Determining an orthonormal set of basis vectors for the linear space

The following is example C.4 from Appendix C (Linear Spaces Review) of Introduction to Laplace Transforms and Fourier Series, Second Edition, by Phil Dyke:

Example C.4 Determine an orthonormal set of vectors for the linear space that consists of all real linear functions: $$\{a+bx:a,b\in\mathbb{R}\ 0\leq x\leq1\}$$ using an inner product $$\langle f,g\rangle=\int_0^1f g\,dx.$$ Solution The set $\{1,x\}$ forms a basis, but is is not orthogonal. Let $a+bx$ and $c+dx$ be two vectors. In order to be orthogonal we must have $$\langle a+bx,c+dx\rangle=\int_0^1(a+bx)(c+dx)\,dx=0.$$ Performing the elementary integration gives the following condition on the constants $a,b,c$ and $d$ $$ac+\frac{1}{2}(bc+ad)+\frac{1}{3}bd=0.$$ In order to be orthonormal too we also need $$\|a+bc\|=1\,\text{ and }\,\|c+dx\|=1$$ and these give, additionally, $$a^2+b^2=1,\ c^2+d^2=1.$$ There are four unknowns and three equations here, so we can make a convenient choice. Let us set $$a=-b=\frac{1}{\sqrt{2}}$$ which gives $$\frac{1}{\sqrt{2}}(1-x)$$

as one vector. The first equation now gives $3c=-d$ from which $$c=\frac{1}{\sqrt{10}},\ \ d=-\frac{3}{\sqrt{10}}.$$ Hence the set $\{(1-x)/\sqrt{10},(1-3x)/\sqrt{10}\}$ is a possible orthonormal one.
$\$ $\$ Of course there are infinitely many possible orthonormal sets, the above was one simple choice. The next definition follows naturally.

I have the following questions:

1. How do we determine $a^2 + b^2 = 1$ and $c^2 + d^2 = 1$ from $\|a + bx\| = 1$ and $\|c + dx\| = 1$? This seems similar to the norm of a complex number $a + bi$, but we're not dealing with complex numbers in this case, since we're dealing with the space of all real linear functions, so I'm not sure how these are being derived?

2. If we have $a = -b = \dfrac{1}{\sqrt{2}}$ and $3c = -d$, then we have the following: \begin{align} \dfrac{1}{\sqrt{2}}c + \dfrac{1}{2} \left[ \left( \dfrac{-1}{\sqrt{2}} \right)c + \left( \dfrac{1}{\sqrt{2}} \right) (-3c) \right] + \dfrac{1}{3} \left( \dfrac{-1}{\sqrt{2}} \right)(-3c) = 0 \\ \rightarrow \dfrac{c}{\sqrt{2}} - \dfrac{c}{2 \sqrt{2}} - \dfrac{3c}{2\sqrt{2}} + \dfrac{c}{\sqrt{2}} = 0 \\ \rightarrow \dfrac{2c}{\sqrt{2}} - \dfrac{4c}{2\sqrt{2}} = 0 \\ \rightarrow 0 = 0\ ?\end{align}

Have I made an error? Where does the $c = \dfrac{1}{\sqrt{10}}$ and $-\dfrac{3}{\sqrt{10}}$ come from?

I would greatly appreciate it if people could please take the time to clarify these.

EDIT:

The following is proved in the textbook:

Example C.3 Prove that $\|a\|=\sqrt{\langle\mathbf{a}.\mathbf{a}\rangle}\in V$ is indeed a norm for the vector space $V$ with inner product $\langle,\,\rangle$.

Which seems to suggest that $\|x\| = \sqrt{\langle x,x\rangle}$?

• The norm of $a + bx$ is defined to be $$\sqrt{\langle a + bx, a + bx \rangle} = \sqrt{\int_0^1 (a + bx)^2 \, \mathrm{d}x} = \sqrt{a^2 + ab + \frac{b^2}{3}}.$$ Assuming the norm of $a + bx$ is $\sqrt{a^2 + b^2}$ is a really amateur error, especially for a textbook author. Certainly $\sqrt{a^2 + b^2}$ is a norm for the space, but it's certainly not the norm derived from the inner product, and using it won't give you an orthonormal basis with respect to the given inner product. – Theo Bendit Jul 3 '18 at 6:26
• Thanks for the response. To be clear, I'm not saying that the author assumed that the norm of $a + bx$ is $\sqrt{a^2 + b^2}$ -- that's just what it seemed like the author was doing, based on my incomplete understanding of the solution. – The Pointer Jul 3 '18 at 7:13

By definition of norm induced from inner product, $||x||^2 = \langle x,x\rangle$.

That is, if $||a+bx|| = 1$ then $\langle a+bx,a+bx\rangle = 1$ ,so writing this down: $$\int_{0}^1 (a+bx)(a+bx) dx = 1 \implies \int_0^1 (a^2 + 2abx + b^2x^2) dx = 1 \\ \implies a^2 + ab + \frac {b^2}3 = 1$$

therefore, the statement that $a^2 + b^2 = 1$ is FALSE.

This is actually quite clear with an example : $a=0 , b=1$ satisfies $a^2+b^2 = 1$, and gives the polynomial $x$, but $||x||^2 = \int_0^1 x^2 = \frac 13$, so $||x|| \neq 1$. Instead, the other statement given is correct. Replacing $a$ and $b$ by $c$ and $d$ gives you the other analogously correct statement.

Once this happens, you may set $a,b$ to any suitable values, and check what happens to $c$ and $d$.

For example, set $b = 0$ : from the above equation, this forces $a = \pm 1$, we will take $a =1$.

From the equation that $\langle a+bx,c+dx\rangle = 0$ that the author has derived in your question above, substituting (and cancelling $b$) and rearranging gives $2c+d = 0$, so $d = -2c$.

This must be combined with $c^2 + cd + \frac{d^2}{3} = 1$. Setting it, we get $c^2(1 - 2 + \frac 43) = 1$, so $c^2 = 3$. Just take $c = + \sqrt 3$, so $d = -2\sqrt 3$.

In this manner, we may verify that the polynomial $a+bx = 1$ and $c + dx = \sqrt 3(1 - 2x)$ form an orthonormal basis for the space.

• Thanks for the response. I'm reading through your answer carefully, to ensure that I understand everything. First question: Are you sure that the definition of norm induced from inner product is $||x||^2 = \langle x,x\rangle$, or should it be $||x|| = \langle x,x\rangle$? – The Pointer Jul 3 '18 at 7:11
• I am sure that $||x||\mathbf{^2} = \langle x,x\rangle$. Since $||x|| = 1$, this could have been confusing. – астон вілла олоф мэллбэрг Jul 3 '18 at 7:28
• See, there are many sorts of norms that you can put on a space. However, given an inner product, the norm induced by it is defined by the relation that $||x||^2 = \langle x,x\rangle$. This norm, is the one that is usually referred to, in an inner product space. It is possible that there may another norm e.g. $\sqrt {a^2 + b^2}$ but this is not induced from the inner product. Referring to the edit, note that $||a|| = \sqrt{\langle a \cdot a\rangle}$, so squaring both sides, $||a||^2 = \langle a \cdot a\rangle$, which is the same as in my post. – астон вілла олоф мэллбэрг Jul 3 '18 at 7:35
• Yes, the last equation you have written is correct. But the left hand side , we want it to equal to $1$ if $a+bx$ is to be part of an orthonormal basis. Then, we set the right hand side equal to $1$. Now, $1^2 = 1$, so the square root simply vanishes after squaring both sides. – астон вілла олоф мэллбэрг Jul 3 '18 at 9:00
• You are welcome! – астон вілла олоф мэллбэрг Jul 3 '18 at 9:07