Consequently, a compact operator "T" cannot have a bounded inverse if "H" is infinite-dimensional. If "ST" = "TS" = "I", then the identity operator would be compact, a contradiction. If sequences of bounded operators "Bn" → "B", "Cn" → "C" in the strong operator topology and "T" is compact, then formula_6 converges to formula_7 in norm. For example, consider the Hilbert space formula_8 with standard basis {"en"}. Let "Pm" be the orthogonal projection on the linear span of {"e"1 ... "em"}. The sequence {"Pm"} converges to the identity operator "I" strongly but not uniformly. Define "T" by formula_9 "T" is compact, and, as claimed above, "PmT" → "IT" = "T" in the uniform operator topology: for all "x",  Notice each "Pm" is a finite-rank operator. Similar reasoning shows that if "T" is compact, then "T" is the uniform limit of some sequence of finite-rank operators. By the norm-closedness of the ideal of compact operators, the converse is also true. The quotient C*-algebra of "L"("H") modulo the compact operators is called the Calkin algebra, in which one can consider properties of an operator up to compact perturbation. A bounded operator "T" on a Hilbert space "H" is said to be self-adjoint if "T" = "T*", or equivalently,  It follows that  is real for every "x" ∈ "H", thus eigenvalues of "T", when they exist, are real. When a closed linear subspace "L" of "H" is invariant under "T", then the restriction of "T" to "L" is a self-adjoint operator on "L", and furthermore, the orthogonal complement "L"⊥ of "L" is also invariant under "T". For example, the space "H" can be decomposed as orthogonal direct sum of two "T"–invariant closed linear subspaces: the kernel of "T", and the orthogonal complement of the kernel (which is equal to the closure of the range of "T", for any bounded self-adjoint operator).