Let \(T\) be a linear operator on the finite-dimensional complex inner product space \(V\). Prove that the following ten statements about \(T\) are equivalent.
\(T\) is normal.
\(||T\alpha|| = ||T^*\alpha||\) for every \(\alpha\) in \(V\).
\(T = T_1 + iT_2\), where \(T_1\) and \(T_2\) are self-adjoint and \(T_1T_2 = T_2T_1\).
If \(\alpha\) is a vector and \(c\) a scalar such that \(T\alpha = c\alpha\), then \(T^*\alpha = \overline{c}\alpha\).
There is an orthonormal basis for \(V\) consisting of characteristic vectors for \(T\).
There is an orthonormal basis \(\mathfrak{B}\) such that \([T]_{\mathfrak{B}}\) is diagonal.
There is a polynomial \(g\) with complex coefficients such that \(T^* = g(T)\).
Every subspace which is invariant under \(T\) is also invariant under \(T^*\).
\(T = NU\), where \(N\) is non-negative, \(U\) is unitary, and \(N\) commutes with \(U\).
\(T = c_1E_1 + \cdots + c_kE_k\), where \(I = E_1 + \cdots + E_k\), \(E_iE_j = 0\) for \(i \neq j\), and \(E_i^2 = E_i = E_i^*\).
Solution
We prove that each condition is equivalent to (a).
1. Conditions (a) and (b) are equivalent.
If \(T\) is normal, then for every \(\alpha\), \[ \begin{aligned} \|T\alpha\|^2 &=(\alpha\mid T^*T\alpha)\\ &=(\alpha\mid TT^*\alpha)\\ &=\|T^*\alpha\|^2. \end{aligned} \] Thus (a) implies (b).
Conversely assume (b) and put \[ H=T^*T-TT^*. \] The operator \(H\) is self-adjoint, and the norm identity says \[ (\alpha\mid H\alpha)=0 \] for every \(\alpha\). By the polarization identity for Hermitian forms, the sesquilinear form \[ (\alpha,\beta)\longmapsto(\alpha\mid H\beta) \] is identically zero. Hence \(H=0\), so \(T^*T=TT^*\) and \(T\) is normal.
2. Conditions (a) and (c) are equivalent.
For every operator define \[ T_1=\frac{T+T^*}{2}, \qquad T_2=\frac{T-T^*}{2i}. \] Then \(T_1,T_2\) are self-adjoint and \[ T=T_1+iT_2. \] A direct calculation gives \[ T_1T_2-T_2T_1 =\frac{1}{2i}(T^*T-TT^*). \] Thus \(T\) is normal exactly when \(T_1\) and \(T_2\) commute, proving (a)\(\iff\)(c).
3. Condition (a) implies (e), and (e) is equivalent to (f).
The finite-dimensional spectral theorem says precisely that a normal operator on a complex inner-product space has an orthonormal basis of characteristic vectors. Hence (a) implies (e) [@HK71].
Given an ordered orthonormal basis, its vectors are characteristic vectors for \(T\) exactly when the matrix of \(T\) in that basis is diagonal. Thus (e) and (f) are equivalent. Conversely, a diagonal matrix commutes with its adjoint, so (f) implies (a).
4. Condition (a) implies (d).
By step 3, decompose \(V\) as the orthogonal direct sum of the eigenspaces of \(T\). On the eigenspace \(E_c=\ker(T-cI)\), the adjoint is multiplication by \(\overline c\). Therefore \[ T\alpha=c\alpha \quad\Longrightarrow\quad T^*\alpha=\overline c\,\alpha. \] Thus (d) holds.
5. Condition (d) implies (e), hence implies normality.
We argue by induction on \(\dim V\). The claim is trivial in dimension zero. Since the scalar field is \(\mathbb C\), a nonzero finite-dimensional space has a characteristic vector \(\alpha\) for \(T\); normalize it so that \(\|\alpha\|=1\), and write \[ T\alpha=c\alpha. \] By (d), \[ T^*\alpha=\overline c\,\alpha. \]
Let \(W=\alpha^\perp\). If \(x\in W\), then \[ (Tx\mid\alpha)=(x\mid T^*\alpha)=0, \] so \(W\) is \(T\)-invariant. It is also \(T^*\)-invariant, since \[ (T^*x\mid\alpha)=(x\mid T\alpha)=0. \] The restriction \(T|_W\) again satisfies (d): every characteristic vector of \(T|_W\) is one for \(T\), and the corresponding adjoint action remains inside \(W\). By induction, \(W\) has an orthonormal basis of characteristic vectors for \(T|_W\). Adjoining \(\alpha\) gives an orthonormal basis of characteristic vectors for \(T\). Thus (e), and hence (a), holds.
6. Conditions (a) and (g) are equivalent.
Assume \(T\) is normal and let \(c_1,\ldots,c_r\) be its distinct characteristic values. By Lagrange interpolation there is a polynomial \(g\in\mathbb C[x]\) such that \[ g(c_j)=\overline{c_j} \qquad(1\le j\le r). \] On each eigenspace \(E_{c_j}\), the operator \(g(T)\) acts by \(\overline{c_j}\), which is exactly the action of \(T^*\). Hence \[ T^*=g(T). \] Thus (a) implies (g).
Conversely, if \(T^*=g(T)\), then \(T\) commutes with \(T^*\) because every operator commutes with every polynomial in itself. Therefore \(T\) is normal.
7. Conditions (a) and (h) are equivalent.
If \(T\) is normal, step 6 gives \(T^*=g(T)\). Every \(T\)-invariant subspace is invariant under every polynomial in \(T\), hence under \(T^*\). Thus (a) implies (h).
Conversely assume (h). Let \(T\alpha=c\alpha\) with \(\alpha\ne0\). The line \(L=\mathbb C\alpha\) is \(T\)-invariant, hence by (h) it is \(T^*\)-invariant. So \[ T^*\alpha=d\alpha \] for some \(d\). The adjoint identity gives \[ c\|\alpha\|^2 =(T\alpha\mid\alpha) =(\alpha\mid T^*\alpha) =\overline d\,\|\alpha\|^2, \] and therefore \(d=\overline c\). Thus (d) holds, and step 5 implies that \(T\) is normal.
8. Condition (a) implies (i).
Let \[ T=\sum_{j=1}^r c_jE_j \] be the spectral resolution of the normal operator \(T\), where the \(E_j\) are the mutually orthogonal self-adjoint projections onto its distinct eigenspaces. Define \[ N=\sum_{j=1}^r |c_j|E_j \] and choose \[ u_j= \begin{cases} c_j/|c_j|,&c_j\ne0,\\ 1,&c_j=0, \end{cases} \qquad U=\sum_{j=1}^r u_jE_j. \] Then \(N\) is non-negative, \(U\) is unitary, and both are linear combinations of the same mutually orthogonal projections, so they commute. Moreover \[ NU=\sum_j |c_j|u_jE_j =\sum_j c_jE_j =T. \] Hence (i) holds.
9. Condition (i) implies normality.
Assume \[ T=NU, \] where \(N=N^*\) is non-negative, \(U\) is unitary, and \(NU=UN\). Since \(U\) is invertible, \(N\) also commutes with \(U^*\). Therefore \[ T^*=U^*N, \] and \[ TT^*=NUU^*N=N^2, \] while \[ T^*T=U^*N^2U=N^2. \] Thus \(T\) is normal.
10. Conditions (a) and (j) are equivalent.
If \(T\) is normal, take the distinct characteristic values \(c_1,\ldots,c_k\) and let \(E_i\) be the orthogonal projection onto the \(c_i\)-eigenspace. The spectral theorem gives \[ I=E_1+\cdots+E_k, \qquad E_iE_j=0\quad(i\ne j), \] with \[ E_i^2=E_i=E_i^*, \] and \[ T=c_1E_1+\cdots+c_kE_k. \] Thus (j) holds.
Conversely, from (j), \[ T^*=\overline{c_1}E_1+\cdots+\overline{c_k}E_k. \] Using the orthogonality of the projections, \[ TT^*=\sum_i |c_i|^2E_i=T^*T. \] Hence \(T\) is normal.
Since each of (b)–(j) is equivalent to (a), all ten statements are equivalent.