Some results on n-power operators

Introduction

In this section, we define terms, terminologies and the notations used. We denote Hilbert space with 𝐻 and 𝐵(𝐻) to mean a Banach algebra of all bounded linear operators. Suppose 𝑇 ∈ 𝐵(𝐻), then 𝑇 * is the adjoint of 𝑇. An operator 𝑇 is a self-adjoint operator if 𝑇 = 𝑇 * . T ∈ 𝐵(𝐻) is normal if 𝑇𝑇 * = 𝑇 * 𝑇 and n-power normal if 𝑇 𝑛 𝑇 * = 𝑇 * 𝑇 𝑛 . Then 𝑇 2 𝑇 * = 𝑇 * 𝑇 2 implies that

We have verified that 𝑇 2 𝑇 * = 𝑇 * 𝑇 2 for the matrix T given above.Therefore, T is a 2-power normal operator.

Theorem

Let T be a diagonal matrix. Then T is n-power normal for all integers 𝑛 ≥ 2

Hence T is 2-power normal.

Now suppose that T is 3-power normal, then 𝑇 3 = ( 𝑎 3 0 0 𝑏 3 ) and 𝑇 * = ( 𝑎 0 0 𝑏 )

Hence T is n-power normal for all 𝑛 ≥ 2

Theorem

The product of two commuting normal operators T and S is normal if T commutes with 𝑆 * and S commutes with 𝑇 * . Example Lets illustrate the theorem with a numerical example using 2x2 complex matrix

Proof

First, we verify that 𝑇 and 𝑆 are 2-power normal

Since 𝑇 2 𝑇 * = 𝑇 * 𝑇 2 , 𝑇 then T is 2-power normal.

Since 𝑇𝑆 = 𝑆𝑇, 𝑇 and 𝑆 commute.

We verify that 𝑇𝑆 is 2-power normal.

Since (𝑇𝑆) 2 (𝑇𝑆) * = (𝑇𝑆) * (𝑇𝑆) 2 , 𝑇𝑆 is 2-power normal.

Theorem

The product of two operators S, T is 2-normal if ST + TS = 0 and 𝑆 2 commutes with 𝑇 * and 𝑆 * commutes with 𝑇 2 .

Proof

We note that 𝑆𝑇 = -𝑇𝑆 from the above equation. ) and T = ( 0 𝑖 𝑖 0 )

We show that 𝑆 2 commutes with

Since -𝐼 is a scalar multiple of the identity matrix, it commutes with any matrix, including 𝑇 * .

We now show that 𝑆 * commutes with 𝑇 2 𝑆 * = 𝑆 𝑇 = ( 0 -1 1 0 ) and 𝑇 2 = ( 0 𝑖 𝑖 0

Again, since -𝐼 is a scalar multiple of the identity matrix, it commutes with any matrix, including 𝑆 * .

Theorem

If T is a 2-power normal operator and also partial isometry, then T is n-power normal for all integers n ∈ 𝐵(𝐻)

Proof

Let T be 2-normal and partial isometry. This implies that Hence T is 3-normal Now suppose that 𝑇 𝑘 is k-normal for all integers k = 2,3...

Hence T is (k+1)-normal for all 𝑘 ≥ 2

Theorem

An operator T ∈ B(H) is n-normal if and only if 𝑇 𝑛 is normal for some integer n.

Proof

If T is n-power normal, then this implies that 𝑇 𝑛 𝑇 * = 𝑇 * 𝑇 𝑛 .

This shows that 𝑇 𝑛 (𝑇 * ) 𝑛 = 𝑇 * 𝑇 𝑛 (𝑇 * ) 𝑛-1

=𝑇 * (𝑇 𝑛 𝑇 * )(𝑇 * ) 𝑛-1 =𝑇 * 𝑇 𝑛 𝑇 * )(𝑇 * ) 𝑛-2 =(𝑇 * ) 𝑛 𝑇 𝑛 .

Hence T is n-power normal.

Conversely, let 𝑇 𝑛 be normal, then 𝑇 𝑛 𝑇 = 𝑇𝑇 𝑛 . This implies that by Fuglede theorem that (𝑇 𝑛 ) * 𝑇 = 𝑇(𝑇 𝑛 ) * that is 𝑇 * 𝑇 𝑛 = 𝑇 𝑛 𝑇 * . Hence T is n-normal.

Theorem

If S, T are n-power normal operators which commute, then ST is an n-power normal operator

Proof

Since S, T are commuting n-power normal operators, 𝑆 𝑛 , 𝑇 𝑛 are commuting normal operators. So 𝑆 𝑛 𝑇 𝑛 is a normal operator. Since 𝑆 𝑛 𝑇 𝑛 = (𝑆𝑇) 𝑛 , (𝑆𝑇) 𝑛 is normal. ST is n-normal.

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