§ 30. Normal and Self-Adjoint Operators

We assume that all vector spaces are over the field \(F\), where \(F\) denotes either \(\mathbb{R}\) or \(\mathbb{C}\).

We have seen the importance of diagonalizable operators in Chapter 5. For these operators, it is necessary and sufficient for the vector space \(V\) to possess a basis of eigenvectors. As \(V\) is an inner product space in this chapter, it is reasonable to seek conditions that guarantee that \(V\) has an orthonormal basis of eigenvectors.

Lemma 30.1

Let \(T\) be a linear operator on a finite-dimensional inner product space \(V\). If \(T\) has an eigenvector, then so does \(T^{*}\).

Proof

Suppose that \(v\) is an eigenvector of \(T\) with corresponding eigenvalue \(\lambda\). Then for any \(x \in V\),

\[ 0=\langle 0, x\rangle=\langle(T-\lambda I)(v), x\rangle=\left\langle v,(T-\lambda I)^{*}(x)\right\rangle=\left\langle v,\left(T^{*}-\overline{\lambda} I\right)(x)\right\rangle, \]

and hence \(v\) is orthogonal to the range of \(T^{*}-\overline{\lambda} I\). So \(T^{*}-\overline{\lambda} I\) is not onto and hence is not one-to-one. Thus \(T^{*}-\overline{\lambda} I\) has a nonzero null space, and any nonzero vector in this null space is an eigenvector of \(T^{*}\) with corresponding eigenvalue \(\overline{\lambda}\).

Theorem 30.2 : Schur theorem

Let \(T\) be a linear operator on a finite-dimensional inner product space \(V\). Suppose that the characteristic polynomial of \(T\) splits. Then there exists an orthonormal basis \(\beta\) for \(V\) such that the matrix \([T]_{\beta}\) is upper triangular.

Proof

The proof is by mathematical induction on the dimension \(n\) of \(V\). The result is immediate if \(n=1\). So suppose that the result is true for linear operators on \((n-1)\)-dimensional inner product spaces whose characteristic polynomials split. By Lemma 30.1, we can assume that \(T^{*}\) has a unit eigenvector \(z\). Suppose that \(T^{*}(z)=\lambda z\) and that \(W=\operatorname{span}(\{z\})\).

We show that \(W^{\perp}\) is \(T\)-invariant. If \(y \in W^{\perp}\) and \(x=c z \in W\), then

\[ \begin{aligned} \langle T(y), x\rangle & =\langle T(y), c z\rangle=\left\langle y, T^{*}(c z)\right\rangle=\left\langle y, c T^{*}(z)\right\rangle=\langle y, c \lambda z\rangle \\ & =\overline{c \lambda}\langle y, z\rangle=\overline{c \lambda}(0)=0 . \end{aligned} \]

So \(T(y) \in W^{\perp}\). It is easy to show (see Theorem 26.6) that the characteristic polynomial of \(T_{W^{\perp}}\) divides the characteristic polynomial of \(T\) and hence splits.

By Theorem 28.20(c), \(\operatorname{dim}\left(W^{\perp}\right)=n-1\), so we may apply the induction hypothesis to \(T_{W^{\perp}}\) and obtain an orthonormal basis \(\gamma\) of \(W^{\perp}\) such that \(\left[T_{W^{\perp}}\right]_{\gamma}\) is upper triangular. Clearly, \(\beta=\gamma \cup\{z\}\) is an orthonormal basis for \(V\) such that \([T]_{\beta}\) is upper triangular.

Normality and Orthonormal Diagonalizability in Complex Inner Product Spaces

Definition 30.3 : Normal Operator / Normal Matrix

Let \(V\) be an inner product space, and let \(T\) be a linear operator on \(V\). We say that \(T\) is normal if \(T T^{*}=T^{*} T\).

An \(n \times n\) real or complex matrix \(A\) is normal if \(A A^{*}=A^{*} A\).

Theorem 30.4 : Normality and orthonormal matrix representations

Let \(V\) be a finite-dimensional inner product space, let \(\beta\) be an orthonormal basis for \(V\), and let \(T\) be a linear operator on \(V\). Then \(T\) is normal if and only if \([T]_{\beta}\) is normal.

Proof

Let \(A=[T]_{\beta}\). By Theorem 29.6, we have \(\left[T^{*}\right]_{\beta}=A^{*}\).

Now \(T\) is normal if and only if \(T T^{*}=T^{*} T\). Taking matrix representations with respect to the orthonormal basis \(\beta\) and using the fact that matrix representations respect composition, we obtain

\[ [T T^{*}]_{\beta}=[T]_{\beta}\left[T^{*}\right]_{\beta}=A A^{*} \quad \text{and} \quad [T^{*} T]_{\beta}=\left[T^{*}\right]_{\beta}[T]_{\beta}=A^{*} A . \]

Hence \(T T^{*}=T^{*} T\) if and only if \(A A^{*}=A^{*} A\), which is exactly the condition that \(A=[T]_{\beta}\) is normal.

Theorem 30.5 : Basic properties of normal operators

Let \(V\) be an inner product space, and let \(T\) be a normal operator on \(V\). Then the following statements are true.

• (a) \(\|T(x)\|=\left\|T^{*}(x)\right\|\) for all \(x \in V\).
• (b) \(T-c I\) is normal for every \(c \in F\).
• (c) If \(x\) is an eigenvector of \(T\), then \(x\) is also an eigenvector of \(T^{*}\). In fact, if \(T(x)=\lambda x\), then \(T^{*}(x)=\overline{\lambda} x\).
• (d) If \(\lambda_{1}\) and \(\lambda_{2}\) are distinct eigenvalues of \(T\) with corresponding eigenvectors \(x_{1}\) and \(x_{2}\), then \(x_{1}\) and \(x_{2}\) are orthogonal.

Proof

• (a) For any \(x \in V\), we have

  \[ \begin{aligned} \|T(x)\|^{2} & =\langle T(x), T(x)\rangle=\left\langle T^{*} T(x), x\right\rangle=\left\langle T T^{*}(x), x\right\rangle \\ & =\left\langle T^{*}(x), T^{*}(x)\right\rangle=\left\|T^{*}(x)\right\|^{2} . \end{aligned} \]

• (b) Let \(c \in F\) and set \(S=T-cI\). Then \(S^{*}=T^{*}-\overline{c}\,I\). Compute both products:

  \[ \begin{aligned} S^{*}S &=(T^{*}-\overline{c}\,I)(T-cI) =T^{*}T-cT^{*}-\overline{c}\,T+|c|^{2}I, \\ SS^{*} &=(T-cI)(T^{*}-\overline{c}\,I) =TT^{*}-\overline{c}\,T-cT^{*}+|c|^{2}I . \end{aligned} \]

  Since \(T\) is normal, \(T^{*}T=TT^{*}\), and the remaining terms in the two expansions are identical. Hence \(S^{*}S=SS^{*}\), so \(T-cI\) is normal.

• (c) Suppose that \(T(x)=\lambda x\) for some \(x \in V\). Let \(U=T-\lambda I\). Then \(U(x)=0\), and \(U\) is normal by (b). Thus (a) implies that

  \[ 0=\|U(x)\|=\left\|U^{*}(x)\right\|=\left\|\left(T^{*}-\overline{\lambda} I\right)(x)\right\|=\left\|T^{*}(x)-\overline{\lambda} x\right\| . \]

  Hence \(T^{*}(x)=\overline{\lambda} x\). So \(x\) is an eigenvector of \(T^{*}\).

• (d) Let \(\lambda_{1}\) and \(\lambda_{2}\) be distinct eigenvalues of \(T\) with corresponding eigenvectors \(x_{1}\) and \(x_{2}\). Then, using (c), we have

  \[ \lambda_{1}\left\langle x_{1}, x_{2}\right\rangle=\left\langle\lambda_{1} x_{1}, x_{2}\right\rangle=\left\langle T\left(x_{1}\right), x_{2}\right\rangle=\left\langle x_{1}, T^{*}\left(x_{2}\right)\right\rangle \]
  \[ =\left\langle x_{1}, \overline{\lambda_{2}} x_{2}\right\rangle=\lambda_{2}\left\langle x_{1}, x_{2}\right\rangle . \]

  Since \(\lambda_{1} \neq \lambda_{2}\), we conclude that \(\left\langle x_{1}, x_{2}\right\rangle=0\).

Theorem 30.6 : Orthonormal diagonalizability is equivalent to normality in finite-dimensional complex inner product spaces.

Let \(T\) be a linear operator on a finite-dimensional complex inner product space \(V\). Then \(T\) is normal if and only if there exists an orthonormal basis for \(V\) consisting of eigenvectors of \(T\).

Proof

• (\(\Rightarrow\))

  Suppose that \(T\) is normal. By the fundamental theorem of algebra, the characteristic polynomial of \(T\) splits. So we may apply Theorem 30.2 to obtain an orthonormal basis \(\beta=\left\{v_{1}, v_{2}, \ldots, v_{n}\right\}\) for \(V\) such that \([T]_{\beta}=A\) is upper triangular.

  We know that \(v_{1}\) is an eigenvector of \(T\) because \(A\) is upper triangular. Assume that \(v_{1}, v_{2}, \ldots, v_{k-1}\) are eigenvectors of \(T\). We claim that \(v_{k}\) is also an eigenvector of \(T\). It then follows by mathematical induction on \(k\) that all of the \(v_{i}\)'s are eigenvectors of \(T\). Consider any \(j<k\), and let \(\lambda_{j}\) denote the eigenvalue of \(T\) corresponding to \(v_{j}\). By Theorem 30.5(c), \(T^{*}\left(v_{j}\right)=\overline{\lambda}_{j} v_{j}\). Since \(A\) is upper triangular,

  \[ T\left(v_{k}\right)=A_{1 k} v_{1}+A_{2 k} v_{2}+\cdots+A_{j k} v_{j}+\cdots+A_{k k} v_{k} . \]

  Furthermore, by Corollary 28.10,

  \[ A_{j k}=\left\langle T\left(v_{k}\right), v_{j}\right\rangle=\left\langle v_{k}, T^{*}\left(v_{j}\right)\right\rangle=\left\langle v_{k}, \overline{\lambda}_{j} v_{j}\right\rangle=\lambda_{j}\left\langle v_{k}, v_{j}\right\rangle=0 . \]

  It follows that \(T\left(v_{k}\right)=A_{k k} v_{k}\), and hence \(v_{k}\) is an eigenvector of \(T\). So by induction, all the vectors in \(\beta\) are eigenvectors of \(T\).

• (\(\Leftarrow\))

  Conversely, suppose there exists an orthonormal basis \(\beta=\{v_{1},\ldots,v_{n}\}\) of \(V\) consisting of eigenvectors of \(T\). Then \([T]_{\beta}\) is diagonal; write \([T]_{\beta}=D\). By Theorem 29.6, \(\left[T^{*}\right]_{\beta}=[T]_{\beta}^{*}=D^{*}\), which is also diagonal. Diagonal matrices commute, so \(DD^{*}=D^{*}D\). Translating back to operators (again using that matrix multiplication corresponds to composition in the same basis), we obtain

  \[ [TT^{*}]_{\beta}=[T]_{\beta}[T^{*}]_{\beta}=DD^{*} \quad\text{and}\quad [T^{*}T]_{\beta}=[T^{*}]_{\beta}[T]_{\beta}=D^{*}D, \]

  hence \([TT^{*}]_{\beta}=[T^{*}T]_{\beta}\). Therefore \(TT^{*}=T^{*}T\), and \(T\) is normal.

Example 30.7 : Normality is not enough for orthonormal diagonalizability in infinite-dimensional complex inner product spaces.

Consider the inner product space \(H\) with the orthonormal set \(S\) from Example 27.18. Let \(V=\operatorname{span}(S)\), and let \(T\) and \(U\) be the linear operators on \(V\) defined by \(T(f)=f_{1} f\) and \(U(f)=f_{-1} f\). Then

\[ T\left(f_{n}\right)=f_{n+1} \quad \text { and } \quad U\left(f_{n}\right)=f_{n-1} \]

for all integers \(n\). Thus

\[ \left\langle T\left(f_{m}\right), f_{n}\right\rangle=\left\langle f_{m+1}, f_{n}\right\rangle=\delta_{(m+1), n}=\delta_{m,(n-1)}=\left\langle f_{m}, f_{n-1}\right\rangle=\left\langle f_{m}, U\left(f_{n}\right)\right\rangle . \]

It follows that \(U=T^{*}\). Furthermore, \(T T^{*}=I=T^{*} T\), so \(T\) is normal.

We show that \(T\) has no eigenvectors. Suppose that \(f\) is an eigenvector of \(T\), say, \(T(f)=\lambda f\) for some \(\lambda\). Since \(V\) equals the span of \(S\), we may write

\[ f=\sum_{i=n}^{m} a_{i} f_{i}, \quad \text { where } a_{m} \neq 0 . \]

Hence

\[ \sum_{i=n}^{m} a_{i} f_{i+1}=T(f)=\lambda f=\sum_{i=n}^{m} \lambda a_{i} f_{i} . \]

Since \(a_{m} \neq 0\), we can write \(f_{m+1}\) as a linear combination of \(f_{n}, f_{n+1}, \ldots, f_{m}\). But this is a contradiction because \(S\) is linearly independent.

Example 30.8 : Normality is not enough for orthonormal diagonalizability in real inner product spaces.

Let \(T: \mathbb{R}^{2} \rightarrow \mathbb{R}^{2}\) be rotation by \(\theta\), where \(0<\theta<\pi\). The matrix representation of \(T\) in the standard ordered basis is given by

\[ A=\left(\begin{array}{rr} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right) . \]

Note that \(A A^{*}=I=A^{*} A\), so \(A\), and hence \(T\), is normal.

However, the rotation transform does not have any eigenvectors.

This example illustrates that normality is not sufficient to guarantee the existence of an orthonormal basis of eigenvectors for real inner product spaces.

Self-Adjointness and Orthonormal Diagonalizability in Real Inner Product Spaces

Definition 30.9 : Self-Adjoint (Hermitian) Operator / Self-Adjoint (Hermitian) Matrix

Let \(T\) be a linear operator on an inner product space \(V\). We say that \(T\) is self-adjoint (Hermitian) if \(T=T^{*}\).

An \(n \times n\) real or complex matrix \(A\) is self-adjoint (Hermitian) if \(A=A^{*}\).

Theorem 30.10 : Self-adjointness and orthonormal matrix representations

Let \(V\) be a finite-dimensional inner product space, let \(\beta\) be an orthonormal basis for \(V\), and let \(T\) be a linear operator on \(V\). Then \(T\) is self-adjoint if and only if \([T]_{\beta}\) is self-adjoint.

If \(F=\mathbb{R}\), this condition is equivalent to \([T]_{\beta}\) being symmetric.

Proof

Let \(A=[T]_{\beta}\). By Theorem 29.6, we have \(\left[T^{*}\right]_{\beta}=A^{*}\).

Now \(T\) is self-adjoint if and only if \(T=T^{*}\). Taking matrix representations with respect to \(\beta\) gives

\[ [T]_{\beta}=\left[T^{*}\right]_{\beta} \quad \Longleftrightarrow \quad A=A^{*}. \]

Thus \(T\) is self-adjoint if and only if \([T]_{\beta}\) is self-adjoint.

Lemma 30.11 : Eigenvalues and splitting for self-adjoint operators

Let \(T\) be a self-adjoint operator on a finite-dimensional inner product space \(V\). Then

• (a) Every eigenvalue of \(T\) is real.
• (b) Suppose that \(V\) is a real inner product space. Then the characteristic polynomial of \(T\) splits.

Proof

• (a)
  Suppose that \(T(x)=\lambda x\) for \(x \neq 0\). Because a self-adjoint operator is also normal, we can apply Theorem 30.5(c) to obtain

  \[ \lambda x=T(x)=T^{*}(x)=\overline{\lambda} x . \]

  So \(\lambda=\overline{\lambda}\). That is, \(\lambda\) is real.

• (b)
  Let \(n=\operatorname{dim}(V)\), let \(\beta\) be an orthonormal basis for \(V\), and let \(A=[T]_{\beta}\). Then \(A\) is self-adjoint. Let \(T_{A}\) be the linear operator on \(\mathbb{C}^{n}\) defined by \(T_{A}(x)=A x\) for all \(x \in \mathbb{C}^{n}\). Note that \(T_{A}\) is self-adjoint because \(\left[T_{A}\right]_{\gamma}=A\), where \(\gamma\) is the standard ordered (orthonormal) basis for \(\mathbb{C}^{n}\). So, by (a), the eigenvalues of \(T_{A}\) are real. By the fundamental theorem of algebra, the characteristic polynomial of \(T_{A}\) splits into factors of the form \(t-\lambda\). Since each \(\lambda\) is real, the characteristic polynomial splits over \(\mathbb{R}\). But \(T_{A}\) has the same characteristic polynomial as \(A\), which has the same characteristic polynomial as \(T\). Therefore the characteristic polynomial of \(T\) splits.

Theorem 30.12 : Orthonormal diagonalizability is equivalent to self-adjointness in finite-dimensional real inner product spaces.

Let \(T\) be a linear operator on a finite-dimensional real inner product space \(V\). Then \(T\) is self-adjoint if and only if there exists an orthonormal basis \(\beta\) for \(V\) consisting of eigenvectors of \(T\).

Proof

• (\(\Rightarrow\))
  Suppose that \(T\) is self-adjoint. By Lemma 30.11, we may apply Theorem 30.2 to obtain an orthonormal basis \(\beta\) for \(V\) such that the matrix \(A=[T]_{\beta}\) is upper triangular. But

  \[ A^{*}=[T]_{\beta}^{*}=\left[T^{*}\right]_{\beta}=[T]_{\beta}=A . \]

  So \(A\) and \(A^{*}\) are both upper triangular, and therefore \(A\) is a diagonal matrix. Thus \(\beta\) must consist of eigenvectors of \(T\).

• (\(\Leftarrow\))
  Conversely, suppose there exists an orthonormal basis \(\beta=\{v_{1},\ldots,v_{n}\}\) of \(V\) consisting of eigenvectors of \(T\). Then \(A=[T]_{\beta}\) is diagonal (its diagonal entries are the corresponding eigenvalues). Over \(\mathbb{R}\), a diagonal matrix satisfies \(A^{*}=A^{t}=A\).

  Using Theorem 29.6 again (because \(\beta\) is orthonormal),

  \[ [T^{*}]_{\beta}=[T]_{\beta}^{*}=A^{*}=A=[T]_{\beta}. \]

  By uniqueness of the matrix representation of an operator relative to a fixed basis, this implies \(T^{*}=T\). Hence \(T\) is self-adjoint.

Exercise

Exercise 30.4

Let \(T\) and \(U\) be self-adjoint operators on an inner product space \(V\). Prove that \(TU\) is self-adjoint if and only if \(TU=UT\).

Exercise 30.6

Let \(V\) be a complex inner product space, and let \(T\) be a linear operator on \(V\). Define

\[ T_{1}=\frac{1}{2}\left(T+T^{*}\right) \quad \text { and } \quad T_{2}=\frac{1}{2 i}\left(T-T^{*}\right) . \]

• (a) Prove that \(T_{1}\) and \(T_{2}\) are self-adjoint and that \(T=T_{1}+i T_{2}\).
• (b) Suppose also that \(T=U_{1}+i U_{2}\), where \(U_{1}\) and \(U_{2}\) are self-adjoint. Prove that \(U_{1}=T_{1}\) and \(U_{2}=T_{2}\).
• (c) Prove that \(T\) is normal if and only if \(T_{1} T_{2}=T_{2} T_{1}\).

Exercise 30.7

Let \(T\) be a linear operator on an inner product space \(V\), and let \(W\) be a \(T\)-invariant subspace of \(V\). Prove the following results.

• (a) If \(T\) is self-adjoint, then \(T_{W}\) is self-adjoint.
• (b) \(W^{\perp}\) is \(T^{*}\)-invariant.
• (c) If \(W\) is both \(T\)- and \(T^{*}\)-invariant, then \(\left(T_{W}\right)^{*}=\left(T^{*}\right)_{W}\).
• (d) If \(W\) is both \(T\)- and \(T^{*}\)-invariant and \(T\) is normal, then \(T_{W}\) is normal.

Exercise 30.8

Let \(T\) be a normal operator on a finite-dimensional complex inner product space \(V\), and let \(W\) be a subspace of \(V\). Prove that if \(W\) is \(T\)-invariant, then \(W\) is also \(T^{*}\)-invariant.

Hint: Use Exercise 26.24.

Exercise 30.9

Let \(T\) be a normal operator on a finite-dimensional inner product space \(V\). Prove that \(N(T)=N\left(T^{*}\right)\) and \(R(T)=R\left(T^{*}\right)\).

Hint: Use Theorem 30.5 and Exercise 29.12.

Exercise 30.10

Let \(T\) be a self-adjoint operator on a finite-dimensional inner product space \(V\). Prove that for all \(x \in V\)

\[ \|T(x) \pm i x\|^{2}=\|T(x)\|^{2}+\|x\|^{2} . \]

Deduce that \(T-i I\) is invertible and that \(\left[(T-i I)^{-1}\right]^{*}=(T+i I)^{-1}\).

Exercise 30.11

Assume that \(T\) is a linear operator on a complex (not necessarily finite-dimensional) inner product space \(V\) with an adjoint \(T^{*}\). Prove the following results.

• (a) If \(T\) is self-adjoint, then \(\langle T(x), x\rangle\) is real for all \(x \in V\).

• (b) If \(T\) satisfies \(\langle T(x), x\rangle=0\) for all \(x \in V\), then \(T=T_{0}\).

  Hint: Replace \(x\) by \(x+y\) and then by \(x+i y\), and expand the resulting inner products.

• (c) If \(\langle T(x), x\rangle\) is real for all \(x \in V\), then \(T=T^{*}\).

Exercise 30.12

Let \(T\) be a normal operator on a finite-dimensional real inner product space \(V\) whose characteristic polynomial splits. Prove that \(V\) has an orthonormal basis of eigenvectors of \(T\). Hence prove that \(T\) is self-adjoint.

Exercise 30.13

An \(n \times n\) real matrix \(A\) is said to be a Gramian matrix if there exists a real (square) matrix \(B\) such that \(A=B^{t} B\). Prove that \(A\) is a Gramian matrix if and only if \(A\) is symmetric and all of its eigenvalues are nonnegative.

Hint: Apply Theorem 30.12 to \(T=L_{A}\) to obtain an orthonormal basis \(\left\{v_{1}, v_{2}, \ldots, v_{n}\right\}\) of eigenvectors with the associated eigenvalues \(\lambda_{1}, \lambda_{2}, \ldots, \lambda_{n}\). Define the linear operator \(U\) by \(U\left(v_{i}\right)=\sqrt{\lambda_{i}} v_{i}\).

Exercise 30.14

Simultaneous diagonalization.

Let \(V\) be a finite-dimensional real inner product space, and let \(U\) and \(T\) be self-adjoint linear operators on \(V\) such that \(UT=TU\). Prove that there exists an orthonormal basis for \(V\) consisting of vectors that are eigenvectors of both \(U\) and \(T\). (The complex version of this result appears as Exercise 32.10.)

Hint: For any eigenspace \(W=E_{\lambda}\) of \(T\), we have that \(W\) is both \(T\)- and \(U\)-invariant. By Exercise 30.7, we have that \(W^{\perp}\) is both \(T\)- and \(U\)-invariant. Apply Theorem 30.12 and Theorem 26.6.

Exercise 30.15

Let \(A\) and \(B\) be symmetric \(n \times n\) matrices such that \(AB=BA\). Use Exercise 30.14 to prove that there exists an orthogonal matrix \(P\) such that \(P^{t} A P\) and \(P^{t} B P\) are both diagonal matrices.

Exercise 30.16

Prove the Cayley-Hamilton theorem for a complex \(n \times n\) matrix \(A\). That is, if \(f(t)\) is the characteristic polynomial of \(A\), prove that \(f(A)=0\).

Hint: Use Theorem 30.2 to show that \(A\) may be assumed to be upper triangular, in which case

\[ f(t)=\prod_{i=1}^{n}\left(A_{i i}-t\right) . \]

Now if \(T=L_{A}\), we have \(\left(A_{j j} I-T\right)\left(e_{j}\right) \in \operatorname{span}\left(\left\{e_{1}, e_{2}, \ldots, e_{j-1}\right\}\right)\) for \(j \geq 2\), where \(\left\{e_{1}, e_{2}, \ldots, e_{n}\right\}\) is the standard ordered basis for \(\mathbb{C}^{n}\). (The general case is proved in Section 26.)

Definition 30.13 : Positive Definite / Positive Semidefinite

A linear operator \(T\) on a finite-dimensional inner product space is called positive definite [positive semidefinite] if \(T\) is self-adjoint and \(\langle T(x), x\rangle>0\) [\(\langle T(x), x\rangle \geq 0\)] for all \(x \neq 0\).

An \(n \times n\) matrix \(A\) with entries from \(\mathbb{R}\) or \(\mathbb{C}\) is called positive definite [positive semidefinite] if \(L_{A}\) is positive definite [positive semidefinite].

Exercise 30.17

Let \(T\) and \(U\) be self-adjoint linear operators on an \(n\)-dimensional inner product space \(V\), and let \(A=[T]_{\beta}\), where \(\beta\) is an orthonormal basis for \(V\). Prove the following results.

• (a) \(T\) is positive definite [semidefinite] if and only if all of its eigenvalues are positive [nonnegative].

• (b) \(T\) is positive definite if and only if

  \[ \sum_{i, j} A_{i j} a_{j} \bar{a}_{i}>0 \text { for all nonzero } n \text {-tuples }\left(a_{1}, a_{2}, \ldots, a_{n}\right) . \]

• (c) \(T\) is positive semidefinite if and only if \(A=B^{*} B\) for some square matrix \(B\).

• (d) If \(T\) and \(U\) are positive semidefinite operators such that \(T^{2}=U^{2}\), then \(T=U\).
• (e) If \(T\) and \(U\) are positive definite operators such that \(TU=UT\), then \(TU\) is positive definite.
• (f) \(T\) is positive definite [semidefinite] if and only if \(A\) is positive definite [semidefinite].

Because of (f), results analogous to items (a) through (d) hold for matrices as well as operators.

Exercise 30.18

Let \(T: V \rightarrow W\) be a linear transformation, where \(V\) and \(W\) are finite-dimensional inner product spaces. Prove the following results.

• (a) \(T^{*}T\) and \(TT^{*}\) are positive semidefinite.

  (See Exercise 29.15.)

• (b) \(\operatorname{rank}\left(T^{*} T\right)=\operatorname{rank}\left(TT^{*}\right)=\operatorname{rank}(T)\).

Exercise 30.19

Let \(T\) and \(U\) be positive definite operators on an inner product space \(V\). Prove the following results.

• (a) \(T+U\) is positive definite.
• (b) If \(c>0\), then \(c T\) is positive definite.
• (c) \(T^{-1}\) is positive definite.

Exercise 30.20

Let \(V\) be an inner product space with inner product \(\langle\cdot, \cdot\rangle\), and let \(T\) be a positive definite linear operator on \(V\). Prove that \(\langle x, y\rangle^{\prime}=\langle T(x), y\rangle\) defines another inner product on \(V\).

Exercise 30.21

Let \(V\) be a finite-dimensional inner product space, and let \(T\) and \(U\) be self-adjoint operators on \(V\) such that \(T\) is positive definite. Prove that both \(TU\) and \(UT\) are diagonalizable linear operators that have only real eigenvalues.

Hint: Show that \(UT\) is self-adjoint with respect to the inner product \(\langle x, y\rangle^{\prime}=\langle T(x), y\rangle\). To show that \(TU\) is self-adjoint, repeat the argument with \(T^{-1}\) in place of \(T\).

Exercise 30.22

This exercise provides a converse to Exercise 30.20. Let \(V\) be a finite-dimensional inner product space with inner product \(\langle\cdot, \cdot\rangle\), and let \(\langle\cdot, \cdot\rangle^{\prime}\) be any other inner product on \(V\).

• (a) Prove that there exists a unique linear operator \(T\) on \(V\) such that \(\langle x, y\rangle^{\prime}=\langle T(x), y\rangle\) for all \(x\) and \(y\) in \(V\).

  Hint: Let \(\beta=\left\{v_{1}, v_{2}, \ldots, v_{n}\right\}\) be an orthonormal basis for \(V\) with respect to \(\langle\cdot, \cdot\rangle\), and define a matrix \(A\) by \(A_{i j}=\left\langle v_{j}, v_{i}\right\rangle^{\prime}\) for all \(i\) and \(j\). Let \(T\) be the unique linear operator on \(V\) such that \([T]_{\beta}=A\).

• (b) Prove that the operator \(T\) of (a) is positive definite with respect to both inner products.

Exercise 30.23

Let \(U\) be a diagonalizable linear operator on a finite-dimensional inner product space \(V\) such that all of the eigenvalues of \(U\) are real. Prove that there exist positive definite linear operators \(T_{1}\) and \(T_{1}^{\prime}\) and self-adjoint linear operators \(T_{2}\) and \(T_{2}^{\prime}\) such that \(U=T_{2} T_{1}=T_{1}^{\prime} T_{2}^{\prime}\).

Hint: Let \(\langle\cdot, \cdot\rangle\) be the inner product associated with \(V\), \(\beta\) a basis of eigenvectors for \(U\), \(\langle\cdot, \cdot\rangle^{\prime}\) the inner product on \(V\) with respect to which \(\beta\) is orthonormal (see Exercise 27.22(a)), and \(T_{1}\) the positive definite operator according to Exercise 30.22. Show that \(U\) is self-adjoint with respect to \(\langle\cdot, \cdot\rangle^{\prime}\) and \(U=T_{1}^{-1} U^{*} T_{1}\) (the adjoint is with respect to \(\langle\cdot, \cdot\rangle\)). Let \(T_{2}=T_{1}^{-1} U^{*}\).

Exercise 30.24

This argument gives another proof of Schur's theorem. Let \(T\) be a linear operator on a finite-dimensional inner product space \(V\).

• (a) Suppose that \(\beta\) is an ordered basis for \(V\) such that \([T]_{\beta}\) is an upper triangular matrix. Let \(\gamma\) be the orthonormal basis for \(V\) obtained by applying the Gram-Schmidt orthogonalization process to \(\beta\) and then normalizing the resulting vectors. Prove that \([T]_{\gamma}\) is an upper triangular matrix.
• (b) Use Exercise 26.32 and (a) to obtain an alternate proof of Schur's theorem.