## 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 $L=L_1+L_2=v_2. \ \Box$