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Let C∞ denote a set of all sequences of complex numbers (z1,z2,z3, . . .) and define `p := {(z1,z2,z3, . . .) ∈ C∞ |∑
∞n=1 |zn|p < ∞} for all 1≤ p < ∞. Then `p is a linear subspace of C∞.
Define L : `p→ `p by L(z1,z2,z3, . . .) = (z2,z3, . . .). This is called the left shift operator. It has the ∞×∞ matrixrepresentation
L =
0 1 0 0 · · ·0 0 1 0 · · ·0 0 0 1 · · ·0 0 0 0 · · ·...
......
.... . .
.
One can easily verify as follows:
L(z1,z2,z3, . . .) =
0 1 0 0 · · ·0 0 1 0 · · ·0 0 0 1 · · ·0 0 0 0 · · ·...
......
.... . .
z1z2z3z4...
=
z2z3z4z5...
= (z2,z3, . . .).
Define R : `p → `p by R(z1,z2,z3, . . .) = (0,z1,z2, . . .). This is called the right shift operator. It has the ∞×∞
matrix representation
R =
0 0 0 0 · · ·1 0 0 0 · · ·0 1 0 0 · · ·0 0 1 0 · · ·...
......
.... . .
.
(We notice from this matrix that R = LT .) One can easily verify as follows:
R(z1,z2,z3, . . .) =
0 0 0 0 · · ·1 0 0 0 · · ·0 1 0 0 · · ·0 0 1 0 · · ·...
......
.... . .
z1z2z3z4...
=
0z1z2z3...
= (0,z1,z2, . . .).
We notice that L has a right inverse, which is R, because
LR =
0 1 0 0 · · ·0 0 1 0 · · ·0 0 0 1 · · ·0 0 0 0 · · ·...
......
.... . .
0 0 0 0 · · ·1 0 0 0 · · ·0 1 0 0 · · ·0 0 1 0 · · ·...
......
.... . .
=
1 0 0 0 · · ·0 1 0 0 · · ·0 0 1 0 · · ·0 0 0 1 · · ·...
......
.... . .
= I.
But L does not have a left inverse because
RL =
0 0 0 0 · · ·1 0 0 0 · · ·0 1 0 0 · · ·0 0 1 0 · · ·...
......
.... . .
0 1 0 0 · · ·0 0 1 0 · · ·0 0 0 1 · · ·0 0 0 0 · · ·...
......
.... . .
=
0 0 0 0 · · ·0 1 0 0 · · ·0 0 1 0 · · ·0 0 0 1 · · ·...
......
.... . .
6= I.
Hence, L is not invertible and is only semi-orthogonal.We would like to note that invariant factors for ∞×∞ matrices do not exist. In other words, there are no minimal
or characteristic polynomials for these kinds of matrices. However, we can still find through other means the matrix’seigenvalues as well as its spectrum.
Theorem 1. If λ lies inside the unit disk in the complex plane, then λ is an eigenvalue of the left-shift operatorL : `p→ `p for all 1≤ p < ∞; that is, σ(L) = {λ ∈C | |λ |< 1}. (Technically, this argument works for 0 < p < ∞; thechoice of 1≤ p < ∞ is for other reasons in graduate real analysis.)
Proof. Let λ ∈C be an eigenvalue of L, so that L(z1,z2,z3, . . .)= λ (z1,z2,z3, . . .) for some (z1,z2,z3, . . .) 6=(0,0,0, . . .).Then
Lz1 = λ z1,
Lz2 = λ z2 = λ2z1,
Lz3 = λ z3 = λ3z1,
...Lzn = λ zn = λ
nz1,
...
So L(z1,z2,z3, . . .) = (λ z1,λ2z1,λ
3z1, . . .).Since again L : `p→ `p, we must have Lzn ∈ `p, which means by its definition we must have ∑
∞n=1 |Lzn|p < ∞. But
we also have∞
∑n=1|Lzn|p =
∞
∑n=1|λ nz1|p = |z1|p
∞
∑n=1|λ |pn.
So we require that |z1|p ∑∞n=1 |λ |pn < ∞, and this holds only if |λ | < 1. Otherwise, if |λ | ≥ 1, then Lzn 6∈ `2 because
|z1|p ∑∞n=1 |λ |2n = ∞. Hence, σ(L) = {λ ∈ C | |λ |< 1} if L : `p→ `p.
Theorem 2. The right-shift operator R : `p→ `p for all 1≤ p < ∞ has no eigenvalues; that is, σ(R) =∅.
Proof. Suppose by way of contradiction that λ is an eigenvalue of R, so that R(z1,z2,z3, . . .) = λ (z1,z2,z3, . . .) forsome (z1,z2,z3, . . .) 6= (0,0,0, . . .). Then
0 = Rz1 = λ z1,
z1 = Rz2 = λ z2,
z2 = Rz3 = λ z3,
...zn+1 = Rzn = λ zn,
...
So (λ z1,λ z2,λ z3, . . .) = (0,z1,z2, . . .).Case 1: Suppose that λ = 0. Then R(z1,z2,z3, . . .) = 0(z1,z2,z3, . . .), and so (0,z1,z2, . . .) = (0,0,0, . . .). This
equation implies that zn = 0 for all n ∈ {1,2, . . .}, and so (z1,z2,z3, . . .) = (0,0,0, . . .), which is a contradiction (aneigenvector corresponding to an eigenvalue must be non-zero).
Case 2: Suppose that λ 6= 0. Then λ z1 = 0 implies that z1 = 0. In turn, λ z2 = z1 = 0 implies z2 = 0, λ z3 = z2 = 0implies z2 = 0, and so on. An induction proof will establish that, in general, zn = 0 for all n ∈ {1,2, . . .}. Thus,(z1,z2,z3, . . .) = (0,0,0, . . .), which is a contradiction.
Hence, λ ∈ C is not an eigenvalue of R. So R : `p→ `p has no eigenvalues, and thus σ(R) =∅.
