Fix tr for Symmetric/Hermitian block matrices (#55522)

Since `Symmetric` and `Hermitian` symmetrize the diagonal elements of
the parent, we can't forward `tr` to the parent unless it is already
symmetric. This limits the existing `tr` methods to matrices of
`Number`s, which is the common use-case. `tr` for `Symmetric` block
matrices would now use the fallback implementation that explicitly
computes the `diag`.
This resolves the following discrepancy:
```julia
julia> S = Symmetric(fill([1 2; 3 4], 3, 3))
3×3 Symmetric{AbstractMatrix, Matrix{Matrix{Int64}}}:
 [1 2; 2 4]  [1 2; 3 4]  [1 2; 3 4]
 [1 3; 2 4]  [1 2; 2 4]  [1 2; 3 4]
 [1 3; 2 4]  [1 3; 2 4]  [1 2; 2 4]

julia> tr(S)
2×2 Matrix{Int64}:
 3   6
 9  12

julia> sum(diag(S))
2×2 Symmetric{Int64, Matrix{Int64}}:
 3   6
 6  12
```

stdlib/LinearAlgebra/src/symmetric.jl

@@ -449,8 +449,8 @@ Base.copy(A::Adjoint{<:Any,<:Symmetric}) =
 Base.copy(A::Transpose{<:Any,<:Hermitian}) =
     Hermitian(copy(transpose(A.parent.data)), ifelse(A.parent.uplo == 'U', :L, :U))
 
-tr(A::Symmetric) = tr(A.data) # to avoid AbstractMatrix fallback (incl. allocations)
-tr(A::Hermitian) = real(tr(A.data))
+tr(A::Symmetric{<:Number}) = tr(A.data) # to avoid AbstractMatrix fallback (incl. allocations)
+tr(A::Hermitian{<:Number}) = real(tr(A.data))
 
 Base.conj(A::Symmetric) = Symmetric(parentof_applytri(conj, A), sym_uplo(A.uplo))
 Base.conj(A::Hermitian) = Hermitian(parentof_applytri(conj, A), sym_uplo(A.uplo))

stdlib/LinearAlgebra/test/symmetric.jl

@@ -1116,4 +1116,15 @@ end
     end
 end
 
+@testset "tr for block matrices" begin
+    m = [1 2; 3 4]
+    for b in (m, m * (1 + im))
+        M = fill(b, 3, 3)
+        for ST in (Symmetric, Hermitian)
+            S = ST(M)
+            @test tr(S) == sum(diag(S))
+        end
+    end
+end
+
 end # module TestSymmetric