10.1E: Inner Products and Norms Exercises
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In each case, determine which of axioms P1-P5 fail to hold.
- \(V=\mathbb{R}^2,\left\langle\left(x_1, y_1\right),\left(x_2, y_2\right)\right\rangle=x_1 y_1 x_2 y_2\)
- \( V=\mathbb{R}^3,\left\langle\left(x_1, x_2, x_3\right), \quad\left(y_1, y_2, y_3\right)\right\rangle=x_1 y_1- \\x_2 y_2+x_3 y_3\)
- \(V=\mathbb{C},\langle z, w\rangle=z \bar{w}\), where \(\bar{w}\) is complex conjugation
- \(V=\mathbf{P}_3,\langle p(x), q(x)\rangle=p(1) q(1)\)
- \(V=\mathbf{M}_{22},\langle A, B\rangle=\operatorname{det}(A B)\)
- \(V=\mathbf{F}[0,1],\langle f, g\rangle=f(1) g(0)+f(0) g(1)\)
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- Pf fails
- P5 fails
- P5 fails
Let \(V\) be an inner product space. If \(U \subseteq V\) is a subspace, show that \(U\) is an inner product space using the same inner product.
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Axioms P1–P5 hold in U because they hold in \(V\).
In each case, find a scalar multiple of \(\mathbf{v}\) that is a unit vector.
- \(\mathbf{v}=f\) in \(\mathbf{C}[0,1]\) where \(f(x)=x^2\) \(\langle\mathbf{f}, \mathbf{g}\rangle \int_0^1 f(x) g(x) d x\)
- \(\mathbf{v}=f\) in \(\mathbf{C}[-\pi, \pi]\) where \(f(x)=\cos x\) \(\langle\mathbf{f}, \mathbf{g}\rangle \int_{-\pi}^\pi f(x) g(x) d x\)
- \(\mathbf{v}=\left[\begin{array}{l}1 \\ 3\end{array}\right]\) in \(\mathbb{R}^2\) where \(\langle\mathbf{v}, \quad \mathbf{w}\rangle=\) \(\mathbf{v}^T\left[\begin{array}{ll}1 & 1 \\ 1 & 2\end{array}\right] \mathbf{w}\)
- \(\mathbf{v}=\left[\begin{array}{r}3 \\ -1\end{array}\right]\) in \(\mathbb{R}^2,\langle\mathbf{v}, \mathbf{w}\rangle=\mathbf{v}^T\left[\begin{array}{rr}1 & -1 \\ -1 & 2\end{array}\right] \mathbf{w}\)
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- \(\frac{1}{\sqrt{\pi}}f \)
- \(\frac{1}{\sqrt{17}}\left[\begin{array}{r}3 \\ -1\end{array}\right]\)
In each case, find the distance between \(\mathbf{u}\) and \(\mathbf{v}\).
- \(\mathbf{u}=(3,-1,2,0), \mathbf{v}=(1,1,1,3) ;\langle\mathbf{u}, \mathbf{v}\rangle=\mathbf{u} \cdot \mathbf{v}\)
- \(\mathbf{u}=(1,2,-1,2), \mathbf{v}=(2,1,-1,3) ;\langle\mathbf{u}, \mathbf{v}\rangle=\mathbf{u}\). \(\mathbf{v}\)
- \(\mathbf{u}=f, \mathbf{v}=g\) in \(\mathbf{C}[0,1]\) where \(f(x)=x^2\) and \(g(x)=1-x ;\langle f, g\rangle=\int_0^1 f(x) g(x) d x\)
- \(\mathbf{u}=f, \mathbf{v}=g\) in \(\mathbf{C}[-\pi, \pi]\) where \(f(x)=1\) and \(g(x)=\cos x ;\langle f, g\rangle=\int_{-\pi}^\pi f(x) g(x) d x\)
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- \( $\sqrt{3}\)
- \($\sqrt{3 \pi}\)
Let \(a_1, a_2, \ldots, a_n\) be positive numbers. Given \(\mathbf{v}=\left(v_1, v_2, \ldots, v_n\right)\) and \(\mathbf{w}=\left(w_1\right.\), \(\left.w_2, \ldots, w_n\right)\), define \(\langle\mathbf{v}, \mathbf{w}\rangle=a_1 v_1 w_1+\ldots+a_n v_n w_n\). Show that this is an inner product on \(\mathbb{R}^n\).
If \(\left\{\mathbf{b}_1, \ldots, \mathbf{b}_n\right\}\) is a basis of \(V\) and if \(\mathbf{v}=v_1 \mathbf{b}_1+\cdots+v_n \mathbf{b}_n\) and \(\mathbf{w}=w_1 \mathbf{b}_1+\cdots+w_n \mathbf{b}_n\) are vectors in \(V\), define
\[ \langle\mathbf{v}, \mathbf{w}\rangle=v_1 w_1+\cdots+v_n w_n .\nonumber \]
Show that this is an inner product on \(V\).
If \(p=p(x)\) and \(q=q(x)\) are polynomials in \(\mathbf{P}_n\), define
\[ \langle p, q\rangle=p(0) q(0)+p(1) q(1)+\cdots+p(n) q(n) \nonumber \]
Show that this is an inner product on \(\mathbf{P}_n\). [Hint for P5: Theorem 6.5.4 or Appendix D.]
Let \(\mathbf{D}_n\) denote the space of all functions from the set \(\{1,2,3, \ldots, n\}\) to \(\mathbb{R}\) with pointwise addition and scalar multiplication (see Exercise 35 Section 6.3). Show that \(\langle\),\(\rangle \) is an inner product on \(\mathbf{D}_n\) if \(\langle f, g\rangle=f(1) g(1)+f(2) g(2)+\) \(\ldots+f(n) g(n)\).
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P1 and P2 are clear since \(f(i)\) and \(g(i)\) are real numbers.
\[\text { P3: } \begin{aligned}
\langle f+g, h\rangle & =\sum_i(f+g)(i) \cdot h(i) \\
& =\sum_i(f(i)+g(i)) \cdot h(i) \\
& =\sum_i[f(i) h(i)+g(i) h(i)] \\
& =\sum_i f(i) h(i)+\sum_i g(i) h(i) \\
& =\langle f, h\rangle+\langle g, h\rangle .
\end{aligned}\]\[\text { P4: } \begin{aligned}
\langle r f, g\rangle & =\sum_i(r f)(i) \cdot g(i) \\
& =\sum_i r f(i) \cdot g(i) \\
& =r \sum_i f(i) \cdot g(i) \\
& =r\langle f, g\rangle
\end{aligned}\]
P5: If \(f \neq 0\), then \(\langle f, f\rangle=\sum_i f(i)^2>0\) because some \(f(i) \neq 0\).
Let re \((z)\) denote the real part of the complex number \(z\). Show that \(\langle\),\(\rangle\) is an inner product on \(\mathbb{C}\) if \(\langle z, w\rangle=r e(z \bar{w})\).
If \(T: V \rightarrow V\) is an isomorphism of the inner product space \(V\), show that
\[ \langle\mathbf{v}, \mathbf{w}\rangle_1=\langle T(\mathbf{v}), T(\mathbf{w})\rangle
\nonumber \]
defines a new inner product \(\langle,\rangle_1\) on \(V\).
Show that every inner product \(\langle\),\(\rangle\) on \(\mathbb{R}^n\) has the form \(\langle\mathbf{x}, \mathbf{y}\rangle=(U \mathbf{x}) \cdot(U \mathbf{y})\) for some upper triangular matrix \(U\) with positive diagonal entries. [Hint: Theorem 8.3.3.]
In each case, show that \(\langle\mathbf{v}, \mathbf{w}\rangle\) \(=\mathbf{v}^T A \mathbf{w}\) defines an inner product on \(\mathbb{R}^2\) and hence show that \(A\) is positive definite.
- \(A=\left[\begin{array}{ll}2 & 1 \\ 1 & 1\end{array}\right]\)
- \(A=\left[\begin{array}{rr}5 & -3 \\ -3 & 2\end{array}\right]\)
- \(A=\left[\begin{array}{ll}3 & 2 \\ 2 & 3\end{array}\right]\)
- \(A=\left[\begin{array}{ll}3 & 4 \\ 4 & 6\end{array}\right]\)
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- \(\langle\mathbf{v}, \quad \mathbf{v}\rangle=5 v_1^2-6 v_1 v_2+2 v_2^2= \frac{1}{5}\left[\left(5 v_1-3 v_2\right)^2+v_2^2\right]\)
- \(\langle\mathbf{v}, \mathbf{v}\rangle=3 v_1^2+8 v_1 v_2+6 v_2^2=\frac{1}{3}\left[\left(3 v_1+4 v_2\right)^2+\right. \left.2 v_2^2\right]\)
In each case, find a symmetric matrix \(A\) such that \(\langle\mathbf{v}, \mathbf{w}\rangle=\mathbf{v}^T A \mathbf{w}\).
- \(\left.\left\langle\begin{array}{l}v_1 \\ v_2\end{array}\right],\left[\begin{array}{l}w_1 \\ w_2\end{array}\right]\right\rangle=v_1 w_1+2 v_1 w_2+\)
- \(\left\langle\left[\begin{array}{l}v_1 \\ v_2\end{array}\right],\left[\begin{array}{l}w_1 \\ w_2\end{array}\right]\right\rangle=v_1 w_1-v_1 w_2-v_2 w_1+\)
- \(\left\langle\left[\begin{array}{l}v_1 \\ v_2 \\ v_3\end{array}\right],\left[\begin{array}{l}w_1 \\ w_2 \\ w_3\end{array}\right]\right\rangle=2 v_1 w_1+v_2 w_2+\)
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- \[\left[\begin{array}{rr}
1 & -2 \\
-2 & 1
\end{array}\right]\] - \[\left[\begin{array}{rrr}
1 & 0 & -2 \\
0 & 2 & 0 \\
-2 & 0 & 5
\end{array}\right]\]
- \[\left[\begin{array}{rr}
If \(A\) is symmetric and \(\mathbf{x}^T A \mathbf{x}=\) 0 for all columns \(\mathbf{x}\) in \(\mathbb{R}^n\), show that \(A=0\). [Hint: Consider \(\langle\mathbf{x}+\mathbf{y}, \mathbf{x}+\mathbf{y}\rangle\) where \(\langle\mathbf{x}, \mathbf{y}\rangle=\mathbf{x}^T A \mathbf{y}\).]
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By the condition, \(\langle\mathbf{x}, \mathbf{y}\rangle=\frac{1}{2}\langle\mathbf{x}+\mathbf{y}, \mathbf{x}+ \mathbf{y}\rangle=0\) for all \(\mathbf{x}, \mathbf{y}\). Let \(\mathbf{e}_i\) denote column \(i\) of \(I\). If \(A =\left[a_{i j}\right]\), then \(a_{i j}=\mathbf{e}_i^T A \mathbf{e}_j=\left\{\mathbf{e}_i, \mathbf{e}_j\right\}=0\) for all \(i\) and \(j\).
Show that the sum of two inner products on \(V\) is again an inner product.
Let \(\|\mathbf{u}\|=1,\|\mathbf{v}\|=2,\|\mathbf{w}\|=\sqrt{3}\), \(\langle\mathbf{u}, \mathbf{v}\rangle=-1,\langle\mathbf{u}, \mathbf{w}\rangle=0\) and \(\langle\mathbf{v}, \mathbf{w}\rangle=3\). Compute:
- \(\langle\mathbf{v}+\mathbf{w}, 2 \mathbf{u}-\mathbf{v}\rangle\)
- \(\langle\mathbf{u}-2 \mathbf{v}-\mathbf{w}, 3 \mathbf{w}-\mathbf{v}\rangle\)
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- -15
Given the data in Exercise 16, show that \(\mathbf{u}+\mathbf{v}=\mathbf{w}\).
Show that no vectors exist such that \(\|\mathbf{u}\|=1,\|\mathbf{v}\|=2\), and \(\langle\mathbf{u}, \mathbf{v}\rangle=-3\).
Complete Example 10.1.2.
Prove Theorem 10.1.1.
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- Using P2: \(\langle\mathbf{u}, \mathbf{v}+\mathbf{w}\rangle=\langle\mathbf{v}+\mathbf{w}, \mathbf{u}\rangle= \langle\mathbf{v}, \mathbf{u}\rangle+\langle\mathbf{w}, \mathbf{u}\rangle=\langle\mathbf{u}, \mathbf{v}\rangle+\langle\mathbf{u}, \mathbf{w}\rangle\).
- Using P2 and P4: \(\langle\mathbf{v}, r \mathbf{w}\rangle=\langle r \mathbf{w}, \mathbf{v}\rangle= r\langle\mathbf{w}, \mathbf{v}\rangle=r\langle\mathbf{v}, \mathbf{w}\rangle.\)
- Using P3: \(\langle\mathbf{0}, \mathbf{v}\rangle=\langle\mathbf{0}+\mathbf{0}, \mathbf{v}\rangle=\langle\mathbf{0}, \mathbf{v}\rangle+\langle\mathbf{0}, \mathbf{v}\rangle\), so \(\langle\mathbf{0}, \mathbf{v}\rangle=0\). The rest is P2.
- Assume that \(\langle\mathbf{v}, \mathbf{v}\rangle=0\). If \(\mathbf{v} \neq \mathbf{0}\) this contradicts P5, so \(\mathbf{v}=\mathbf{0}\). Conversely, if \(\mathbf{v}=\mathbf{0}\), then \(\langle\mathbf{v}, \mathbf{v}\rangle =0\) by Part 3 of this theorem.
Prove Theorem 10.1.6.
Let \(\mathbf{u}\) and \(\mathbf{v}\) be vectors in an inner product space \(V\).
- Expand \(\langle 2 \mathbf{u}-7 \mathbf{v}, 3 \mathbf{u}+5 \mathbf{v}\rangle\).
- Expand \(\langle 3 \mathbf{u}-4 \mathbf{v}, 5 \mathbf{u}+\mathbf{v}\rangle\).
- Show that \(\|\mathbf{u}+\mathbf{v}\|^2=\|\mathbf{u}\|^2+2\langle\mathbf{u}, \mathbf{v}\rangle+\|\mathbf{v}\|^2\).
- Show that \(\|\mathbf{u}-\mathbf{v}\|^2=\|\mathbf{u}\|^2-2\langle\mathbf{u}, \mathbf{v}\rangle+\|\mathbf{v}\|^2\).
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- \(15\|\mathbf{u}\|^2-17\langle\mathbf{u}, \mathbf{v}\rangle-4\|\mathbf{v}\|^2\)
- \(\|\mathbf{u}+\mathbf{v}\|^2=\langle\mathbf{u}+\mathbf{v}, \mathbf{u}+\mathbf{v}\rangle=\|\mathbf{u}\|^2+2\langle\mathbf{u}, \mathbf{v}\rangle +\|\mathbf{v}\|^2\)
Show that \(\|\mathbf{v}\|^2+\|\mathbf{w}\|^2=\) \(\frac{1}{2}\left\{\|\mathbf{v}+\mathbf{w}\|^2+\|\mathbf{v}-\mathbf{w}\|^2\right\}\) for any \(\mathbf{v}\) and \(\mathbf{w}\) 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 \(\mathrm{d}(\mathbf{v}, \mathbf{w})=\mathrm{d}(\mathbf{v}+\mathbf{u}, \mathbf{w}+\mathbf{u})\) for all \(\mathbf{v}, \mathbf{w}\), and \(\mathbf{u}\) in \(V\).
- Show that \(\langle\mathbf{u}, \mathbf{v}\rangle=\frac{1}{4}\left[\|\mathbf{u}+\mathbf{v}\|^2=\|\mathbf{u}-\mathbf{v}\|^2\right]\) for all \(\mathbf{u}, \mathbf{v}\) in an inner product space \(V\).
- If \(\langle\),\(\rangle and \langle,\rangle^{\prime}\) are two inner products on \(V\) that have equal associated norm functions, show that \(\langle\mathbf{u}, \mathbf{v}\rangle=\langle\mathbf{u}, \mathbf{v}\rangle^{\prime}\) holds for all \(\mathbf{u}\) and \(v\).
Let \(\mathbf{v}\) denote a vector in an inner product space \(V\).
- Show that \(W=\{\mathbf{w} \mid \mathbf{w}\) in \(V,\langle\mathbf{v}, \mathbf{w}\rangle=0\}\) is a subspace of \(V\).
- If \(V=\mathbb{R}^3\) with the dot product, and if \(\mathbf{v}=(1\), \(-1,2)\), find a basis for \(W(W\) as in (a)).
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- \(\{(1,1,0),(0,2,1)\}\)
Given vectors \(\mathbf{w}_1, \mathbf{w}_2, \ldots, \mathbf{w}_n\) and \(\mathbf{v}\), assume that \(\left\langle\mathbf{v}, \mathbf{w}_i\right\rangle=0\) for each \(i\). Show that \(\langle\mathbf{v}, \mathbf{w}\rangle=0\) for all \(\mathbf{w}\) in \(\operatorname{span}\left\{\mathbf{w}_1, \mathbf{w}_2, \ldots, \mathbf{w}_n\right\}\).
If \(V=\operatorname{span}\left\{\mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_n\right\}\) and \(\left\langle\mathbf{v}, \mathbf{v}_i\right\rangle=\left\langle\mathbf{w}, \mathbf{v}_i\right\rangle\) holds for each \(i\). Show that \(\mathbf{v}=\mathbf{w}\).
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\(\left\langle\mathbf{v}-\mathbf{w}, \mathbf{v}_i\right\rangle=\left\langle\mathbf{v}, \mathbf{v}_i\right\rangle-\left\langle\mathbf{w}, \mathbf{v}_i\right\rangle=0$ for each $i$, so $\mathbf{v}=\mathbf{w}\) by Exercise 27.
Use the Cauchy-Schwarz inequality in an inner product space to show that:
- If \(\|\mathbf{u}\| \leq 1\), then \(\langle\mathbf{u}, \mathbf{v}\rangle^2 \leq\|\mathbf{v}\|^2\) for all \(\mathbf{v}\) in \(V\).
- \((x \cos \theta+y \sin \theta)^2 \leq x^2+y^2\) for all real \(x, y\), and \(\theta\).
- \(\left\|r_1 \mathbf{v}_1+\cdots+r_n \mathbf{v}_n\right\|^2 \leq\left[r_1\left\|\mathbf{v}_1\right\|+\cdots+\right.\) \(\left.r_n\left\|\mathbf{v}_n\right\|\right]^2\) for all vectors \(\mathbf{v}_i\), and all \(r_i>0\) in \(\mathbb{R}\).
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- If \(\mathbf{u}=(\cos \theta, \sin \theta)\) in \(\mathbb{R}^2\) (with the dot product) then \(\|\mathbf{u}\|=1\). Use (a) with \(\mathbf{v}= (x, y)\).
If \(A\) is a \(2 \times n\) matrix, let \(\mathbf{u}\) and \(\mathbf{v}\) denote the rows of \(A\).
- Show that \(A A^T=\left[\begin{array}{rr}\|\mathbf{u}\|^2 & \mathbf{u} \cdot \mathbf{v} \\ \mathbf{u} \cdot \mathbf{v} & \|\mathbf{v}\|^2\end{array}\right]\).
- Show that \(\text{det}\left(A A^T\right) \geq 0\).
- If \(\mathbf{v}\) and \(\mathbf{w}\) are nonzero vectors in an inner product space \(V\), show that \(-1 \leq \frac{\langle\mathbf{v}, \mathbf{w}\rangle}{\|\mathbf{v}\|\|\mathbf{w}\|} \leq 1\), and hence that a unique angle \(\theta\) exists such that \(\frac{\langle\mathbf{v}, \mathbf{w}\rangle}{\|\mathbf{v}\|\|\mathbf{w}\|}=\cos \theta\) and \(0 \leq \theta \leq \pi\). This angle \(\theta\) is called the angle between \(\mathbf{v}\) and \(\mathbf{w}\).
- Find the angle between \(\mathbf{v}=(1,2,-1,1,3)\) and \(\mathbf{w}=(2,1,0,2,0)\) in \(\mathbb{R}^5\) with the dot product.
- If \(\theta\) is the angle between \(\mathbf{v}\) and \(\mathbf{w}\), show that the law of cosines is valid: \[ \|\mathbf{v}-\mathbf{w}\|=\|\mathbf{v}\|^2+\|\mathbf{w}\|^2-2\|\mathbf{v}\|\|\mathbf{w}\| \cos \theta .\nonumber \]
If \(V=\mathbb{R}^2\), define \(\|(x, y)\|=|x|+\) \(|y|\).
- Show that \(\|\cdot\|\) satisfies the conditions in Theorem 10.1.5.
- Show that \(\|\cdot\|\) does not arise from an inner product on \(\mathbb{R}^2\) given by a matrix \(A\). [Hint: If it did, use Theorem 10.1.2 to find numbers \(a\), \(b\), and \(c\) such that \(\|(x, y)\|^2=a x^2+b x y+c y^2\) for all \(x\) and \(y\).]