable under f.

We need to show that if g(x) = μx then g(f(x)) = μf(x) but we know that g(f(x)) = f(g(x)) and therefore g(f(x)) = f(g(x)) = f(μx) = μf(x).

1.3. Prove that f and g are simultaneously diagonalisable.

Since g is diagonalisable, we can write V = ⊕V_{μ} where V_{μ} = ker(g − μ Id) and μ runs over the eigenvalues of g. We know by 1.2 that each V_{μ} is stable under f and we know by 1.1 that f is diagonalisable on each V_{μ}. Each V_{μ} therefore has a basis consisting of eigenvectors of f and they are also eigenvectors of g since any element of V_{μ} is one.

The union of those bases over μ gives a basis of V consisting of vectors which are eigenvectors for both f and g so that f and g are simultaneously diagonalisable.

2. Topology of some spaces of matrices

In the remainder of this exam we will study the topology of certain spaces of matrices with complex coefficients. The absolute value | · | is the usual one on C. We will need to use three topological spaces:

• The space C^{n} is the space of n-uples (x_{1}, ⋯, x_{n}) with the distance defined by d(x, y) = sup_{1≤i≤n} |x_{i} − y_{i}| if x = (x_{1}, ⋯, x_{n}) and y = (y_{1}, ⋯, y_{n}).

• The space C[X]_{n} is the set of polynomials P (X) = X^{n} + p_{n−1}X^{n−1} + ⋯ + p_{0} with the distance defined by d(P, Q) = sup_{0≤i≤n−1} |p_{i} −q_{i}| if P (X) = X^{n} +p_{n−1}X^{n−1} +⋯+p_{0} and Q(X) = X^{n} + q_{n−1}X^{n−1} + ⋯ + q_{0}.

• The space M(n, C) is the space of all n × n matrices with complex coefficients. If A = (a_{ij})_{1≤i,j≤n} and B = (b_{ij})_{1≤i,j≤n}, we define d(A, B) = sup_{1≤i,j≤n} |a_{ij} − b_{ij}|.

2.1. Explain briefly why the following maps are continuous. You should not use ϵ/δ arguments; no more than a few lines of justification are necessary:

(1) det : M(n, C) → C.

(2) the “characteristic polynomial” map χ : M(n, C) → C[X]_{n} which maps M to χ_{M} = det(X Id −M).

(3) the map f : C^{n} → C[X]_{n} which maps (z_{1}, ⋯, z_{n}) to ∏_{i=1}^{n}(X − z_{i}).

(4) the “conjugation by P ” map M ↦ P MP^{−1} from M(n, C) to itself, where P is an invertible matrix.

All those maps are basically given by polynomials in the variables which occur and are therefore continuous.

2.2. Prove that the set of invertible matrices M ∈ M(n, C) is open in M(n, C).

We’ve seen in class that M is invertible if and only if det(M) ≠ 0. This means that the set of invertible matrices in M(n, C) is det^{−1}(C^{∗}) which is open in M(n, C) because C^{∗} is open in C and det is a continuous map by 2.1.

3. Diagonalisable matrices

Let D(n, C) be the subset of M(n, C) consisting of diagonalisable matrices. We will show that D(n, C) is dense in M(n, C) and we will compute its interior.

3.1. Let M be a matrix which is upper triangular. Prove that each diagonal entry is an eigenvalue of M.

If the diagonal entries are the m_{ii}’s then det(X Id −M) = ∏_{i}(X − m_{ii}) and we’ve seen in class that the eigenvalues are exactly the roots of the characteristic polynomial, so they are exactly the m_{ii}’s.

3.2. Prove that D(n, C) is dense in M(n, C).

Let M be a matrix in M(n, C). We know that there exists an invertible matrix P such that Q = P^{−1}MP is upper triangular. Let Q_{i} be a matrix which is the same as Q except that the coefficients q_{1}, ⋯, q_{n} on the diagonal are changed to some complex numbers q_{1}^{i}, ⋯, q_{n}^{i} such that: (1) for j = 1 ⋯ n, |q_{j}^{i} − q_{j}| < 1/i and (2) the numbers q_{1}^{i}, ⋯, q_{n}^{i} are pairwise distinct.

Condition (1) shows that Q_{i} → Q. Condition (2) shows that Q_{i} has n distinct eigenvalues (by 2.1), and therefore it is diagonalisable. Since the map X ↦ P XP^{−1} is continuous by 2.1, Q_{i} → Q implies that P Q_{i}P^{−1} → M and each P Q_{i}P^{−1} is in D(n, C). Therefore, D(n, C) is dense in M(n, C).

3.3. Prove that if E is a non-empty subset of C^{n} then E is bounded if and only if f(E) ⊂ C[X]_{n} is bounded (here f is the map from (3) of exercise 2.1).

It is clear that if E is bounded, then f(E) is bounded - indeed, if |z_{i}| < M and P (X) = ∏(X − z_{i}) = X^{n} + p_{n−1}X^{n−1} + ⋯ + p_{0}, then p_{j} is the sum of all possible products of n − j distinct z_{i}’s and so |p_{j}| < M^{n−j}\binom{n}{j}.

The reverse implication is your homework number 9, problem A.3. It says that if you bound the coefficients of P, then you can bound its roots, so that f(E) bounded implies E bounded.

3.4. Use this to prove that the set of polynomials having only simple roots is open in C[X]_{n}.

This is the same as proving that the set of polynomials having a double (or worse) root is closed. Let the P_{i}’s be polynomials having a double root z_{i} and such that P_{i} → P where P is a polynomial. We will prove that P also has a double root.

Since P_{i} → P, there is some closed and bounded ball E of C[X]_{n} containing the P_{i}’s and P. By 3.3, f(E) is bounded so \overline{f(E)} is compact, and some subsequence z_{φ(i)} of the z_{i}’s will have a limit z.

Since z_{φ(i)} is a double root of P_{φ(i)} and P_{φ(i)} → P and z_{φ(i)} → z, we see that z is a double root of P (to see this, use the fact that z is a double root of P if and only if P (z) = P′(z) = 0 and that if P_{φ(i)} → P, then P′_{φ(i)} → P′).

3.5. Prove that the interior of D(n, C) in M(n, C) is the set of matrices having n distinct eigenvalues.

First we show that if M has n distinct eigenvalues, then M is in the interior of D(n, C). We know by 2.1 that the map M ↦ χ_{M} is continuous. If M has n distinct eigenvalues, then χ_{M} has simple roots and by 3.4 there is some open ball U around χ_{M} which contains only polynomials with simple roots. The set χ^{−1}(U) is an open subset of M(n, C) (since χ is continuous) containing M and only matrices with n distinct eigenvalues which are therefore in D(n, C) also. This shows that M is in the interior of D(n, C).

Now suppose that M has fewer than n distinct eigenvalues, so that there exists P such that Q = P^{−1}MP is diagonal with Q_{11} = Q_{22}. Let Q_{ϵ} be the matrix Q but with Q_{12} = ϵ. It is clear that Q_{ϵ} → Q as ϵ → 0 and that Q_{ϵ} is not in D(n, C). We then have P Q_{ϵ}P^{−1} → M and P Q_{ϵ}P^{−1} is not in D(n, C) so that M is not in the interior of D(n, C).