§ 27. Inner Products and Norms

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

Inner Product

Definition 27.1 : Inner Product

Let \(V\) be a vector space over \(F\). An inner product on \(V\) is a function that assigns, to every ordered pair of vectors \(x\) and \(y\) in \(V\), a scalar in \(F\), denoted \(\langle x, y\rangle\), such that for all \(x, y\), and \(z\) in \(V\) and all \(c\) in \(F\), the following hold.

• (a) \(\langle x+z, y\rangle=\langle x, y\rangle+\langle z, y\rangle\).
• (b) \(\langle c x, y\rangle=c\langle x, y\rangle\).
• (c) \(\overline{\langle x, y\rangle}=\langle y, x\rangle\), where the bar denotes complex conjugation.
• (d) \(\langle x, x\rangle>0\) if \(x \neq 0\).

Note that (c) reduces to \(\langle x, y\rangle=\langle y, x\rangle\) if \(F=\mathbb{R}\).

Corollary 27.2 : Linearity in the first component

Conditions (a) and (b) simply require that the inner product be linear in the first component. It is easily shown that if \(a_{1}, a_{2}, \ldots, a_{n} \in F\) and \(y, v_{1}, v_{2}, \ldots, v_{n} \in V\), then

\[ \left\langle\sum_{i=1}^{n} a_{i} v_{i}, y\right\rangle=\sum_{i=1}^{n} a_{i}\left\langle v_{i}, y\right\rangle . \]

Definition 27.3 : Standard Inner Product on \(F^{n}\)

For \(x=\left(a_{1}, a_{2}, \ldots, a_{n}\right)\) and \(y=\left(b_{1}, b_{2}, \ldots, b_{n}\right)\) in \(F^{n}\), define

\[ \langle x, y\rangle=\sum_{i=1}^{n} a_{i} \overline{b_{i}} . \]

Proof

We verify that \(\langle\cdot,\cdot\rangle\) satisfies conditions (a) through (d) of Definition 27.1.

Let \(x=(a_{1},\ldots,a_{n})\), \(y=(b_{1},\ldots,b_{n})\), and \(z=(c_{1},\ldots,c_{n})\) be vectors in \(F^{n}\), and let \(c\in F\).

• (a)

  \[ \begin{aligned} \langle x+z, y\rangle &=\sum_{i=1}^{n}(a_i+c_i)\overline{b_i} =\sum_{i=1}^{n}a_i\overline{b_i}+\sum_{i=1}^{n}c_i\overline{b_i} =\langle x, y\rangle+\langle z, y\rangle . \end{aligned} \]

• (b)

  \[ \langle c x, y\rangle =\sum_{i=1}^{n}(c a_i)\overline{b_i} =c\sum_{i=1}^{n}a_i\overline{b_i} =c\langle x, y\rangle . \]

• (c)

  \[ \overline{\langle x, y\rangle} =\overline{\sum_{i=1}^{n}a_i\overline{b_i}} =\sum_{i=1}^{n}\overline{a_i\overline{b_i}} =\sum_{i=1}^{n}\overline{a_i}\,b_i =\sum_{i=1}^{n}b_i\overline{a_i} =\langle y, x\rangle . \]

• (d)

  \[ \langle x, x\rangle=\sum_{i=1}^{n}a_i\overline{a_i}=\sum_{i=1}^{n}|a_i|^{2}\ge 0. \]

  If \(x\neq 0\), then \(a_j\neq 0\) for some \(j\), so \(|a_j|^2>0\), and therefore \(\langle x,x\rangle>0\).

This inner product is called the standard inner product on \(F^{n}\).

When \(F=\mathbb{R}\) the conjugations are not needed, and in early courses this standard inner product is usually called the dot product and is denoted by \(x \cdot y\) instead of \(\langle x, y\rangle\).

Example 27.4 : Inner product on \(\mathcal{C}([0,1])\)

Let \(V=\mathcal{C}([0,1])\), the vector space of real-valued continuous functions on \([0,1]\). For \(f, g \in V\), define \(\langle f, g\rangle=\int_{0}^{1} f(t) g(t) d t\). Since the preceding integral is linear in \(f\), (a) and (b) are immediate, and (c) is trivial. If \(f \neq 0\), then \(f^{2}\) is bounded away from zero on some subinterval of \([0,1]\) (continuity is used here), and hence \(\langle f, f\rangle=\int_{0}^{1}[f(t)]^{2} d t>0\).

Definition 27.5 : Conjugate Transpose (Adjoint)

Let \(A \in \mathrm{M}_{m \times n}(F)\). We define the conjugate transpose or adjoint of \(A\) to be the \(n \times m\) matrix \(A^{*}\) such that \(\left(A^{*}\right)_{i j}=\overline{A_{j i}}\) for all \(i, j\).

Definition 27.6 : Frobenius Inner Product

Let \(V=\mathrm{M}_{n \times n}(F)\), and define \(\langle A, B\rangle=\operatorname{tr}\left(B^{*} A\right)\) for \(A, B \in V\). (Recall that the trace of a matrix \(A\) is defined by \(\operatorname{tr}(A)=\sum_{i=1}^{n} A_{i i}\).)

Proof

We verify that \(\langle\cdot,\cdot\rangle\) satisfies conditions (a) through (d) of Definition 27.1.

• (a)
  Let \(A,B,C \in V\). Using \(C^{*}(A+B)=C^{*}A+C^{*}B\) and \(\operatorname{tr}(X+Y)=\operatorname{tr}(X)+\operatorname{tr}(Y)\), we have

  \[ \begin{aligned} \langle A+B, C\rangle & =\operatorname{tr}\left(C^{*}(A+B)\right) =\operatorname{tr}\left(C^{*}A+C^{*}B\right) \\ & =\operatorname{tr}\left(C^{*}A\right)+\operatorname{tr}\left(C^{*}B\right) =\langle A, C\rangle+\langle B, C\rangle . \end{aligned} \]

• (b)
  Let \(c \in F\) and \(A,B \in V\). Since \(B^{*}(cA)=c(B^{*}A)\) and \(\operatorname{tr}(cX)=c\,\operatorname{tr}(X)\), we obtain

  \[ \langle cA, B\rangle =\operatorname{tr}\left(B^{*}(cA)\right) =\operatorname{tr}\left(c\,B^{*}A\right) =c\,\operatorname{tr}\left(B^{*}A\right) =c\langle A,B\rangle . \]

• (c)
  Let \(A,B \in V\). First note that

  \[ \overline{\operatorname{tr}(X)}=\operatorname{tr}(X^{*}) \]

  for any square matrix \(X\), because \(\overline{\sum_i X_{ii}}=\sum_i \overline{X_{ii}}=\sum_i (X^{*})_{ii}\). Hence, using \((B^{*}A)^{*}=A^{*}B\) and \(\operatorname{tr}(Y)=\operatorname{tr}(Y^{t})\), we have

  \[ \begin{aligned} \overline{\langle A,B\rangle} & =\overline{\operatorname{tr}(B^{*}A)} =\operatorname{tr}\left((B^{*}A)^{*}\right) =\operatorname{tr}(A^{*}B) =\langle B,A\rangle . \end{aligned} \]

• (d)
  Let \(A \in V\). We compute

  \[ \begin{aligned} \langle A,A\rangle & =\operatorname{tr}(A^{*}A) =\sum_{i=1}^{n}(A^{*}A)_{ii} =\sum_{i=1}^{n}\sum_{k=1}^{n}(A^{*})_{ik}A_{ki} \\ & =\sum_{i=1}^{n}\sum_{k=1}^{n}\overline{A_{ki}}\,A_{ki} =\sum_{i=1}^{n}\sum_{k=1}^{n}|A_{ki}|^{2}. \end{aligned} \]

  If \(A\neq O\), then \(A_{ki}\neq 0\) for some \(k,i\), so at least one term \(|A_{ki}|^{2}\) is positive, and therefore \(\langle A,A\rangle>0\).

This inner product on \(\mathrm{M}_{n \times n}(F)\) is called the Frobenius inner product.

Definition 27.7 : Inner Product Space

A vector space \(V\) over \(F\) endowed with a specific inner product is called an inner product space. If \(F=\mathbb{C}\), we call \(V\) a complex inner product space, whereas if \(F=\mathbb{R}\), we call \(V\) a real inner product space.

Concept 27.8 : Integration of complex-valued functions

At this point, we mention a few facts about integration of complex-valued functions. First, the imaginary number \(i\) can be treated as a constant under the integration sign. Second, every complex-valued function \(f\) may be written as \(f=f_{1}+i f_{2}\), where \(f_{1}\) and \(f_{2}\) are real-valued functions. Thus we have

\[ \int f=\int f_{1}+i \int f_{2} \quad \text { and } \quad \overline{\int f}=\int \overline{f} \]

Example 27.9 : Inner product on \(H\)

A very important inner product space that resembles \(\mathcal{C}([0,1])\) is the space \(H\) of continuous complex-valued functions defined on the interval \([0,2 \pi]\) with the inner product

\[ \langle f, g\rangle=\frac{1}{2 \pi} \int_{0}^{2 \pi} f(t) \overline{g(t)} d t \]

The reason for the constant \(1 / 2 \pi\) will become evident later. This inner product space, which arises often in the context of physical situations, is examined more closely in later sections.

Proof

We verify that \(\langle\cdot,\cdot\rangle\) satisfies conditions (a) through (d) of Definition 27.1.

• (a)
  Let \(f,g,h \in H\). Using linearity of the integral and the identity \((f+h)\overline{g}=f\overline{g}+h\overline{g}\), we have

  \[ \begin{aligned} \langle f+h, g\rangle & =\frac{1}{2\pi}\int_{0}^{2\pi}(f(t)+h(t))\overline{g(t)}\,dt \\ & =\frac{1}{2\pi}\int_{0}^{2\pi}f(t)\overline{g(t)}\,dt +\frac{1}{2\pi}\int_{0}^{2\pi}h(t)\overline{g(t)}\,dt \\ & =\langle f,g\rangle+\langle h,g\rangle . \end{aligned} \]

• (b)
  Let \(c \in \mathbb{C}\) and \(f,g \in H\). Since \((cf)\overline{g}=c(f\overline{g})\) and the integral is homogeneous, we obtain

  \[ \begin{aligned} \langle cf, g\rangle & =\frac{1}{2\pi}\int_{0}^{2\pi}(cf(t))\overline{g(t)}\,dt =c\cdot \frac{1}{2\pi}\int_{0}^{2\pi}f(t)\overline{g(t)}\,dt \\ & =c\,\langle f,g\rangle . \end{aligned} \]

• (c)
  Let \(f,g \in H\). Using \(\overline{\int \phi}=\int \overline{\phi}\) and \(\overline{f\overline{g}}=\overline{f}\,g\), we have

  \[ \begin{aligned} \overline{\langle f,g\rangle} & =\overline{\frac{1}{2\pi}\int_{0}^{2\pi} f(t)\overline{g(t)}\,dt} =\frac{1}{2\pi}\int_{0}^{2\pi}\overline{f(t)\overline{g(t)}}\,dt \\ & =\frac{1}{2\pi}\int_{0}^{2\pi}\overline{f(t)}\,g(t)\,dt =\frac{1}{2\pi}\int_{0}^{2\pi} g(t)\overline{f(t)}\,dt =\langle g,f\rangle . \end{aligned} \]

• (d)
  Let \(f \in H\). Then

  \[ \langle f,f\rangle =\frac{1}{2\pi}\int_{0}^{2\pi} f(t)\overline{f(t)}\,dt =\frac{1}{2\pi}\int_{0}^{2\pi} |f(t)|^{2}\,dt . \]

  Since \(|f(t)|^{2}\ge 0\) for all \(t\), we have \(\langle f,f\rangle\ge 0\). If \(f\neq 0\), then there exists \(t_{0}\in[0,2\pi]\) with \(f(t_{0})\neq 0\). By continuity, \(|f(t)|^{2}>0\) on some interval around \(t_{0}\), so the integral is strictly positive. Hence \(\langle f,f\rangle>0\) when \(f\neq 0\).

Therefore \(\langle f,g\rangle=\frac{1}{2\pi}\int_{0}^{2\pi} f(t)\overline{g(t)}\,dt\) defines an inner product on \(H\).

Theorem 27.10 : Basic properties of inner products

Let \(V\) be an inner product space. Then for \(x, y, z \in V\) and \(c \in F\), the following statements are true.

• (a) \(\langle x, y+z\rangle=\langle x, y\rangle+\langle x, z\rangle\).
• (b) \(\langle x, c y\rangle=\overline{c}\langle x, y\rangle\).
• (c) \(\langle x, 0\rangle=\langle 0, x\rangle=0\).
• (d) \(\langle x, x\rangle=0\) if and only if \(x=0\).
• (e) If \(\langle x, y\rangle=\langle x, z\rangle\) for all \(x \in V\), then \(y=z\).

Proof

• (a)
  We have

  \[ \begin{aligned} \langle x, y+z\rangle & =\overline{\langle y+z, x\rangle} =\overline{\langle y, x\rangle+\langle z, x\rangle} \\ & =\overline{\langle y, x\rangle}+\overline{\langle z, x\rangle} =\langle x, y\rangle+\langle x, z\rangle . \end{aligned} \]

• (b)
  Using conjugate symmetry and linearity in the first variable,

  \[ \begin{aligned} \langle x, c y\rangle & =\overline{\langle c y, x\rangle} =\overline{c\langle y, x\rangle} =\overline{c}\,\overline{\langle y, x\rangle} =\overline{c}\,\langle x, y\rangle . \end{aligned} \]

• (c)
  Since \(0=0x\) for any scalar \(0\in F\), by linearity in the first variable we have \(\langle 0, x\rangle=\langle 0x, x\rangle=0\langle x,x\rangle=0\). Then by conjugate symmetry,

  \[ \langle x, 0\rangle=\overline{\langle 0, x\rangle}=\overline{0}=0 . \]

• (d)
  If \(x=0\), then \(\langle x,x\rangle=\langle 0,0\rangle=0\) by (c).

  Conversely, if \(\langle x,x\rangle=0\), then \(x\) cannot be nonzero, since the positivity axiom of an inner product says that \(\langle x,x\rangle>0\) whenever \(x\neq 0\). Hence \(x=0\).

• (e)
  Assume that \(\langle x, y\rangle=\langle x, z\rangle\) for all \(x\in V\). Then for all \(x\in V\),

  \[ \langle x, y-z\rangle=\langle x,y\rangle-\langle x,z\rangle=0, \]

  using part (a) and the fact that \(y+(-z)=y-z\). Taking \(x=y-z\) gives \(\langle y-z, y-z\rangle=0\). By part (d), this implies \(y-z=0\), and hence \(y=z\).

Corollary 27.11 : Conjugate linearity in the second component

The reader should observe that (a) and (b) of Theorem 27.10 show that the inner product is conjugate linear in the second component.

Norm

Definition 27.12 : Norm

Let \(V\) be an inner product space. For \(x \in V\), we define the norm or length of \(x\) by \(\|x\|=\sqrt{\langle x, x\rangle}\).

Example 27.13 : Euclidean norm on \(F^{n}\)

Let \(V=F^{n}\). If \(x=\left(a_{1}, a_{2}, \ldots, a_{n}\right)\), then

\[ \|x\|=\left\|\left(a_{1}, a_{2}, \ldots, a_{n}\right)\right\|=\left[\sum_{i=1}^{n}\left|a_{i}\right|^{2}\right]^{1 / 2} \]

is the Euclidean definition of length. Note that if \(n=1\), we have \(\|a\|=|a|\).

Theorem 27.14 : Cauchy-Schwarz and triangle inequalities

Let \(V\) be an inner product space over \(F\). Then for all \(x, y \in V\) and \(c \in F\), the following statements are true.

• (a) \(\|c x\|=|c| \cdot\|x\|\).
• (b) \(\|x\|=0\) if and only if \(x=0\). In any case, \(\|x\| \geq 0\).
• (c) (Cauchy-Schwarz Inequality) \(|\langle x, y\rangle| \leq\|x\| \cdot\|y\|\).
• (d) (Triangle Inequality) \(\|x+y\| \leq\|x\|+\|y\|\).

Proof

• (a)
  By definition, \(\|x\|=\sqrt{\langle x,x\rangle}\). Using linearity in the first variable and conjugate symmetry,

  \[ \begin{aligned} \|cx\|^{2} & =\langle cx, cx\rangle =c\langle x, cx\rangle =c\,\overline{\langle cx, x\rangle} =c\,\overline{c\langle x,x\rangle} \\ & =c\,\overline{c}\,\overline{\langle x,x\rangle} =|c|^{2}\langle x,x\rangle =|c|^{2}\|x\|^{2}. \end{aligned} \]

  Taking square roots (and noting both sides are \(\ge 0\)) gives \(\|cx\|=|c|\,\|x\|\).

• (b)
  Since \(\|x\|=\sqrt{\langle x,x\rangle}\) and \(\langle x,x\rangle\ge 0\) for all \(x\), we have \(\|x\|\ge 0\).

  If \(x=0\), then \(\|x\|^{2}=\langle 0,0\rangle=0\), so \(\|x\|=0\).

  Conversely, if \(\|x\|=0\), then \(\|x\|^{2}=\langle x,x\rangle=0\). By positive definiteness of the inner product, this implies \(x=0\).

• (c)
  If \(y=0\), then \(|\langle x,y\rangle|=0\le \|x\|\|y\|\). Assume \(y\neq 0\). For any scalar \(c\in F\),

  \[ 0\le \|x-cy\|^{2}=\langle x-cy,\,x-cy\rangle. \]

  Expand using linearity in the first variable, part (b) of Theorem 27.10, and conjugate symmetry:

  \[ \begin{aligned} \langle x-cy,\,x-cy\rangle & =\langle x, x-cy\rangle-\langle cy, x-cy\rangle \\ & =\langle x,x\rangle-\overline{c}\langle x,y\rangle-c\langle y,x\rangle+|c|^{2}\langle y,y\rangle. \end{aligned} \]

  Choose

  \[ c=\frac{\langle x,y\rangle}{\langle y,y\rangle}, \]

  which is valid because \(\langle y,y\rangle>0\) when \(y\neq 0\). Then \(\overline{c}=\dfrac{\langle y,x\rangle}{\langle y,y\rangle}\) and \(|c|^{2}=\dfrac{|\langle x,y\rangle|^{2}}{\langle y,y\rangle^{2}}\). Substituting gives

  \[ \begin{aligned} 0 & \le \langle x,x\rangle -\frac{|\langle x,y\rangle|^{2}}{\langle y,y\rangle} -\frac{|\langle x,y\rangle|^{2}}{\langle y,y\rangle} +\frac{|\langle x,y\rangle|^{2}}{\langle y,y\rangle} \\ & =\langle x,x\rangle-\frac{|\langle x,y\rangle|^{2}}{\langle y,y\rangle}. \end{aligned} \]

  Hence

  \[ |\langle x,y\rangle|^{2}\le \langle x,x\rangle\,\langle y,y\rangle=\|x\|^{2}\|y\|^{2}, \]

  and taking square roots yields \(|\langle x,y\rangle|\le \|x\|\,\|y\|\).

• (d)
  Compute

  \[ \begin{aligned} \|x+y\|^{2} & =\langle x+y, x+y\rangle \\ & =\langle x,x\rangle+\langle x,y\rangle+\langle y,x\rangle+\langle y,y\rangle \\ & =\|x\|^{2}+\|y\|^{2}+\langle x,y\rangle+\overline{\langle x,y\rangle}. \end{aligned} \]

  Since \(\langle x,y\rangle+\overline{\langle x,y\rangle}=2\Re\langle x,y\rangle\) and \(\Re\langle x,y\rangle\le |\langle x,y\rangle|\), we have

  \[ \|x+y\|^{2}\le \|x\|^{2}+2|\langle x,y\rangle|+\|y\|^{2}. \]

  By (c), \(|\langle x,y\rangle|\le \|x\|\,\|y\|\), so

  \[ \|x+y\|^{2}\le \|x\|^{2}+2\|x\|\,\|y\|+\|y\|^{2}=(\|x\|+\|y\|)^{2}. \]

  Taking square roots (both sides are \(\ge 0\)) gives \(\|x+y\|\le \|x\|+\|y\|\).

Example 27.15 : Cauchy-Schwarz and triangle inequalities in \(F^{n}\)

For \(F^{n}\), we may apply (c) and (d) of Theorem 27.14 to the standard inner product to obtain the following well-known inequalities:

\[ \left|\sum_{i=1}^{n} a_{i} \overline{b_{i}}\right| \leq\left[\sum_{i=1}^{n}\left|a_{i}\right|^{2}\right]^{1 / 2}\left[\sum_{i=1}^{n}\left|b_{i}\right|^{2}\right]^{1 / 2} \]

and

\[ \left[\sum_{i=1}^{n}\left|a_{i}+b_{i}\right|^{2}\right]^{1 / 2} \leq\left[\sum_{i=1}^{n}\left|a_{i}\right|^{2}\right]^{1 / 2}+\left[\sum_{i=1}^{n}\left|b_{i}\right|^{2}\right]^{1 / 2} \]

Orthogonality and Orthonormality

Definition 27.16 : Orthogonality and Orthonormality

Let \(V\) be an inner product space. Vectors \(x\) and \(y\) in \(V\) are orthogonal (perpendicular) if \(\langle x, y\rangle=0\). A subset \(S\) of \(V\) is orthogonal if any two distinct vectors in \(S\) are orthogonal. A vector \(x\) in \(V\) is a unit vector if \(\|x\|=1\).

Finally, a subset \(S\) of \(V\) is orthonormal if \(S\) is orthogonal and consists entirely of unit vectors. Note that if \(S=\left\{v_{1}, v_{2}, \ldots\right\}\), then \(S\) is orthonormal if and only if \(\left\langle v_{i}, v_{j}\right\rangle=\delta_{i j}\), where \(\delta_{i j}\) denotes the Kronecker delta.

Definition 27.17 : Normalizing

The process of multiplying a nonzero vector by the reciprocal of its length is called normalizing.

Example 27.18 : An orthonormal subset in \(H\)

Recall the inner product space \(H\). We introduce an important orthonormal subset \(S\) of \(H\). For any integer \(n\), let \(f_{n}(t)=e^{i n t}\), where \(0 \leq t \leq 2 \pi\). (Recall that \(e^{i n t}=\cos n t+i \sin n t\).) Now define \(S=\left\{f_{n}: n \text { is an integer }\right\}\). Clearly \(S\) is a subset of \(H\). Using the property that \(\overline{e^{i t}}=e^{-i t}\) for every real number \(t\), we have, for \(m \neq n\),

\[ \begin{aligned} \left\langle f_{m}, f_{n}\right\rangle & =\frac{1}{2 \pi} \int_{0}^{2 \pi} e^{i m t} \overline{e^{i n t}} d t=\frac{1}{2 \pi} \int_{0}^{2 \pi} e^{i(m-n) t} d t \\ & =\left.\frac{1}{2 \pi i(m-n)} e^{i(m-n) t}\right|_{0} ^{2 \pi}=0 \end{aligned} \]

Also,

\[ \left\langle f_{n}, f_{n}\right\rangle=\frac{1}{2 \pi} \int_{0}^{2 \pi} e^{i(n-n) t} d t=\frac{1}{2 \pi} \int_{0}^{2 \pi} 1 d t=1 \]

In other words, \(\left\langle f_{m}, f_{n}\right\rangle=\delta_{m n}\).

Exercise

Exercise 27.10

Let \(V\) be an inner product space, and suppose that \(x\) and \(y\) are orthogonal vectors in \(V\). Prove that \(\|x+y\|^{2}=\|x\|^{2}+\|y\|^{2}\). Deduce the Pythagorean theorem in \(\mathbb{R}^{2}\).

Exercise 27.11

Prove the parallelogram law on an inner product space \(V\); that is, show that

\[ \|x+y\|^{2}+\|x-y\|^{2}=2\|x\|^{2}+2\|y\|^{2} \quad \text { for all } x, y \in V \]

What does this equation state about parallelograms in \(\mathbb{R}^{2}\)?

Exercise 27.12

Let \(\left\{v_{1}, v_{2}, \ldots, v_{k}\right\}\) be an orthogonal set in \(V\), and let \(a_{1}, a_{2}, \ldots, a_{k}\) be scalars. Prove that

\[ \left\|\sum_{i=1}^{k} a_{i} v_{i}\right\|^{2}=\sum_{i=1}^{k}\left|a_{i}\right|^{2}\left\|v_{i}\right\|^{2} . \]

Exercise 27.15

• (a) Prove that if \(V\) is an inner product space, then \(|\langle x, y\rangle|=\|x\| \cdot\|y\|\) if and only if one of the vectors \(x\) or \(y\) is a multiple of the other.

  Hint: If the identity holds and \(y \neq 0\), let

  \[ a=\frac{\langle x, y\rangle}{\|y\|^{2}} \]

  and let \(z=x-a y\). Prove that \(y\) and \(z\) are orthogonal and

  \[ |a|=\frac{\|x\|}{\|y\|} . \]

  Then apply Exercise 27.10 to \(\|x\|^{2}=\|a y+z\|^{2}\) to obtain \(\|z\|=0\).

• (b) Derive a similar result for the equality \(\|x+y\|=\|x\|+\|y\|\). Generalize it to the case of \(n\) vectors.

Exercise 27.17

Let \(T\) be a linear operator on an inner product space \(V\), and suppose that \(\|T(x)\|=\|x\|\) for all \(x\). Prove that \(T\) is one-to-one.

Exercise 27.18

Let \(V\) be a vector space over \(F\), where \(F=\mathbb{R}\) or \(F=\mathbb{C}\), and let \(W\) be an inner product space over \(F\) with inner product \(\langle\cdot, \cdot\rangle\). If \(T: V \rightarrow W\) is linear, prove that \(\langle x, y\rangle^{\prime}=\langle T(x), T(y)\rangle\) defines an inner product on \(V\) if and only if \(T\) is one-to-one.

Exercise 27.19

Let \(V\) be an inner product space. Prove that

• (a) \(\|x \pm y\|^{2}=\|x\|^{2} \pm 2 \Re\langle x, y\rangle+\|y\|^{2}\) for all \(x, y \in V\), where \(\Re\langle x, y\rangle\) denotes the real part of the complex number \(\langle x, y\rangle\).
• (b) \(|\|x\|-\|y\|| \leq\|x-y\|\) for all \(x, y \in V\).

Exercise 27.20

Let \(V\) be an inner product space over \(F\). Prove the polar identities: For all \(x, y \in V\),

• (a) \(\langle x, y\rangle=\frac{1}{4}\|x+y\|^{2}-\frac{1}{4}\|x-y\|^{2} \quad\) if \(F=\mathbb{R}\).
• (b) \(\langle x, y\rangle=\frac{1}{4} \sum_{k=1}^{4} i^{k}\left\|x+i^{k} y\right\|^{2} \quad\) if \(F=\mathbb{C}\), where \(i^{2}=-1\).

Exercise 27.21

Let \(A\) be an \(n \times n\) matrix. Define

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

• (a) Prove that \(A_{1}^{*}=A_{1}\), \(A_{2}^{*}=A_{2}\), and \(A=A_{1}+i A_{2}\). Would it be reasonable to define \(A_{1}\) and \(A_{2}\) to be the real and imaginary parts, respectively, of the matrix \(A\)?
• (b) Let \(A\) be an \(n \times n\) matrix. Prove that the representation in (a) is unique. That is, prove that if \(A=B_{1}+i B_{2}\), where \(B_{1}^{*}=B_{1}\) and \(B_{2}^{*}=B_{2}\), then \(B_{1}=A_{1}\) and \(B_{2}=A_{2}\).

Exercise 27.22

Let \(V\) be a real or complex vector space (possibly infinite-dimensional), and let \(\beta\) be a basis for \(V\). For \(x, y \in V\) there exist \(v_{1}, v_{2}, \ldots, v_{n} \in \beta\) such that

\[ x=\sum_{i=1}^{n} a_{i} v_{i} \quad \text { and } \quad y=\sum_{i=1}^{n} b_{i} v_{i} . \]

Define

\[ \langle x, y\rangle=\sum_{i=1}^{n} a_{i} \overline{b}_{i} . \]

• (a) Prove that \(\langle\cdot, \cdot\rangle\) is an inner product on \(V\) and that \(\beta\) is an orthonormal basis for \(V\). Thus every real or complex vector space may be regarded as an inner product space.
• (b) Prove that if \(V=\mathbb{R}^{n}\) or \(V=\mathbb{C}^{n}\) and \(\beta\) is the standard ordered basis, then the inner product defined above is the standard inner product.

Exercise 27.23

Let \(V=F^{n}\), and let \(A \in \mathrm{M}_{n \times n}(F)\).

• (a) Prove that \(\langle x, A y\rangle=\left\langle A^{*} x, y\right\rangle\) for all \(x, y \in V\).
• (b) Suppose that for some \(B \in \mathrm{M}_{n \times n}(F)\), we have \(\langle x, A y\rangle=\langle B x, y\rangle\) for all \(x, y \in V\). Prove that \(B=A^{*}\).
• (c) Let \(\alpha\) be the standard ordered basis for \(V\). For any orthonormal basis \(\beta\) for \(V\), let \(Q\) be the \(n \times n\) matrix whose columns are the vectors in \(\beta\). Prove that \(Q^{*}=Q^{-1}\).
• (d) Define linear operators \(T\) and \(U\) on \(V\) by \(T(x)=A x\) and \(U(x)=A^{*} x\). Show that \([U]_{\beta}=[T]_{\beta}^{*}\) for any orthonormal basis \(\beta\) for \(V\).

Definition 27.19 : Norm

Let \(V\) be a vector space over \(F\), where \(F\) is either \(\mathbb{R}\) or \(\mathbb{C}\). Regardless of whether \(V\) is or is not an inner product space, we may still define a norm \(\|\cdot\|\) as a real-valued function on \(V\) satisfying the following three conditions for all \(x, y \in V\) and \(a \in F\).

• (1) \(\|x\| \geq 0\), and \(\|x\|=0\) if and only if \(x=0\).
• (2) \(\|a x\|=|a| \cdot\|x\|\).
• (3) \(\|x+y\| \leq\|x\|+\|y\|\).

Exercise 27.26

Let \(\|\cdot\|\) be a norm on a vector space \(V\), and define, for each ordered pair of vectors, the scalar \(d(x, y)=\|x-y\|\), called the distance between \(x\) and \(y\). Prove the following results for all \(x, y, z \in V\).

• (a) \(d(x, y) \geq 0\).
• (b) \(d(x, y)=d(y, x)\).
• (c) \(d(x, y) \leq d(x, z)+d(z, y)\).
• (d) \(d(x, x)=0\).
• (e) \(d(x, y) \neq 0\) if \(x \neq y\).

Exercise 27.27

Let \(\|\cdot\|\) be a norm on a real vector space \(V\) satisfying the parallelogram law given in Exercise 27.11. Define

\[ \langle x, y\rangle=\frac{1}{4}\left[\|x+y\|^{2}-\|x-y\|^{2}\right] . \]

Prove that \(\langle\cdot, \cdot\rangle\) defines an inner product on \(V\) such that \(\|x\|^{2}=\langle x, x\rangle\) for all \(x \in V\).

Hints:

• (a) Prove \(\langle x, 2 y\rangle=2\langle x, y\rangle\) for all \(x, y \in V\).
• (b) Prove \(\langle x+u, y\rangle=\langle x, y\rangle+\langle u, y\rangle\) for all \(x, u, y \in V\).
• (c) Prove \(\langle n x, y\rangle=n\langle x, y\rangle\) for every positive integer \(n\) and every \(x, y \in V\).
• (d) Prove \(m\left\langle\frac{1}{m} x, y\right\rangle=\langle x, y\rangle\) for every positive integer \(m\) and every \(x, y \in V\).
• (e) Prove \(\langle r x, y\rangle=r\langle x, y\rangle\) for every rational number \(r\) and every \(x, y \in V\).

• (f) Prove \(|\langle x, y\rangle| \leq\|x\| \|y\|\) for every \(x, y \in V\).

  Hint: Condition (3) in the definition of norm can be helpful.

• (g) Prove that for every \(c \in \mathbb{R}\), every rational number \(r\), and every \(x, y \in V\),

  \[ |c\langle x, y\rangle-\langle c x, y\rangle|=|(c-r)\langle x, y\rangle-\langle(c-r) x, y\rangle| \leq 2|c-r|\|x\| \|y\| \]

• (h) Use the fact that for any \(c \in \mathbb{R}\), \(|c-r|\) can be made arbitrarily small, where \(r\) varies over the set of rational numbers, to establish item (b) of the definition of inner product.

Exercise 27.28

Let \(V\) be a complex inner product space with an inner product \(\langle\cdot, \cdot\rangle\). Let \([\cdot, \cdot]\) be the real-valued function such that \([x, y]\) is the real part of the complex number \(\langle x, y\rangle\) for all \(x, y \in V\). Prove that \([\cdot, \cdot]\) is an inner product for \(V\), where \(V\) is regarded as a vector space over \(\mathbb{R}\). Prove, furthermore, that \([x, i x]=0\) for all \(x \in V\).

Exercise 27.29

Let \(V\) be a vector space over \(\mathbb{C}\), and suppose that \([\cdot, \cdot]\) is a real inner product on \(V\), where \(V\) is regarded as a vector space over \(\mathbb{R}\), such that \([x, i x]=0\) for all \(x \in V\). Let \(\langle\cdot, \cdot\rangle\) be the complex-valued function defined by

\[ \langle x, y\rangle=[x, y]+i[x, i y] \quad \text { for } x, y \in V . \]

Prove that \(\langle\cdot, \cdot\rangle\) is a complex inner product on \(V\).

Exercise 27.30

Let \(\|\cdot\|\) be a norm (as defined in Definition 27.19) on a complex vector space \(V\) satisfying the parallelogram law given in Exercise 27.11. Prove that there is an inner product \(\langle\cdot, \cdot\rangle\) on \(V\) such that \(\|x\|^{2}=\langle x, x\rangle\) for all \(x \in V\).

Hint: Apply Exercise 27.27 to \(V\) regarded as a vector space over \(\mathbb{R}\). Then apply Exercise 27.29.