Homework 9
1. Consider R 4 \R^4 R 4 ⟨ u , v ⟩ \langle\mathbf{u},\mathbf{v}\rangle ⟨ u , v ⟩
(i) Find the norm of the vectors u = ( 1 , 2 , 3 , 2 ) \mathbf{u} = (1, 2, 3, 2) u = ( 1 , 2 , 3 , 2 ) v = ( 2 , 1 , − 1 , 0 ) \mathbf{v} = (2, 1, −1, 0) v = ( 2 , 1 , − 1 , 0 )
∣ ∣ u ∣ ∣ = 1 2 + 2 2 + 3 2 + 2 2 = 3 2 ∣ ∣ v ∣ ∣ = 2 2 + 1 2 + ( − 1 ) 2 + 0 2 = 6 ||\mathbf{u}|| = \sqrt{1^2+2^2+3^2+2^2} = 3\sqrt{2}\\
||\mathbf{v}|| = \sqrt{2^2+1^2+(−1)^2+0^2} = \sqrt{6} ∣∣ u ∣∣ = 1 2 + 2 2 + 3 2 + 2 2 = 3 2 ∣∣ v ∣∣ = 2 2 + 1 2 + ( − 1 ) 2 + 0 2 = 6
(ii) What is the angle between u \mathbf{u} u v \mathbf{v} v
θ = cos − 1 ⟨ u , v ⟩ ∣ ∣ u ∣ ∣ ⋅ ∣ ∣ v ∣ ∣ = cos − 1 1 ( 2 ) + 2 ( 1 ) + 3 ( − 1 ) + 2 ( 0 ) 3 2 6 = cos − 1 1 6 3 ≈ 84.4782 ° \theta = \cos^{-1}\frac{\langle\mathbf{u},\mathbf{v}\rangle}{||\mathbf{u}||\cdot||\mathbf{v}||}
= \cos^{-1}\frac{1(2)+2(1)+3(-1)+2(0)}{3\sqrt{2}\sqrt{6}}
= \cos^{-1}\frac{1}{6\sqrt{3}}
\approx 84.4782\degree θ = cos − 1 ∣∣ u ∣∣ ⋅ ∣∣ v ∣∣ ⟨ u , v ⟩ = cos − 1 3 2 6 1 ( 2 ) + 2 ( 1 ) + 3 ( − 1 ) + 2 ( 0 ) = cos − 1 6 3 1 ≈ 84.4782°
2. Consider v 1 = [ − 1 2 1 ] , v 2 = [ 1 0 1 ] , v 3 = [ − 1 − 1 1 ] \mathbf{v}_1=\begin{bmatrix} -1 \\ 2 \\ 1\end{bmatrix},\mathbf{v}_2=\begin{bmatrix} 1 \\ 0 \\ 1\end{bmatrix},\mathbf{v}_3=\begin{bmatrix} -1 \\ -1 \\ 1\end{bmatrix} v 1 = − 1 2 1 , v 2 = 1 0 1 , v 3 = − 1 − 1 1
The inner product for all pairs of v 1 , v 2 , v 3 \mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3 v 1 , v 2 , v 3
⟨ v 1 , v 2 ⟩ = ( − 1 ) ( 1 ) + 2 ( 0 ) + 1 ( 1 ) = 0 = ⟨ v 2 , v 3 ⟩ = 1 ( − 1 ) + 0 ( − 1 ) + 1 ( 1 ) = 0 = ⟨ v 1 , v 3 ⟩ = ( − 1 ) ( − 1 ) + 2 ( − 1 ) + 1 ( 1 ) = 0 \begin{align*}
\langle\mathbf{v}_1,\mathbf{v}_2\rangle
&= (-1)(1) + 2(0) + 1(1)
&= 0 \\
= \langle\mathbf{v}_2,\mathbf{v}_3\rangle
&= 1(-1) + 0(-1) + 1(1)
&= 0 \\
= \langle\mathbf{v}_1,\mathbf{v}_3\rangle
&= (-1)(-1) + 2(-1) + 1(1)
&= 0
\end{align*} ⟨ v 1 , v 2 ⟩ = ⟨ v 2 , v 3 ⟩ = ⟨ v 1 , v 3 ⟩ = ( − 1 ) ( 1 ) + 2 ( 0 ) + 1 ( 1 ) = 1 ( − 1 ) + 0 ( − 1 ) + 1 ( 1 ) = ( − 1 ) ( − 1 ) + 2 ( − 1 ) + 1 ( 1 ) = 0 = 0 = 0
And a matrix composed of these vectors is full rank.
rref [ ∣ ∣ ∣ v 1 v 2 v 3 ∣ ∣ ∣ ] = rref [ − 1 1 − 1 2 0 − 1 1 1 1 ] = [ 1 0 0 0 1 0 0 0 1 ] \operatorname{rref}\begin{bmatrix}
| & | & | \\
\mathbf{v}_1 & \mathbf{v}_2 & \mathbf{v}_3 \\
| & | & | \\
\end{bmatrix} =
\operatorname{rref}\begin{bmatrix}
-1 & 1 & -1 \\
2 & 0 & -1 \\
1 & 1 & 1
\end{bmatrix} =
\begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{bmatrix} rref ∣ v 1 ∣ ∣ v 2 ∣ ∣ v 3 ∣ = rref − 1 2 1 1 0 1 − 1 − 1 1 = 1 0 0 0 1 0 0 0 1
Hence, v 1 , v 2 , v 3 \mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3 v 1 , v 2 , v 3 R 3 \R^3 R 3
(ii) Find the orthonormal basis generated by v 1 , v 2 , v 3 \mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3 v 1 , v 2 , v 3
v ^ 1 = v 1 ∣ ∣ v 1 ∣ ∣ = 1 ( − 1 ) 2 + 2 2 + 1 2 [ − 1 2 1 ] = [ − 1 6 2 6 1 6 ] v ^ 2 = v 2 ∣ ∣ v 2 ∣ ∣ = 1 1 2 + 0 2 + 1 2 [ 1 0 1 ] = [ 1 2 0 1 2 ] v ^ 3 = v 3 ∣ ∣ v 3 ∣ ∣ = 1 ( − 1 ) 2 + ( − 1 ) 2 + 1 2 [ − 1 − 1 1 ] = [ − 1 3 − 1 3 1 3 ] \mathbf{\hat{v}}_1
= \frac{\mathbf{v}_1}{||\mathbf{v}_1||}
= \frac{1}{\sqrt{(-1)^2+2^2+1^2}}\begin{bmatrix}
-1 \\ 2 \\ 1
\end{bmatrix}
= \begin{bmatrix}
-\frac{1}{\sqrt{6}} \\[.5em] \frac{2}{\sqrt{6}} \\[.5em] \frac{1}{\sqrt{6}}
\end{bmatrix}
\\[1em]
\mathbf{\hat{v}}_2
= \frac{\mathbf{v}_2}{||\mathbf{v}_2||}
= \frac{1}{\sqrt{1^2+0^2+1^2}}\begin{bmatrix}
1 \\ 0 \\ 1
\end{bmatrix}
= \begin{bmatrix}
\frac{1}{\sqrt{2}} \\[.5em] 0 \\ \frac{1}{\sqrt{2}}
\end{bmatrix}
\\[1em]
\mathbf{\hat{v}}_3
= \frac{\mathbf{v}_3}{||\mathbf{v}_3||}
= \frac{1}{\sqrt{(-1)^2+(-1)^2+1^2}}\begin{bmatrix}
-1 \\ -1 \\ 1
\end{bmatrix}
= \begin{bmatrix}
-\frac{1}{\sqrt{3}} \\[.5em] -\frac{1}{\sqrt{3}} \\[.5em] \frac{1}{\sqrt{3}}
\end{bmatrix} v ^ 1 = ∣∣ v 1 ∣∣ v 1 = ( − 1 ) 2 + 2 2 + 1 2 1 − 1 2 1 = − 6 1 6 2 6 1 v ^ 2 = ∣∣ v 2 ∣∣ v 2 = 1 2 + 0 2 + 1 2 1 1 0 1 = 2 1 0 2 1 v ^ 3 = ∣∣ v 3 ∣∣ v 3 = ( − 1 ) 2 + ( − 1 ) 2 + 1 2 1 − 1 − 1 1 = − 3 1 − 3 1 3 1
The orthonormal basis { v ^ 1 , v ^ 2 , v ^ 3 } \set{\mathbf{\hat{v}}_1,\mathbf{\hat{v}}_2,\mathbf{\hat{v}}_3} { v ^ 1 , v ^ 2 , v ^ 3 } v 1 , v 2 , v 3 \mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3 v 1 , v 2 , v 3
{ [ − 1 6 2 6 1 6 ] , [ 1 2 0 1 2 ] , [ − 1 3 − 1 3 1 3 ] } . \Set{
\begin{bmatrix}
-\frac{1}{\sqrt{6}} \\[.5em] \frac{2}{\sqrt{6}} \\[.5em] \frac{1}{\sqrt{6}}
\end{bmatrix},
\begin{bmatrix}
\frac{1}{\sqrt{2}} \\[.5em] 0 \\[.5em] \frac{1}{\sqrt{2}}
\end{bmatrix},
\begin{bmatrix}
-\frac{1}{\sqrt{3}} \\[.5em] -\frac{1}{\sqrt{3}} \\[.5em] \frac{1}{\sqrt{3}}
\end{bmatrix}
}. ⎩ ⎨ ⎧ − 6 1 6 2 6 1 , 2 1 0 2 1 , − 3 1 − 3 1 3 1 ⎭ ⎬ ⎫ .
(iii) Express v 4 = [ 1 2 0 ] \mathbf{v}_4=\begin{bmatrix}1 \\ 2 \\ 0\end{bmatrix} v 4 = 1 2 0 v 1 , v 2 , v 3 \mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3 v 1 , v 2 , v 3
rref [ ∣ ∣ ∣ ∣ v 1 v 2 v 3 v 4 ∣ ∣ ∣ ∣ ] = rref [ − 1 1 − 1 1 2 0 − 1 2 1 1 1 0 ] = [ 1 0 0 1 2 0 1 0 1 2 0 0 1 − 1 ] ∴ v 4 = 1 2 v 1 + 1 2 v 2 − v 3 \operatorname{rref}\left[
\begin{array}{ccc|c}
| & | & | & | \\
\mathbf{v}_1 & \mathbf{v}_2 & \mathbf{v}_3 & \mathbf{v}_4 \\
| & | & | & | \\
\end{array}
\right] =
\operatorname{rref}\left[
\begin{array}{ccc|c}
-1 & 1 & -1 & 1 \\
2 & 0 & -1 & 2 \\
1 & 1 & 1 & 0
\end{array}
\right] = \left[
\begin{array}{ccc|c}
1 & 0 & 0 & \frac{1}{2} \\[.3em]
0 & 1 & 0 & \frac{1}{2} \\[.3em]
0 & 0 & 1 & -1
\end{array}
\right]
\\[1em]
\therefore\mathbf{v}_4 = \frac{1}{2}\mathbf{v}_1 + \frac{1}{2}\mathbf{v}_2 - \mathbf{v}_3 rref ∣ v 1 ∣ ∣ v 2 ∣ ∣ v 3 ∣ ∣ v 4 ∣ = rref − 1 2 1 1 0 1 − 1 − 1 1 1 2 0 = 1 0 0 0 1 0 0 0 1 2 1 2 1 − 1 ∴ v 4 = 2 1 v 1 + 2 1 v 2 − v 3
3. Consider the space of all continuous functions on [ 0 , 1 ] [0, 1] [ 0 , 1 ] C [ 0 , 1 ] C[0, 1] C [ 0 , 1 ] ⟨ f , g ⟩ = ∫ 0 1 f ( x ) g ( x ) d x \langle f, g\rangle = \int_0^1 f(x)g(x)\d x ⟨ f , g ⟩ = ∫ 0 1 f ( x ) g ( x ) d x
(i) Find the norm of f ( x ) = x n f(x) = x^n f ( x ) = x n n n n
For n ∈ Z + n\in\Z^+ n ∈ Z +
⟨ x n , x n ⟩ = ∫ 0 1 x 2 n d x = x 2 n 2 n + 1 ∣ 0 1 = 1 2 n 2 n + 1 = 1 2 n + 1 ∣ ∣ x n ∣ ∣ = ⟨ x n , x n ⟩ = 1 2 n + 1 \langle x^n , x^n\rangle
= \int_0^1 x^{2n} \d x
= \left.\frac{x^{2n}}{2n+1}\right|_0^1
= \frac{1^{2n}}{2n+1}
= \frac{1}{2n+1}\\[1em]
||x^n|| = \sqrt{\langle x^n , x^n\rangle} = \frac{1}{\sqrt{2n+1}} ⟨ x n , x n ⟩ = ∫ 0 1 x 2 n d x = 2 n + 1 x 2 n 0 1 = 2 n + 1 1 2 n = 2 n + 1 1 ∣∣ x n ∣∣ = ⟨ x n , x n ⟩ = 2 n + 1 1
(ii) Find the angle between x n x^n x n x m x^m x m
Assuming n , m ∈ Z + n,m\in\Z^+ n , m ∈ Z +
θ = ⟨ x n , x m ⟩ ∣ ∣ x n ∣ ∣ ⋅ ∣ ∣ x m ∣ ∣ = cos − 1 ∫ 0 1 x n m d x ∫ 0 1 x 2 n d x ∫ 0 1 x 2 m d x = cos − 1 1 n m n m + 1 1 2 n 2 n + 1 1 2 m 2 m + 1 = cos − 1 1 ( n m + 1 ) 1 ( 2 n + 1 ) ( 2 m + 1 ) = cos − 1 ( 2 n + 1 ) ( 2 m + 1 ) n m + 1 \begin{align*}
\theta = \frac{\langle x^n, x^m\rangle}{||x^n||\cdot||x^m||}
&=\cos^{-1} \frac{\displaystyle\int_0^1 x^{nm}\d x}
{\displaystyle\int_0^1 x^{2n}\d x\int_0^1 x^{2m}\d x} \\[2.5em]
&=\cos^{-1} \frac{\displaystyle\frac{1^{nm}}{nm+1}}
{\sqrt{\displaystyle\frac{1^{2n}}{2n+1}\frac{1^{2m}}{2m+1}}} \\[2.5em]
&=\cos^{-1} \frac{1}{(nm+1)\displaystyle\frac{1}{\sqrt{(2n+1)(2m+1)}}} \\[2.5em]
&=\cos^{-1} \frac{\sqrt{(2n+1)(2m+1)}}{nm+1}
\end{align*} θ = ∣∣ x n ∣∣ ⋅ ∣∣ x m ∣∣ ⟨ x n , x m ⟩ = cos − 1 ∫ 0 1 x 2 n d x ∫ 0 1 x 2 m d x ∫ 0 1 x nm d x = cos − 1 2 n + 1 1 2 n 2 m + 1 1 2 m nm + 1 1 nm = cos − 1 ( nm + 1 ) ( 2 n + 1 ) ( 2 m + 1 ) 1 1 = cos − 1 nm + 1 ( 2 n + 1 ) ( 2 m + 1 )
Since it wasn’t specified in the question, if n n n m m m n n n m m m n m ≠ − 1 ∧ ( 2 n + 1 ) ( 2 m + 1 ) ≥ 0 nm\neq-1\land(2n+1)(2m+1)\ge0 nm = − 1 ∧ ( 2 n + 1 ) ( 2 m + 1 ) ≥ 0
Again, assuming m , n ∈ Z + m,n\in\Z^+ m , n ∈ Z + ⟨ sin 2 π m x , sin 2 π n x ⟩ = 0 ∀ n ≠ m \langle\sin2\pi mx, \sin2\pi nx\rangle = 0 \; \forall n\neq m ⟨ sin 2 πm x , sin 2 πn x ⟩ = 0 ∀ n = m
Again, since it wasn’t specified in the question, we assume m , n ∈ Z + m,n\in\Z^+ m , n ∈ Z + m m m n n n
For example, take m = − 1 m=-1 m = − 1 n = 1 n=1 n = 1
⟨ sin 2 π m x , sin 2 π n x ⟩ = ⟨ sin ( − 2 π x ) , sin 2 π x ⟩ = ∫ 0 1 sin ( − 2 π x ) sin 2 π x d x = 1 2 ∫ 0 1 cos ( − 4 π x ) − cos ( 0 ) d x = 1 2 ∫ 0 1 cos ( − 4 π x ) − 1 2 ∫ 0 1 cos ( 0 ) d x = − 1 2 \begin{align*}
\langle\sin2\pi mx, \sin2\pi nx\rangle
&= \langle\sin(-2\pi x), \sin2\pi x\rangle \\
&= \int_0^1 \sin(-2\pi x) \sin2\pi x \d x \\
&= \frac{1}{2}\int_0^1\cos(-4\pi x) - \cos(0) \d x \\
&= \frac{1}{2}\int_0^1\cos(-4\pi x) - \frac{1}{2}\int_0^1\cos(0) \d x \\
&= -\frac{1}{2}
\end{align*} ⟨ sin 2 πm x , sin 2 πn x ⟩ = ⟨ sin ( − 2 π x ) , sin 2 π x ⟩ = ∫ 0 1 sin ( − 2 π x ) sin 2 π x d x = 2 1 ∫ 0 1 cos ( − 4 π x ) − cos ( 0 ) d x = 2 1 ∫ 0 1 cos ( − 4 π x ) − 2 1 ∫ 0 1 cos ( 0 ) d x = − 2 1
Then, using the product-to-sum formula
sin α sin β = 1 2 ( cos ( α − β ) − cos ( α + β ) ) , \sin\alpha\sin\beta = \frac{1}{2}(\cos(\alpha-\beta)-\cos(\alpha+\beta)), sin α sin β = 2 1 ( cos ( α − β ) − cos ( α + β )) ,
the inner product can be written as follows:
⟨ sin 2 π m x , sin 2 π n x ⟩ = ∫ 0 1 ( sin 2 π m x ) ( sin 2 π n x ) d x = ∫ 0 1 1 2 ( cos ( 2 π m x − 2 π n x ) − cos ( 2 π m x + 2 π n x ) ) d x = 1 2 ∫ 0 1 cos 2 π x ( m − n ) − cos 2 π x ( m + n ) d x = 1 2 ∫ 0 1 cos 2 π x ( m − n ) d x − 1 2 ∫ 0 1 cos 2 π x ( m + n ) d x \begin{align*}
\langle\sin2\pi mx, \sin2\pi nx\rangle
&= \int_0^1 (\sin2\pi mx)(\sin2\pi nx ) \d x \\
&= \int_0^1 \frac{1}{2}(\cos(2\pi mx - 2\pi nx) - \cos(2\pi mx + 2\pi nx))\d x \\
&= \frac{1}{2}\int_0^1 \cos2\pi x(m-n) - \cos2\pi x(m+n)\d x \\
&= \frac{1}{2}\int_0^1 \cos2\pi x(m-n) \d x - \frac{1}{2}\int_0^1 \cos2\pi x(m+n)\d x
\end{align*} ⟨ sin 2 πm x , sin 2 πn x ⟩ = ∫ 0 1 ( sin 2 πm x ) ( sin 2 πn x ) d x = ∫ 0 1 2 1 ( cos ( 2 πm x − 2 πn x ) − cos ( 2 πm x + 2 πn x )) d x = 2 1 ∫ 0 1 cos 2 π x ( m − n ) − cos 2 π x ( m + n ) d x = 2 1 ∫ 0 1 cos 2 π x ( m − n ) d x − 2 1 ∫ 0 1 cos 2 π x ( m + n ) d x
Notice that if m m m n n n m ≠ n m\neq n m = n x x x 2 π 2\pi 2 π
Since
∫ 0 1 cos 2 π x d x = 0 ∀ x ∈ Z + , \int_0^1 \cos2\pi x \d x = 0 \quad\forall x\in\Z^+, ∫ 0 1 cos 2 π x d x = 0 ∀ x ∈ Z + ,
then the inner product must be zero for all positive integers m ≠ n m\neq n m = n
Or more clearly, if we recall our Calculus II nightmare by performing u u u
⟨ sin 2 π m x , sin 2 π n x ⟩ = 1 2 ∫ 0 1 cos 2 π x ( m − n ) d x − 1 2 ∫ 0 1 cos 2 π x ( m + n ) d x = careful calculations = 1 4 π ( sin 2 π ( m − n ) m − n − sin 2 π ( m + n ) m + n ) \begin{align*}
\langle\sin2\pi mx, \sin2\pi nx\rangle
&= \frac{1}{2}\int_0^1 \cos2\pi x(m-n) \d x - \frac{1}{2}\int_0^1 \cos2\pi x(m+n)\d x \\
&= \href{https://www.wolframalpha.com/input?i=%5Cint_0%5E1+%28%5Csin%282%5Cpi+mx%29%29%28%5Csin%282%5Cpi+nx%29%29+dx+}
{\text{careful calculations}} \\
&= \frac{1}{4\pi}\(
\frac{\sin2\pi(m-n)}{m-n} - \frac{\sin2\pi(m+n)}{m+n}
\)
\end{align*} ⟨ sin 2 πm x , sin 2 πn x ⟩ = 2 1 ∫ 0 1 cos 2 π x ( m − n ) d x − 2 1 ∫ 0 1 cos 2 π x ( m + n ) d x = careful calculations = 4 π 1 ( m − n sin 2 π ( m − n ) − m + n sin 2 π ( m + n ) )
We can see that the argument of sin \sin sin 2 π 2\pi 2 π m = n m=n m = n
4. Prove the identity ⟨ a v + b w , c v + d w ⟩ = a c ∣ ∣ v ∣ ∣ 2 + ( a d + b c ) ⟨ v , w ⟩ + b d ∣ ∣ w ∣ ∣ 2 . \langle a\mathbf{v} + b\mathbf{w}, c\mathbf{v} + d\mathbf{w}\rangle = ac||\mathbf{v}||^2 + (ad + bc)\langle \mathbf{v}, \mathbf{w}\rangle + bd||\mathbf{w}||^2. ⟨ a v + b w , c v + d w ⟩ = a c ∣∣ v ∣ ∣ 2 + ( a d + b c ) ⟨ v , w ⟩ + b d ∣∣ w ∣ ∣ 2 .
⟨ a v + b w , c v + d w ⟩ = ⟨ a v , c v + d w ⟩ + ⟨ b w , c v + d w ⟩ = ⟨ a v , c v ⟩ + ⟨ a v , d w ⟩ + ⟨ b w , c v ⟩ + ⟨ b w , d w ⟩ = a c ⟨ v , v ⟩ + a d ⟨ v , w ⟩ + b c ⟨ w , v ⟩ + b d ⟨ w , w ⟩ = a c ∣ ∣ v ∣ ∣ 2 + ( a d + b c ) ⟨ v , w ⟩ + b d ∣ ∣ w ∣ ∣ 2 \begin{align*}
\langle a\mathbf{v} + b\mathbf{w}, c\mathbf{v} + d\mathbf{w}\rangle
&= \langle a\mathbf{v}, c\mathbf{v} + d\mathbf{w}\rangle
+ \langle b\mathbf{w}, c\mathbf{v} + d\mathbf{w}\rangle \\
&= \langle a\mathbf{v}, c\mathbf{v}\rangle
+ \langle a\mathbf{v},d\mathbf{w}\rangle
+ \langle b\mathbf{w},c\mathbf{v}\rangle
+ \langle b\mathbf{w},d\mathbf{w}\rangle \\
&= ac \langle \mathbf{v},\mathbf{v}\rangle
+ ad \langle \mathbf{v},\mathbf{w}\rangle
+ bc \langle\mathbf{w},\mathbf{v}\rangle
+ bd \langle\mathbf{w},\mathbf{w}\rangle \\
&= ac||\mathbf{v}||^2 + (ad + bc)\langle \mathbf{v}, \mathbf{w}\rangle + bd||\mathbf{w}||^2
\end{align*} ⟨ a v + b w , c v + d w ⟩ = ⟨ a v , c v + d w ⟩ + ⟨ b w , c v + d w ⟩ = ⟨ a v , c v ⟩ + ⟨ a v , d w ⟩ + ⟨ b w , c v ⟩ + ⟨ b w , d w ⟩ = a c ⟨ v , v ⟩ + a d ⟨ v , w ⟩ + b c ⟨ w , v ⟩ + b d ⟨ w , w ⟩ = a c ∣∣ v ∣ ∣ 2 + ( a d + b c ) ⟨ v , w ⟩ + b d ∣∣ w ∣ ∣ 2
5. Given an inner product space V V V
(i) Show that ∣ ∣ x + y ∣ ∣ 2 + ∣ ∣ x − y ∣ ∣ 2 = 2 ( ∣ ∣ x ∣ ∣ 2 + ∣ ∣ y ∣ ∣ 2 ) . ||\mathbf{x} + \mathbf{y}||^2 + ||\mathbf{x} − \mathbf{y}||^2 = 2(||\mathbf{x}||^2 + ||\mathbf{y}||^2). ∣∣ x + y ∣ ∣ 2 + ∣∣ x − y ∣ ∣ 2 = 2 ( ∣∣ x ∣ ∣ 2 + ∣∣ y ∣ ∣ 2 ) .
∣ ∣ x + y ∣ ∣ 2 = ( ∣ ∣ x ∣ ∣ + ∣ ∣ y ∣ ∣ ) 2 = ∣ ∣ x ∣ ∣ 2 + ∣ ∣ y ∣ ∣ 2 + 2 ∣ ∣ x ∣ ∣ ∣ ∣ y ∣ ∣ ∣ ∣ x − y ∣ ∣ 2 = ( ∣ ∣ x ∣ ∣ + ∣ ∣ y ∣ ∣ ) 2 = ∣ ∣ x ∣ ∣ 2 + ∣ ∣ y ∣ ∣ 2 − 2 ∣ ∣ x ∣ ∣ ∣ ∣ y ∣ ∣ ∴ ∣ ∣ x + y ∣ ∣ 2 + ∣ ∣ x − y ∣ ∣ 2 = ∣ ∣ x ∣ ∣ 2 + ∣ ∣ y ∣ ∣ 2 + 2 ∣ ∣ x ∣ ∣ ∣ ∣ y ∣ ∣ + ∣ ∣ x ∣ ∣ 2 + ∣ ∣ y ∣ ∣ 2 − 2 ∣ ∣ x ∣ ∣ ∣ ∣ y ∣ ∣ = ∣ ∣ x ∣ ∣ 2 + ∣ ∣ y ∣ ∣ 2 + ∣ ∣ x ∣ ∣ 2 + ∣ ∣ y ∣ ∣ 2 = 2 ( ∣ ∣ x ∣ ∣ 2 + ∣ ∣ y ∣ ∣ 2 ) ||\mathbf{x}+\mathbf{y}||^2 = (||\mathbf{x}||+||\mathbf{y}||)^2
= ||\mathbf{x}||^2 + ||\mathbf{y}||^2 + 2||\mathbf{x}|| ||\mathbf{y}|| \\
||\mathbf{x}-\mathbf{y}||^2 = (||\mathbf{x}||+||\mathbf{y}||)^2
= ||\mathbf{x}||^2 + ||\mathbf{y}||^2 - 2||\mathbf{x}|| ||\mathbf{y}||
\\[1em]
\begin{align*}
\therefore ||\mathbf{\mathbf{x}} + \mathbf{\mathbf{y}}||^2 + ||\mathbf{\mathbf{x}} − \mathbf{\mathbf{y}}||^2
&= ||\mathbf{x}||^2 + ||\mathbf{y}||^2 \;\cancel{+\; 2||\mathbf{x}|| ||\mathbf{y}||}+ ||\mathbf{x}||^2 + ||\mathbf{y}||^2 \;\cancel{-\; 2||\mathbf{x}|| ||\mathbf{y}||} \\
&= ||\mathbf{x}||^2 + ||\mathbf{y}||^2 + ||\mathbf{x}||^2 + ||\mathbf{y}||^2 \\
&= 2(||\mathbf{x}||^2 + ||\mathbf{y}||^2)
\end{align*} ∣∣ x + y ∣ ∣ 2 = ( ∣∣ x ∣∣ + ∣∣ y ∣∣ ) 2 = ∣∣ x ∣ ∣ 2 + ∣∣ y ∣ ∣ 2 + 2∣∣ x ∣∣∣∣ y ∣∣ ∣∣ x − y ∣ ∣ 2 = ( ∣∣ x ∣∣ + ∣∣ y ∣∣ ) 2 = ∣∣ x ∣ ∣ 2 + ∣∣ y ∣ ∣ 2 − 2∣∣ x ∣∣∣∣ y ∣∣ ∴ ∣∣ x + y ∣ ∣ 2 + ∣∣ x − y ∣ ∣ 2 = ∣∣ x ∣ ∣ 2 + ∣∣ y ∣ ∣ 2 + 2∣∣ x ∣∣∣∣ y ∣∣ + ∣∣ x ∣ ∣ 2 + ∣∣ y ∣ ∣ 2 − 2∣∣ x ∣∣∣∣ y ∣∣ = ∣∣ x ∣ ∣ 2 + ∣∣ y ∣ ∣ 2 + ∣∣ x ∣ ∣ 2 + ∣∣ y ∣ ∣ 2 = 2 ( ∣∣ x ∣ ∣ 2 + ∣∣ y ∣ ∣ 2 )
(ii) Show that ⟨ u , v ⟩ = 1 4 ( ∣ ∣ x + y ∣ ∣ 2 − ∣ ∣ x − y ∣ ∣ 2 ) \langle \mathbf{u}, \mathbf{v}\rangle = \frac{1}{4}(||\mathbf{x} + \mathbf{y}||^2 − ||\mathbf{x} − \mathbf{y}||^2) ⟨ u , v ⟩ = 4 1 ( ∣∣ x + y ∣ ∣ 2 − ∣∣ x − y ∣ ∣ 2 )
Assuming the left-hand side is meant to be ⟨ x , y ⟩ \langle \mathbf{x}, \mathbf{y}\rangle ⟨ x , y ⟩ ⟨ x , y ⟩ = 1 4 ( ∣ ∣ x + y ∣ ∣ 2 − ∣ ∣ x − y ∣ ∣ 2 ) . \langle \mathbf{x}, \mathbf{y}\rangle = \frac{1}{4}(||\mathbf{x} + \mathbf{y}||^2 − ||\mathbf{x} − \mathbf{y}||^2). ⟨ x , y ⟩ = 4 1 ( ∣∣ x + y ∣ ∣ 2 − ∣∣ x − y ∣ ∣ 2 ) .
1 4 ( ∣ ∣ x + y ∣ ∣ 2 − ∣ ∣ x − y ∣ ∣ 2 ) = 1 4 ( ⟨ x + y , x + y ⟩ − ⟨ x − y , x − y ⟩ ) = 1 4 ( ⟨ x + y , x ⟩ + ⟨ x + y , y ⟩ − ( ⟨ x − y , x ⟩ − ⟨ x − y , y ⟩ ) ) = 1 4 ( ⟨ x , x ⟩ + ⟨ x , y ⟩ + ⟨ x , y ⟩ + ⟨ y , y ⟩ − ( ⟨ x , x ⟩ − ⟨ x , y ⟩ − ( ⟨ x , y ⟩ − ⟨ y , y ⟩ ) ) ) = 1 4 ( ⟨ x , x ⟩ + ⟨ x , y ⟩ + ⟨ x , y ⟩ + ⟨ y , y ⟩ − ( ⟨ x , x ⟩ − ⟨ x , y ⟩ − ⟨ x , y ⟩ + ⟨ y , y ⟩ ) ) = 1 4 ( ⟨ x , x ⟩ + ⟨ x , y ⟩ + ⟨ x , y ⟩ + ⟨ y , y ⟩ − ⟨ x , x ⟩ + ⟨ x , y ⟩ + ⟨ x , y ⟩ − ⟨ y , y ⟩ ) = 1 4 ( 4 ⟨ x , y ⟩ ) = ⟨ x , y ⟩ \def<{\langle}\def>{\rangle}
\def X{\mathbf{x}}
\def Y{\mathbf{y}}
\begin{align*}
\frac{1}{4}(||X+Y||^2 - ||X-Y||^2)
&= \frac{1}{4}(<X+Y,X+Y> - <X-Y,X-Y>) \\
&= \frac{1}{4}(<X+Y,X>+<X+Y,Y>-(<X-Y,X>-<X-Y,Y>)) \\
&= \frac{1}{4}(<X,X>+<X,Y>+<X,Y>+<Y,Y>-(<X,X>-<X,Y>-(<X,Y>-<Y,Y>))) \\
&= \frac{1}{4}(<X,X>+<X,Y>+<X,Y>+<Y,Y>-(<X,X>-<X,Y>-<X,Y>+<Y,Y>)) \\
&= \frac{1}{4}(\cancel{<X,X>}+<X,Y>+<X,Y>+\cancel{<Y,Y>}-\cancel{<X,X>}+<X,Y>+<X,Y>-\cancel{<Y,Y>}) \\
&= \frac{1}{4}(4<X,Y>) \\
&= <X,Y>
\end{align*} 4 1 ( ∣∣ x + y ∣ ∣ 2 − ∣∣ x − y ∣ ∣ 2 ) = 4 1 (⟨ x + y , x + y ⟩ − ⟨ x − y , x − y ⟩) = 4 1 (⟨ x + y , x ⟩ + ⟨ x + y , y ⟩ − (⟨ x − y , x ⟩ − ⟨ x − y , y ⟩)) = 4 1 (⟨ x , x ⟩ + ⟨ x , y ⟩ + ⟨ x , y ⟩ + ⟨ y , y ⟩ − (⟨ x , x ⟩ − ⟨ x , y ⟩ − (⟨ x , y ⟩ − ⟨ y , y ⟩))) = 4 1 (⟨ x , x ⟩ + ⟨ x , y ⟩ + ⟨ x , y ⟩ + ⟨ y , y ⟩ − (⟨ x , x ⟩ − ⟨ x , y ⟩ − ⟨ x , y ⟩ + ⟨ y , y ⟩)) = 4 1 ( ⟨ x , x ⟩ + ⟨ x , y ⟩ + ⟨ x , y ⟩ + ⟨ y , y ⟩ − ⟨ x , x ⟩ + ⟨ x , y ⟩ + ⟨ x , y ⟩ − ⟨ y , y ⟩ ) = 4 1 ( 4 ⟨ x , y ⟩) = ⟨ x , y ⟩
(iii) Show that if u \mathbf{u} u v \mathbf{v} v ∣ ∣ u + v ∣ ∣ 2 = ∣ ∣ u ∣ ∣ 2 + ∣ ∣ v ∣ ∣ 2 . ||\mathbf{u}+\mathbf{v}||^2=||\mathbf{u}||^2+||\mathbf{v}||^2. ∣∣ u + v ∣ ∣ 2 = ∣∣ u ∣ ∣ 2 + ∣∣ v ∣ ∣ 2 .
∣ ∣ u + v ∣ ∣ 2 = ⟨ u + v , u + v ⟩ = ⟨ u + v , u ⟩ + ⟨ u + v , v ⟩ = ⟨ u , u ⟩ + ⟨ v , u ⟩ + ⟨ u , v ⟩ + ⟨ v , v ⟩ ∵ u ⊥ v = ∣ ∣ u ∣ ∣ 2 + ∣ ∣ v ∣ ∣ 2 \begin{align*}
||\mathbf{u}+\mathbf{v}||^2
&= \langle\mathbf{u}+\mathbf{v}, \mathbf{u}+\mathbf{v}\rangle \\
&= \langle\mathbf{u}+\mathbf{v}, \mathbf{u}\rangle
+ \langle\mathbf{u}+\mathbf{v}, \mathbf{v}\rangle \\
&= \langle\mathbf{u},\mathbf{u}\rangle
\;\cancel{+\;\langle\mathbf{v},\mathbf{u}\rangle + \langle\mathbf{u},\mathbf{v}\rangle}
+ \langle \mathbf{v},\mathbf{v}\rangle
&\quad\boxed{\because\mathbf{u}\perp \mathbf{v}}\\
&= ||\mathbf{u}||^2+||\mathbf{v}||^2
\end{align*} ∣∣ u + v ∣ ∣ 2 = ⟨ u + v , u + v ⟩ = ⟨ u + v , u ⟩ + ⟨ u + v , v ⟩ = ⟨ u , u ⟩ + ⟨ v , u ⟩ + ⟨ u , v ⟩ + ⟨ v , v ⟩ = ∣∣ u ∣ ∣ 2 + ∣∣ v ∣ ∣ 2 ∵ u ⊥ v