A limit involving unitary matrices

Posted: April 17, 2011 in Elementary Algebra; Problems & Solutions, Linear Algebra
Tags: , ,

Throughout n \geq 1 is an ineteger, V = \mathbb{C}^n, and A is an n \times n matrix with entries in \mathbb{C}.

Definition 1. The complex conjugate of A=[a_{ij}] is the matrix \overline{A}=[\overline{a_{ij}}], where \overline{a_{ij}} is the complex conjugate of a_{ij}. Let A^* be the transpose of \overline{A}.

Remark 1. Recall that the standard inner product over the \mathbb{C}-vector space V = \mathbb{C}^n is defined by \langle u,v \rangle = \sum_{i=1}^n u_i \overline{v_i}, for all vectors u,v \in V, where u_i,v_i are the i-th entry of u,v respectively.

Remark 2. A direct and easy computation shows that \langle Au , v \rangle = \langle u , A^*v \rangle, for all u,v \in V. Since obviously (A^*)^*=A, we also have \langle A^*u, v \rangle = \langle u, Av \rangle, for all u, v \in V.

Lemma. Let T and T^* be the linear transformations corresponding  to A and A^* respectively, i.e. T(v) = Av and T^*(v)=A^*v, for all v \in V = \mathbb{C}^n. Then \text{Im}(T) \oplus \ker(T^*)=V.

Proof. We first remind the reader a couple of things from elementary linear algebra: if W is a subspace of V, then W^{\perp} = \{v \in V : \ \langle v, w \rangle = 0, \ \forall w \in W \}, where the inner product is the one defined in Remark 1. Then W^{\perp} is a subspace of V and W \oplus W^{\perp} = V. So, to prove the lemma, we let W = \text{Im}(T) and we will show that W^{\perp} = \ker(T^*). This is easy to do: v \in W^{\perp} if and only if \langle v,w \rangle = 0, for all w \in W. So, since W=\{ T(u): \ u \in V \}, we have v \in W^{\perp} if and only if \langle v, T(u) \rangle = 0, for all u \in V. But, by Remark 2, we have \langle v, T(u) \rangle = \langle T^*(v), u \rangle. So v \in W^{\perp} if and only if \langle T^*(v), u \rangle = 0, for all u \in V. Thus v \in W^{\perp} if and only if T^*(v) = 0, i.e. v \in \ker(T^*). \ \Box

Definition 2. If A is invertible and A^{-1}=A^*, then A is called unitary.

Remark 3. If A is unitary, then ||Au|| =||u||, for all u \in V = \mathbb{C}^n. The reason is that, by Remark 2 and the above definition, we have ||Au||^2=\langle Au, Au \rangle = \langle u, A^*Au \rangle = \langle u,u \rangle = ||u||^2. Using induction, we get ||A^iu||=||u||, for all integers i and all u \in V.

Problem. Suppose that A is unitary and v \in V = \mathbb{C}^n. Evaluate L=\displaystyle \lim_{m \to \infty} \frac{1}{m} \sum_{i=1}^m A^iv.

Solution. Let T and T^* be the linear transformations corresponding to A and A^* respectively and let I be the identity matrix. Apply the above lemma to the matrix I - A to get \text{Im}(\text{id}-T) \oplus \ker(\text{id}-T^*)=V, where \text{id} is the identity transformation. So there exist (unique) v_1 \in \text{Im}(\text{id}-T) and v_2 \in \ker(\text{id}-T^*) such that v=v_1+v_2.  Let

\displaystyle L_j = \lim_{m\to\infty} \frac{1}{m} \sum_{i=1}^m A^i v_j, \ j=1,2.

Then L=L_1+L_2. Let’s find L_1 first. Since v_1 \in \text{Im}(\text{id}-T), we have v_1 = u - T(u)=u-Au, for some u \in V. Thus \sum_{i=1}^m A^iv_1 = Au-A^{m+1}u and so, by Remark 3,

|| \sum_{i=1}^m A^i v_1|| = ||Au - A^{m+1}u|| \leq ||Au|| + ||A^{m+1}u||=2||u||.

Hence

\displaystyle || \frac{1}{m} \sum_{i=1}^m A^iv_1 || \leq \frac{2||u||}{m}

and therefore L_1=0. We are now going to find L_2. Since v_2 \in \ker(\text{id}-T^*), we have T^*(v_2)=v_2. Thus A^{-1}v_2=A^*v_2=v_2 and so Av_2=v_2. Hence A^iv_2=v_2 for all integers i and so \displaystyle \frac{1}{m} \sum_{i=1}^m A^iv_2 = v_2. Thus L_2=v_2 and hence L=L_1+L_2=v_2. \ \Box

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