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Let the basis $B = \{1,x,x^2\}$ which is orthogonal.

Now, I've seen the following:

$$\|1\| = \sqrt {\langle 1,1\rangle} = \sqrt {4\cdot 1\cdot 1} = 2 $$ $$\|x\| = \sqrt {\langle x,x\rangle} = \sqrt {2\cdot 1\cdot 1} = \sqrt 2 $$ $$\|x^2\| = \sqrt {\langle x^2,x^2\rangle} = \sqrt {1\cdot 1} = \sqrt 1 = 1 $$

And therefore, an orthonormal basis would be: $$B = \{ \frac{1}{2}, \frac{x}{\sqrt 2}, x^2 \}$$

Questions:

  1. Isn't $\{1,x,x^2\}$ already orthonormal?
  2. Isn't the calculation of the norm wrong?

EDIT:
The inner product (for $V=\mathbb{R}_2[x]$) is:
$$\langle a_1 + b_1x + c_1x^2, a_2 + b_2x + c_2x^2\rangle = 4a_1a_2 + 2b_1b_2 + c_1c_2$$

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    $\begingroup$ How is the inner product defined? $\endgroup$ – ajotatxe Jul 27 '15 at 11:06
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    $\begingroup$ Let me edit please $\endgroup$ – jmiller Jul 27 '15 at 11:14
  • $\begingroup$ An orthonormal base means all its elements have norm $1$ and are pairwiseorthogonal. $\endgroup$ – 5xum Jul 27 '15 at 11:18
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    $\begingroup$ @jmiller are you sure you've written the definition of inner product right? I think it should be $\langle a_1 + b_1x + c_1x^2, a_2 + b_2x + c_2x^2\rangle = 4a_1a_2 + 2b_1b_2 + c_1c_2$ $\endgroup$ – Sepideh Abadpour Jul 27 '15 at 11:26
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    $\begingroup$ @5xum, for example, why is it:$$ \|1\| = \sqrt {\langle 1,1\rangle} = \sqrt {4\cdot 1\cdot 1} = 2$$. Where's the $4$ came from? it's unclear to me. $\endgroup$ – jmiller Jul 27 '15 at 11:29
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the definition $\langle a_1 + b_1x + c_1x^2, a_2 + b_2x + c_2x^2\rangle = 4a_1a_2 + 2b_1b_2 + c_1c_2$ is just a definition provided in the problem.For each inner product space we should define the inner product and then define normality and orthogonality based on the definition of the inner product.
So don't say where 2 or 4 comes from. That's just a definition.
For example the common definition of inner product in 3-D vector space is: $\langle (x_1,y_1,z_1),(x_2,y_2,z_2)\rangle=x_1x_2+y_1y_2+z_1z_2$
but I can encounter with a 3-D vector space that has another definition for inner product:

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  • $\begingroup$ Oh I see now.. Thank you! $\endgroup$ – jmiller Jul 27 '15 at 11:49
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    $\begingroup$ An inner product is required to be positive-definite, but $\langle (x_1,y_1,z_1),(x_2,y_2,z_2)\rangle=x_1y_2+y_1x_2+z_1z_2$ isn't. For example $\langle (-1,1,0), (-1,1,0) \rangle = -2 < 0$. $\endgroup$ – A.P. Jul 27 '15 at 12:17
  • $\begingroup$ ok @A.P. it was just an example and I only wanted to state that the inner product should be defined in the problem. and it's not always the same as the common definition you can edit my answer and add positive-definite example $\endgroup$ – Sepideh Abadpour Jul 27 '15 at 12:20
  • $\begingroup$ $\langle (x_1,y_1,z_1),(x_2,y_2,z_2)\rangle=|x_1y_2+y_1x_2+z_1z_2|$ still isn't positive definite because $\langle (1,0,0), (1,0,0) \rangle = 0$ even though $(1,0,0) \neq (0,0,0)$. I added to my answer a characterisation for inner products on a finite-dimensional real vector spaces. You can use it to make up as many examples as you wish... :P $\endgroup$ – A.P. Jul 27 '15 at 14:54
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The computation of the norm is indeed correct, given the inner product you described.

The vectors in $\{1,x,x^2\}$ are easily seen to be orthogonal, but they cannot form an orthonormal basis because they don't have norm $1$. On the other hand, the vectors in $$ \left\{ \frac{1}{\|1\|}, \frac{x}{\|x\|}, \frac{x^2}{\|x^2\|} \right\} = \left\{ \frac{1}{2}, \frac{x}{\sqrt{2}}, x^2 \right\} $$ have norm $1$ and are orthogonal, so indeed they form an orthonormal basis.


Recall that an inner product on a real vector space $V$ is just a function $$ \langle \cdot, \cdot \rangle \colon V \times V \to \Bbb{R} $$ which for every $x,y,z \in V$ and $a,b \in \Bbb{R}$ satisfies:

  • $\langle x,y \rangle = \langle y,x \rangle$
  • $\langle ax + by, z \rangle = a\langle x,z \rangle + b\langle y,z \rangle$
  • $\langle x,x \rangle \geq 0$ and $\langle x,x \rangle = 0$ iff $x = 0$

Given an inner product on $V$, one can define the associated norm $\| x \| = \sqrt{\langle x,x \rangle}$.

Furthermore, one can prove that if $\dim_{\Bbb{R}} V = n$ is finite, then every inner product on $V$ is of the form $$ \langle x,y \rangle_A = x^T A y $$ where $A$ is the symmetric positive-definite $n \times n$ matrix with entries $a_{ij} = \langle e_i,e_j \rangle_A$.

For example the Euclidean (i.e. usual) inner product corresponds to the identity matrix, while the matrix $$ A = \begin{bmatrix} 2 & -1 & 0 \\ -1 & 2 & -1 \\ 0 & -1 & 2 \end{bmatrix} $$ corresponds to the inner product $$ \langle x,y \rangle_A = 2x_1y_1 - x_2y_1 - x_1y_2 + 2x_2y_2 - x_3y_2 - x_2y_3 + 2x_3y_3 $$ (checking that this is indeed an inner product is a good exercise).

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  • $\begingroup$ Fun fact: for any real invertible matrix $A$ the matrix $A^TA$ is positive-definite. $\endgroup$ – A.P. Jul 27 '15 at 14:56

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