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We have used the dot product in \(\RR ^n\) to compute the length of vectors (Corollary cor:length_via_dotprod) and also the angle between vectors. The goal
of this section is to define an inner product on an arbitrary vector space \(V\) over the real numbers. The dot product is an example
an inner product for \(\RR ^n\).
An inner product on a real vector space \(V\) is a function that assigns a real number \(\langle \vec {v}, \vec {w}\rangle \) to every pair \(\vec {v}\), \(\vec {w}\) of vectors in \(V\) in such a way
that the following properties are satisfied.
1.
\(\langle \vec {v}, \vec {w}\rangle \)is a real number for all\(\vec {v}\)and\(\vec {w}\)in\(V\).
A real vector space \(V\) with an inner product \(\langle \ , \rangle \) will be called an inner product space. Note that every subspace of an inner
product space is again an inner product space using the same inner product.
\(\RR ^n\) is an inner product space with the dot product as inner product:
See Theorem th:dotproductproperties. This is also called the euclidean inner
product, and \(\RR ^n\), equipped with the dot product, is called euclidean\(n\)-space.
If \(A\) and \(B\) are \(m \times n\) matrices, define \(\langle A, B\rangle = \mbox {tr}(AB^{T})\) where \(\mbox {tr}(X)\) is the trace of the square matrix \(X\). Show that \(\langle \ , \rangle \) is an inner product in \(\mathbb {M}_{mn}\).
Property prop:inner_prod_1 is clear. Since \(\mbox {tr}(P) = \mbox {tr}(P^{T})\) for every square matrix \(P\), we have Property prop:inner_prod_2:
\begin{equation*} \langle A, B \rangle = \mbox {tr}(AB^T) = \mbox {tr}[(AB^T)^T] = \mbox {tr}(BA^T) = \langle B, A \rangle \end{equation*}
Next, Property prop:inner_prod_3 and Property prop:inner_prod_4 follow because trace is a
linear transformation \(\mathbb {M}_{mn} \to \RR \) (Practice Problem ex:10_1_19). Turning to Property prop:inner_prod_5, let \(\vec {r}_{1}, \vec {r}_{2}, \dots , \vec {r}_{m}\) denote the rows of the matrix \(A\). Then the \((i, j)\)-entry of \(AA^{T}\) is \(\vec {r}_{i} \dotp \vec {r}_{j}\), so
But \(\vec {r}_{j} \dotp \vec {r}_{j}\) is the sum of the squares of the entries of \(\vec {r}_{j}\), so this shows that \(\langle A, A\rangle \) is the sum of the squares of all \(nm\) entries of \(A\). Property prop:inner_prod_5
follows.
The next example is important in analysis.
Let \(\mathcal {F}[a,b]\) be the set of all functions \(f:[a,b]\rightarrow \RR \). Observe that \(\mathcal {F}[a,b]\) is a vector space. Let \(\mathcal {C}[a,b]\) be a subset of \(\mathcal {F}[a,b]\) consisting of all continuous functions.
Why is \(\mathcal {C}[a,b]\) a subspace of \(\mathcal {F}[a,b]\)? Show that
\begin{equation*} \langle f, g \rangle = \int _{a}^{b} f(x)g(x)dx \end{equation*}
defines an inner product on \(\mathcal {C}[a, b]\).
This example (and others later that refer to it) can be omitted with no loss of continuity by students with no calculus
background.
\begin{equation*} \langle rf, g \rangle = \int _{a}^{b} rf(x)g(x)dx = r\int _{a}^{b} f(x)g(x)dx = r\langle f, g \rangle \end{equation*}
Property prop:inner_prod_3 is similar. Finally, theorems of calculus show that \(\langle f, f \rangle = \int _{a}^{b} f(x)^2dx \geq 0\) and, if \(f\) is
continuous, that this is zero if and only if \(f\) is the zero function. This gives Property prop:inner_prod_5.
and it follows that the number \(\langle \vec {0}, \vec {v}\rangle \) must be zero. This observation is
recorded for reference in the following theorem, along with several other properties of inner products. The other proofs are left
as Practice Problem ex:10_1_20.
Let \(\langle \ , \rangle \) be an inner product on a space \(V\); let \(\vec {v}\), \(\vec {u}\), and \(\vec {w}\) denote vectors in \(V\); and let \(r\) denote a real number.
for all \(r\) and \(s\) in \(\RR \) by Property prop:inner_prod_3 and Property prop:inner_prod_4 of Definition def:innerproductspace. Moreover,
there is nothing special about the fact that there are two terms in the linear combination or that it is in the first component:
hold for all \(r_{i}\) and \(s_{i}\) in \(\RR \) and all \(\vec {v}\), \(\vec {w}\), \(\vec {v}_{i}\), and \(\vec {w}_{j}\) in \(V\). These results are described by saying that inner products “preserve” linear
combinations. For example,
If \(A\) is a symmetric \(n \times n\) matrix and \(\vec {x}\) and \(\vec {y}\) are columns in \(\RR ^n\), we regard the \(1 \times 1\) matrix \(\vec {x}^{T}A\vec {y}\) as a number. If we write
\begin{equation*} \langle \vec {x}, \vec {y} \rangle = \vec {x}^TA\vec {y} \quad \mbox { for all columns } \vec {x}, \vec {y} \mbox { in } \RR ^n \end{equation*}
\begin{equation*} \vec {x}^TA \vec {x} > 0 \quad \mbox { for all columns } \vec {x} \neq \vec {0} \mbox { in } \RR ^n \end{equation*}
and this condition characterizes the positive definite matrices (Theorem thm:024830). This proves the first assertion in the next
theorem.
If \(A\) is any \(n \times n\) positive definite matrix, then
\begin{equation*} \langle \vec {x}, \vec {y} \rangle = \vec {x}^TA\vec {y} \mbox { for all columns } \vec {x}, \vec {y} \mbox { in } \RR ^n \end{equation*}
defines an inner product on \(\RR ^n\), and every inner product on \(\RR ^n\) arises in this way.
Given an inner product \(\langle \ , \rangle \) on \(\RR ^n\), let \(\{\vec {e}_{1}, \vec {e}_{2}, \dots , \vec {e}_{n}\}\) be the standard basis of \(\RR ^n\). If \(\vec {x} = \displaystyle \sum _{i = 1}^{n} x_i\vec {e}_i\) and \(\vec {y} = \displaystyle \sum _{j = 1}^{n} y_j\vec {e}_j\) are two vectors in \(\RR ^n\), compute \(\langle \vec {x}, \vec {y}\rangle \) by adding the inner product of
each term \(x_{i}\vec {e}_{i}\) to each term \(y_{j}\vec {e}_{j}\). The result is a double sum.
shows that \(A\) is symmetric. Finally, \(A\) is positive definite by Theorem thm:024830.
Thus, just as every linear operator \(\RR ^n \to \RR ^n\) corresponds to an \(n \times n\) matrix, every inner product on \(\RR ^n\) corresponds to a positive definite \(n \times n\)
matrix. In particular, the dot product corresponds to the identity matrix \(I_{n}\).
If we refer to the inner product space \(\RR ^n\) without specifying the inner product, we mean that the dot product is to be used.
Let the inner product \(\langle \ , \rangle \) be defined on \(\RR ^2\) by
Find a symmetric \(2 \times 2\) matrix \(A\) such that \(\langle \vec {x}, \vec {y}\rangle = \vec {x}^{T}A\vec {y}\) for all \(\vec {x}\), \(\vec {y}\) in \(\RR ^2\).
The \((i, j)\)-entry of the matrix \(A\) is the coefficient of \(v_{i}w_{j}\) in the expression, so \( A = \left [ \begin{array}{rr} 2 & -1 \\ -1 & 1 \end{array} \right ]\). Incidentally, if \(\vec {x} = \left [ \begin{array}{r} x \\ y \end{array} \right ]\), then
for all \(\vec {x}\), so \(\langle \vec {x}, \vec {x}\rangle = 0\) implies \(\vec {x} = \vec {0}\). Hence \(\langle \ , \rangle \) is indeed an inner product, so \(A\) is positive definite.
Let \(\langle \ , \rangle \) be an inner product on \(\RR ^n\) given as in Theorem thm:030372 by a positive definite matrix \(A\). If \(\vec {x} = \left [ \begin{array}{cccc} x_1 & x_2 & \cdots & x_n \end{array} \right ]^T \), then \(\langle \vec {x}, \vec {x}\rangle = \vec {x}^{T}A\vec {x}\) is an expression in the variables \(x_{1}, x_{2}, \dots , x_{n}\)
called a quadratic form. For more on quadratic forms, see Section 8.8 of [Nicholson], pp. 472–482.
Norm and Distance
As in \(\RR ^n\), if \(\langle \ , \rangle \) is an inner product on a space \(V\), the norm\(\norm {\vec {v}}\) of a vector \(\vec {v}\) in \(V\) is defined by
A vector \(\vec {v}\) in an inner product space \(V\) is called a unit vector if \(\norm {\vec {v}} = 1\). The set of all unit vectors in \(V\) is called the unit ball in \(V\). For
example, if \(V = \RR ^2\) (with the dot product) and \(\vec {v} = (x, y)\), then
\begin{equation*} \norm { \vec {v} }^2 = 1 \quad \mbox { if and only if } \quad x^2 + y^2 = 1 \end{equation*}
Hence the unit ball in \(\RR ^2\) is the unit circle\(x^{2} + y^{2} = 1\) with centre at the origin and radius \(1\).
However, the shape of the unit ball varies with the choice of inner product.
Let \(a > 0\) and \(b > 0\). If \(\vec {v} = (x, y)\) and \(\vec {w} = (x_{1}, y_{1})\), define an inner product on \(\RR ^2\) by
The reader can verify (Practice Problem ) that this is indeed an inner product.
In this case
\begin{equation*} \norm { \vec {v} }^2 = 1 \quad \mbox { if and only if } \quad \frac {x^2}{a^2} + \frac {y^2}{b^2} = 1 \end{equation*}
so the unit ball is the ellipse shown in the diagram.
Example 20 graphically illustrates the fact that norms and distances in an inner product space \(V\) vary with the choice of inner
product in \(V\).
If \(\vec {v} \neq \vec {0}\) is any vector in an inner product space \(V\), then \(\frac {1}{\norm { \vec {v} }} \vec {v}\) is the unique unit vector that is a positive multiple of \(\vec {v}\).
The next theorem reveals an important and useful fact about the relationship between norms and inner products, extending
the Cauchy inequality for \(\RR ^n\) (Theorem ??).
Cauchy-Schwarz Inequality If \(\vec {v}\) and \(\vec {w}\) are two vectors in an inner product space \(V\), then
Moreover, equality occurs if and only if
one of \(\vec {v}\) and \(\vec {w}\) is a scalar multiple of the other.
Hermann Amandus Schwarz (1843–1921) was a German mathematician at the
University of Berlin. He had strong geometric intuition, which he applied with great ingenuity to particular problems. A version
of the inequality appeared in 1885.
Proof of Cauchy-Schwarz Inequality Write \(\norm {\vec {v}} = a\) and \(\norm {\vec {w}} = b\). Using Theorem 8 we compute:
It follows that \(ab - \langle \vec {v}, \vec {w}\rangle \geq 0\) and \(ab + \langle \vec {v}, \vec {w}\rangle \geq 0\), and hence that \(-ab \leq \langle \vec {v}, \vec {w}\rangle \leq ab\). But
then \(| \langle \vec {v}, \vec {w}\rangle | \leq ab = \norm {\vec {v}} \norm { \vec {w}}\), as desired.
Conversely, if \(|\langle \vec {v}, \vec {w}\rangle | = \norm {\vec {v}} \norm { \vec {w} } = ab\) then \(\langle \vec {v}, \vec {w}\rangle = \pm ab\). Hence (1) shows that \(b\vec {v} - a\vec {w} = \vec {0}\) or \(b\vec {v} + a\vec {w} = \vec {0}\). It follows that one of \(\vec {v}\) and \(\vec {w}\) is a scalar multiple of the other, even if \(a = 0\) or \(b = 0\).
If \(f\) and \(g\) are continuous functions on the interval \([a, b]\), then (see Example 5)
Another famous inequality, the so-called triangle inequality (See Triangle Inequality in the Appendix), also
comes from the Cauchy-Schwarz inequality. It is included in the following list of basic properties of the norm of a
vector.
If \(V\) is an inner product space, the norm \(\norm { \dotp }\) has the following properties.
1.
\(\norm {\vec {v}} \geq 0\) for every vector \(\vec {v}\) in \(V\).
2.
\(\norm {\vec {v}} = 0\) if and only if \(\vec {v} = \vec {0}\).
3.
\(\norm { r \vec {v}} = |r|\norm {\vec {v}}\) for every \(\vec {v}\) in \(V\) and every \(r\) in \(\RR \).
4.
\(\norm {\vec {v} + \vec {w}} \leq \norm {\vec {v}} + \norm {\vec {w}}\) for all \(\vec {v}\) and \(\vec {w}\) in \(V\) (triangle inequality).
Because \(\norm { \vec {v} } = \sqrt {\langle \vec {v}, \vec {v} \rangle }\), properties 1 and 2 follow immediately from ?? and ?? of Theorem 8. As to 3, compute
It is worth noting that the usual triangle inequality for absolute values,
\begin{equation*} | r + s | \leq |r| + |s| \mbox { for all real numbers } r \mbox { and } s \end{equation*}
is a special case of 4 where \(V = \RR = \RR ^1\) and the dot product \(\langle r, s \rangle = rs\) is
used.
In many calculations in an inner product space, it is required to show that some vector \(\vec {v}\) is zero. This is often accomplished
most easily by showing that its norm \(\norm {\vec {v}}\) is zero. Here is an example.
Let \(\{\vec {v}_{1}, \dots , \vec {v}_{n}\}\) be a spanning set for an inner product space \(V\). If \(\vec {v}\) in \(V\) satisfies \(\langle \vec {v}, \vec {v}_{i}\rangle = 0\) for each \(i = 1, 2, \dots , n\), show that \(\vec {v} = \vec {0}\).
Write \(\vec {v} = r_{1}\vec {v}_{1} + \dots + r_{n}\vec {v}_{n}\), \(r_{i}\) in \(\RR \). To show that \(\vec {v} = \vec {0}\), we show that \(\norm {\vec {v}}^{2} = \langle \vec {v}, \vec {v}\rangle = 0\). Compute:
\(V = \mathbb {C}\), \(\langle z, w \rangle = z\overline {w}\), where \(\overline {w}\) is complex conjugation
Click the arrow to see the answer.
Property 1 fails, as sometimes we get a complex number. However, we will return to
this definition of \(\langle \ , \rangle \) in Complex Matrices – see Definition ??
\(\vec {u} = f\), \(\vec {v} = g \) in \(\mathcal {C}[0, 1]\) where \(f(x) = x^2 \) and \(g(x) = 1 - x\); \(\langle f, g \rangle = \int _{0}^{1} f(x)g(x)dx\)
4.
\(\vec {u} = f\), \(\vec {v} = g \) in \(\mathcal {C}[-\pi , \pi ]\) where \(f(x) = 1\) and \(g(x) = \cos x\); \(\langle f, g \rangle = \int _{-\pi }^{\pi } f(x)g(x)dx\)
Click the arrow to see the answer.
\(\sqrt {3\pi }\)
Let \(a_{1}, a_{2}, \dots , a_{n}\) be positive numbers. Given \(\vec {v} = \begin{bmatrix}v_{1}\\ v_{2}\\ \vdots \\ v_{n}\end{bmatrix}\) and \(\vec {w} = \begin{bmatrix}w_{1}\\ w_{2}\\ \vdots \\ w_{n}\end{bmatrix}\), define \(\langle \vec {v}, \vec {w}\rangle = a_{1}v_{1}w_{1} + \dots + a_{n}v_{n}w_{n}\). Show that this is an inner product on \(\RR ^n\).
If \(\{\vec {b}_{1}, \dots , \vec {b}_{n}\}\) is a basis of \(V\) and if \(\vec {v} = v_1\vec {b}_1 + \dots + v_n\vec {b}_n\) and \(\vec {w} = w_1\vec {b}_1 + \dots + w_n\vec {b}_n\) are vectors in \(V\), define
Let \(\mbox {re}(z)\) denote the real part of the complex number \(z\). Show that \(\langle \ , \rangle \) is an inner product on \(\mathbb {C}\) if \(\langle \vec {z}, \vec {w}\rangle = \mbox {re}(z\overline {w})\).
If \(T : V \to V\) is an isomorphism of the inner product space \(V\), show that
defines a new inner product \(\langle \ , \rangle _{1}\) on \(V\).
Show that every inner product \(\langle \ , \rangle \) on \(\RR ^n\) has the form \(\langle \vec {x}, \vec {y}\rangle = (U\vec {x}) \dotp (U\vec {y})\) for some upper triangular matrix \(U\) with positive diagonal entries.
Theorem ??
In each case, show that \(\langle \vec {v}, \vec {w}\rangle = \vec {v}^{T}A\vec {w}\) defines an inner product on \(\RR ^2\) and hence show that \(A\) is positive definite.
Assume that \(\langle \vec {v}, \vec {v} \rangle = 0\). If \(\vec {v} \neq \vec {0}\) this contradicts Property 5, so \(\vec {v} = \vec {0}\). Conversely, if \(\vec {v} = \vec {0}\), then \(\langle \vec {v}, \vec {v} \rangle = 0\) by Part 3 of this theorem.
for any \(\vec {v}\) and \(\vec {w}\) in an inner product space.
Let \(\langle \ , \rangle \) be an inner product on a vector space \(V\). Show that the corresponding distance function is translation invariant. That is,
show that \(\mbox {d}(\vec {v}, \vec {w}) = \mbox {d}(\vec {v} + \vec {u}, \vec {w} + \vec {u})\) for all \(\vec {v}\), \(\vec {w}\), and \(\vec {u}\) in \(V\).
1.
Show that \(\langle \vec {u}, \vec {v} \rangle = \frac {1}{4}[\norm { \vec {u} + \vec {v} } ^2 - \norm { \vec {u} - \vec {v} } ^2]\) for all \(\vec {u}\), \(\vec {v}\) in an inner product space \(V\).
2.
If \(\langle \ , \rangle \) and \(\langle \ , \rangle ^\prime \) are two inner products on \(V\) that have equal associated norm functions, show that \(\langle \vec {u}, \vec {v}\rangle = \langle \vec {u}, \vec {v}\rangle ^\prime \) holds for all \(\vec {u}\) and \(\vec {v}\).
Let \(\vec {v}\) denote a vector in an inner product space \(V\).
1.
Show that \(W = \{\vec {w} \mid \vec {w} \mbox { in } V, \langle \vec {v}, \vec {w} = 0\}\) is a subspace of \(V\).
2.
Let \(W\) be as in (a). If \(V = \RR ^3\) with the dot product, and if \(\vec {v} = \begin{bmatrix}1\\ -1\\ 2\end{bmatrix}\), find a basis for \(W\).
Given vectors \(\vec {w}_{1}, \vec {w}_{2}, \dots , \vec {w}_{n}\) and \(\vec {v}\), assume that \(\langle \vec {v}, \vec {w}_{i}\rangle = 0\) for each \(i\). Show that \(\langle \vec {v}, \vec {w}\rangle = 0\) for all \(\vec {w}\) in \(\mbox {span}\{\vec {w}_{1}, \vec {w}_{2}, \dots , \vec {w}_{n}\}\).
If \(V = \mbox {span}\{\vec {v}_{1}, \vec {v}_{2}, \dots , \vec {v}_{n}\}\) and \(\langle \vec {v}, \vec {v}_{i}\rangle = \langle \vec {w}, \vec {v}_i\rangle \) holds for each \(i\). Show that \(\vec {v} = \vec {w}\).
\(\langle \vec {v} - \vec {w}, \vec {v}_{i} \rangle = \langle \vec {v}, \vec {v}_{i} \rangle - \langle \vec {w}, \vec {v}_{i} \rangle = 0\) for each \(i\), so \(\vec {v} = \vec {w}\) by Practice Problem .
Use the Cauchy-Schwarz inequality in an inner product space to show that:
1.
If \(\norm {\vec {u}} \leq 1\), then \(\langle \vec {u}, \vec {v}\rangle ^{2} \leq \norm {\vec {v}}^{2}\) for all \(\vec {v}\) in \(V\).
2.
\((x \cos \theta + y \sin \theta )^{2} \leq x^{2} + y^{2}\) for all real \(x\), \(y\), and \(\theta \).
If \(\vec {u} = (\cos \theta , \sin \theta )\) in \(\RR ^2\) (with the dot product) then \(\norm {\vec {u}} = 1\). Use (a) with \(\vec {v} = \begin{bmatrix}x\\ y\end{bmatrix}\).
3.
\(\norm { r_1\vec {v}_1 + \dots + r_n\vec {v}_n } ^2 \leq [r_1 \norm { \vec {v}_1 } + \dots + r_n \norm { \vec {v}_n } ]^2\) for all vectors \(\vec {v}_{i}\), and all \(r_{i} > 0\) in \(\RR \).
If \(A\) is a \(2 \times n\) matrix, let \(\vec {u}\) and \(\vec {v}\) denote the rows of \(A\).
If \(\vec {v}\) and \(\vec {w}\) are nonzero vectors in an inner product space \(V\), show that \(-1 \leq \frac {\langle \vec {v}, \vec {w} \rangle }{\norm { \vec {v} } \norm { \vec {w} }} \leq 1\), and hence that a unique angle \(\theta \) exists such
that \(\frac {\langle \vec {v}, \vec {w} \rangle }{\norm { \vec {v} } \norm { \vec {w} }} = \cos \theta \) and \(0 \leq \theta \leq \pi \). This angle \(\theta \) is called the angle between\(\vec {v}\) and \(\vec {w}\).
2.
Find the angle between \(\vec {v} = \begin{bmatrix}1\\ 2\\ -1\\ 1\\ 3\end{bmatrix}\) and \(\vec {w} = \begin{bmatrix}2\\ 1\\ 0\\ 2\\ 0\end{bmatrix}\) in \(\RR ^5\) with the dot product.
3.
If \(\theta \) is the angle between \(\vec {v}\) and \(\vec {w}\), show that the law of cosines is valid:
Show that \(\norm {\dotp }\) satisfies the conditions in Theorem 26.
2.
Show that \(\norm {\dotp }\) does not arise from an inner product on \(\RR ^2\) given by a matrix \(A\).
If it did, use Theorem 10 to find numbers \(a\), \(b\),
and \(c\) such that \(\norm {\begin{bmatrix}x\\ y\end{bmatrix}}^{2} = ax^{2} + bxy + cy^{2}\) for all \(x\) and \(y\).
Linear Algebra: An Interactive Introduction