Isometries of the Vector p-Norms: Signed and Complex Permutation Matrices
Recall that in linear algebra, the vector p-norm of a vector x ∈ Cn (or x ∈ Rn) is defined to be
where xi is the ith element of x and 1 ≤ p ≤ ∞ (where the p = ∞ case is understood to mean the limit as p approaches ∞, which gives the maximum norm). By far the most well-known of these norms is the Euclidean norm, which arises when p = 2. Another well-known norm arises when p = 1, which gives the “taxicab” norm.
The problem that will be investigated in this post is to characterize what operators preserve the p-norms – i.e., what their isometries are. In the p = 2 case of the Euclidean norm, the answer is well-known: the isometries of the real Euclidean norm are exactly the orthogonal matrices, and the isometries of the complex Euclidean norm are exactly the unitary matrices. It turns out that if p ≠ 2 then the isometry group looks much different. Indeed, Exercise IV.1.3 of [1] asks the reader to show that the isometries of the p = 1 and p = ∞ norms are what are known as complex permutation matrices (to be defined). We will investigate those cases as well as a situation when p ≠ 1, 2, ∞.
p = 1: The “Taxicab” Norm
Recall that a permutation matrix is a matrix with exactly one “1” in each of its rows and columns, and a “0” in every other position. A signed permutation matrix (sometimes called a generalized permutation matrix) is similar – every row and column has exactly one non-zero entry, which is either 1 or -1. Similarly, a complex permutation matrix is a matrix for which every row and column has exactly one non-zero entry, and every non-zero entry is a complex number with modulus 1.
It is not difficult to show that if x ∈ Rn then the taxicab norm of x is preserved by signed permutation matrices, and if x ∈ Cn then the taxicab norm of x is preserved by complex permutation matrices. We will now show that the converse holds:
Theorem 1. Let P ∈ Mn be an n × n matrix. Then
if and only if P is a complex permutation matrix (or a signed permutation matrix, respectively).
Proof. We only prove the “only if” implication, because the “if” implication is trivial (an exercise left for the reader?). So let’s suppose that P is an isometry of the p = 1 vector norm. Let ei denote the ith standard basis vector, let pi denote the ith column of P, and let pij denote the (j,i)-entry of P (i.e., the jth entry of pi). Then Pei = pi for all i, so
Similarly, P(ei + ek) = pi + pk for all i,k, so
However, by the triangle inequality for the absolute value we know that the above equality can only hold if there exist non-negative real constants cijk ≥ 0 such that pij = cijkpkj. However, it is similarly the case that P(ei – ek) = pi – pk for all i,k, so
Using the equality condition for the complex absolute value again we then know that there exist non-negative real constants dijk ≥ 0 such that pij = -dijkpkj. Using the fact that each cijk and each dijk is non-negative, it follows that each row contains at most one non-zero entry (and each row must indeed contain at least one non-zero entry since the isometries of any norm must be nonsingular).
Thus every row has exactly one non-zero entry. By using (again) the fact that isometries must be nonsingular, it follows that each of the non-zero entries must occur in a distinct column (otherwise there would be a zero column). The fact that each non-zero entry has modulus 1 follows from simply noting that P must preserve the p = 1 norm of each ei.
p = ∞: The Maximum Norm
As with the p = 1 case, it is not difficult to show that if x ∈ Rn then the maximum norm of x is preserved by signed permutation matrices, and if x ∈ Cn then the maximum norm of x is preserved by complex permutation matrices. We will now show that the converse holds in this case as well:
Theorem 2. Let P ∈ Mn be an n × n matrix. Then
if and only if P is a complex permutation matrix (or a signed permutation matrix, respectively).
Proof. Again, we only prove the “only if” implication, since the “if” implication is trivial. So suppose that P is an isometry of the p = ∞ vector norm. As before, let ei denote the ith standard basis vector, let pi denote the ith column of P, and let pij denote the (j,i)-entry of P (i.e., the jth entry of pi). Then Pei = pi for all i, so
In other words, each entry of P has modulus at most 1, and each column has at least one element with modulus equal to 1. Also, P(ei ± ek) = pi ± pk for all i,k, so
It follows that if |pij| = 1, then pkj = 0 for all k ≠ i. Because each column has an element with modulus 1, it follows that each row has exactly 1 non-zero entry. Because each column has an entry with modulus 1, it follows that each row and column has exactly 1 non-zero entry, which must have modulus 1, so P is a signed or complex permutation matrix.
Any p ≠ 2
When p = 2, the isometries are orthogonal/unitary matrices. When p = 1 or p = ∞, the isometries are signed/complex permutation matrices, which are a very small subset of the orthogonal/unitary matrices. One might naively expect that the isometries for other values of p somehow interpolate between those two extremes. Alternatively, one might expect that the signed/complex permutation matrices are the only isometries for all other values of p as well. It turns out that the latter conjecture is correct [2,3].
Theorem 3. Let P ∈ Mn be an n × n matrix and let p ∈ [1,2) ∪ (2,∞]. Then
if and only if P is a complex permutation matrix (or a signed permutation matrix, respectively).
References:
- R. Bhatia, Matrix analysis. Volume 169 of Graduate texts in mathematics (1997).
- S. Chang and C. K. Li, Certain Isometries on Rn. Linear Algebra Appl. 165, 251–265 (1992).
- C. K. Li, W. So, Isometries of lp norm. Amer. Math. Monthly 101, 452–453 (1994).
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