189-251B: Algebra 2 Assignment 10

Due: Wednesday, March 26

1. Let T be a normal operator on a finite-dimensional complex inner product space. Show that T has a square root, i.e., there is a normal operator S such that S^{2} = T.

2. Show that a self-adjoint operator T on a finite-dimensional real inner product space need not have a square root. Does it always have a cube root, i.e., is there a self-adjoint linear transformation S such that S^{3} = T?

3. Let V = M_{n}(C) be the complex vector space of n × n matrices with complex entries, and define a complex-valued function on V × V by the rule

⟨A, B⟩ = trace(AB^{∗}),

where B^{∗} denotes the conjugate transpose of the matrix B, and the trace of a matrix is the sum of its diagonal entries.

(a) Show that this function defines a complex inner product on V.

(b) Given a matrix M ∈ M_{n}(C), define a linear transformation T_{M} : V ⟶ V by the rule T_{M}(A) = MA. Show that T_{M}^{∗} = T_{M^{∗}}. (Hint: the identity trace(AB) = trace(BA) may be useful.)

4. A linear operator T on a real inner product space is called an isometry (or an orthogonal transformation) if ||T (v)|| = ||v|| for all v ∈ V. An n×n matrix M is called orthogonal if its columns are an orthonormal set of vectors relative to the standard inner product on R^{n}. Justify this multiple use of the word “orthogonal ” by showing that the matrix of an orthogonal transformation relative to an orthonormal basis for V is an orthogonal matrix. (We already did this, pretty much, in class, but perhaps a bit quickly: this exercise is just to give you a chance to go over your notes...).

5. Show that if T is any linear operator on an inner product space V, then T T^{∗} is self-adjoint.

6. A linear transformation on an inner product space is said to be skewadjoint if T^{∗} = −T. Show that ever linear transformation T on a (real or complex) inner product space can be uniquely writen as the sum of a selfadjoint transformation T_{1} and a skew-adjoint transformation T_{2}, and that T_{1} and T_{2} commute if and only if T is normal.

7. Let V be the vector space of infinite sequences (a_{1}, a_{2}, …) with entries in a field F. The goal of this exercise is to lead you to prove that V never admits a countable basis. In class, we proved this when F is a finite field, by observing that a vector space over F with a countable basis would have to be countable, while V (as a set) is uncountable. As some of you pointed out to me after class, this argument is not completely satisfying: it does not extend to the case where F is uncountable (F = R for example) and besides, such a statement really ought to admit a proof that is “independent of the field”.

a) Show that, if V admits a countable basis, then there is a nested sequence of subspaces

W_{1} ⊂ W_{2} ⊂ ⋯ ⊂ W_{n} ⊂ ⋯

satisfying

dim(W_{j}) = j, ∪_{j=1}^{∞}W_{j} = V.

(The latter equality asserts that every vector v in V belongs to some W_{j}, for j sufficently large and depending of course on v.)

b) It will be convenient to represent a sequence (a_{n})_{n≥1} as a concatenation of sequences (s_{2}, s_{3}, s_{4}, …) where s_{j} is a sequence of length j. For each j ≥ 2, show that there is a sequence S_{j} of length j with the property that

s_{j} = S_{j} ⇒ (s_{2}, …, s_{j}, …) ∉ W_{j−1}.

c) Use b) and a diagonalisation argument analogous to Cantor’s to construct a vector v ∈ V which does not belong to any of the W_{j}’s. Conclude that V does not have a countable basis.