# Orthogonal vectors and the Pythagorean theorem

• Dec 4th 2009, 10:55 PM
Mollier
Orthogonal vectors and the Pythagorean theorem
Problem statement:
The Pythagorean theorem asserts that for a set of $n$ orthogonal vectors ${x_i}$,
$||\sum^{n}_{i=1}x_i||^2 = \sum^{n}_{i=1}||x_i||^2$

(a) Prove this in the case $n=2$ by an explicit computation of $||x_1 + x_2||^2$.
(b) Show that this computation also establishes the general case by induction.

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Attempt at solution:
(a)
$||x_1+x_2||^2 = ||x_1||^2+||x_2||^2$
$x^{T}_{1}x_1+2x^{T}_{1}x_2+x^{T}_2x_2 = x^{T}_{1}x_1 + x^{T}_{2}x_2$

Since the dot product of two orthogonal vectors is zero, we have:

$x^{T}_{1}x_1+x^{T}_2x_2 = x^{T}_{1}x_1 + x^{T}_{2}x_2$

(b)
Basis step, $n=1$: $||x_1||^2 = ||x_1||^2$
I assume that this holds up to $k$ for $1 \leq k < n$.
Now I show that it holds for $k+1$:

$||x_1+x_2+\cdots+x_k+x_{k+1}||^2=||x_1||^2+||x_2|| ^2+\cdots+||x_k||^2+||x_{k+1}||^2$

By orthogonality $x_ix_j=0$ for $i \neq j$:

$x^{T}_{1}x_1 + x^{T}_{2}x_2 + \cdots + x^{T}_{k}x_k + x^{T}_{k+1}x_{k+1} = x^{T}_{1}x_1 + x^{T}_{2}x_2 + \cdots + x^{T}_{k}x_k + x^{T}_{k+1}x_{k+1}$

Since the statement is true for n=1, n=k and n=k+1, it is also true for n.
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I do not have much experience with proofs, and would appreciate some help with this.
Thank you.
• Dec 5th 2009, 01:47 AM
Opalg
Quote:

Originally Posted by Mollier
Problem statement:
The Pythagorean theorem asserts that for a set of $n$ orthogonal vectors ${x_i}$,
$||\sum^{n}_{i=1}x_i||^2 = \sum^{n}_{i=1}||x_i||^2$

(a) Prove this in the case $n=2$ by an explicit computation of $||x_1 + x_2||^2$.
(b) Show that this computation also establishes the general case by induction.

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Attempt at solution:
(a)
$||x_1+x_2||^2 = ||x_1||^2+||x_2||^2$
$x^{T}_{1}x_1+2x^{T}_{1}x_2+x^{T}_2x_2 = x^{T}_{1}x_1 + x^{T}_{2}x_2$

Since the dot product of two orthogonal vectors is zero, we have:

$x^{T}_{1}x_1+x^{T}_2x_2 = x^{T}_{1}x_1 + x^{T}_{2}x_2$

(b)
Basis step, $n=1$: $||x_1||^2 = ||x_1||^2$
I assume that this holds up to $k$ for $1 \leq k < n$.
Now I show that it holds for $k+1$:

$||x_1+x_2+\cdots+x_k+x_{k+1}||^2=||x_1||^2+||x_2|| ^2+\cdots+||x_k||^2+||x_{k+1}||^2$

By orthogonality $x_ix_j=0$ for $i \neq j$:

$x^{T}_{1}x_1 + x^{T}_{2}x_2 + \cdots + x^{T}_{k}x_k + x^{T}_{k+1}x_{k+1} = x^{T}_{1}x_1 + x^{T}_{2}x_2 + \cdots + x^{T}_{k}x_k + x^{T}_{k+1}x_{k+1}$

Since the statement is true for n=1, n=k and n=k+1, it is also true for n.

The proof of (a) is correct. But in (b) you have not used the inductive hypothesis. What you should say is something like this:

Let $y = x_1 + \ldots + x_k$. The inductive hypothesis is that $\|y\|^2 = \|x_1\|^2 + \ldots + \|x_k\|^2$. We want to find $\| x_1 + \ldots + x_k + x_{k+1}\|^2 = \|y + x_{k+1}\|^2$. But by (a) this is equal to $\|y\|^2 + \|x_{k+1}\|^2$. Now use the inductive hypothesis to see that this is equal to $\|x_1\|^2 + \ldots + \|x_{k+1}\|^2$. That completes the inductive step.
• Dec 5th 2009, 06:00 AM
Mollier
I see.
My inductive hypothesis is that the results I get in my basis step (n=1),hold for n=k. Then I check that I get the same result when n = k+1.

Thank you very much!