The Frobenius norm of an n×n identity matrix Ɪn is √n, because ꞮT = Ɪ and then ꞮT · Ɪ = Ɪ. We can therefore conclude that
‖Ɪ‖F = √trace(ꞮT · Ɪ)
‖Ɪ‖F = √trace(Ɪ)
‖Ɪ‖F = √n
as Ɪ consists of only 1s on its diagonal. Read more