Linear Algebra Proof 5

Question

A path between two elements A and B in a vector space V is defined to be a continuous function:

P(t) on 0 ≤ t ≤ 1 with P(0) = A

and P(1) = B.

A set in a vector space is called connected if every two of its members are joined by a path consisting entirely of members from

the set. Prove the following:

(a) Let S₁ be the set of all orthogonal n x n matrices. Show this is not a connected set in R^{n x n} .

(b) Let S₂ be set of all complex-valued n x n unitary matrices. Show that this set is connected in C^{n x n} .

Answer 1

(a) Recall that any orthogonal matrix M has determinant det M = 1 or -1. Let A be an orthogonal matrix with det A = 1 and B an orthogonal matrix with det B = -1. Suppose there exists a path P from A to B in S₁. Then for each P(t) with 0 ≤ t ≤ 1, we have det P(t) = 1 or -1. Since det: S₁→ {-1,1} is a continuous function, det P(t) must also be a continuous function of t (since it is the composition of two continuous functions). However, this is not possible since at some point between t = 0 and t = 1, det P(t) discontinuously changes from 1 to -1. Hence, there cannot exist a path from A to B. This shows that S₁ is not connected.

(b) We will show that, given any unitary matrix U, there exists a path from the identity matrix I (note that I is unitary) to U in S₂. This will imply that there exists a path between any U₁ and U₂ in S₂ since the composition of two paths such that the terminal point of the first is the initial point of the second is another path, proving that S₂ is connected. By the Spectral Theorem applied to unitary matrices, U can be unitarily diagonalized and the eigenvalues of U all have absolute value 1. This means there exists a unitary matrix V such that U takes the form U = V diag{exp[iθ₁], ..., exp[iθn]}V^-1. Now define P(t) for 0 ≤ t ≤ 1 by

P(t) = V diag{exp[itθ₁], ..., exp[itθn]}V^-1. This is a continuous function of t such that P(0) = I and P(1) = U, hence P(t) is a path from I to U.