How to prove the Pythagoras theorem using vectors

I've got a question concerning how to proof the Pythagoras theorem using the following assumption:

$x$ is perpendicular to $y$ (if and only if) $||x+y||^2 = ||x||^2 + ||y||^2$, where $x$ and $y$ are vectors.

I have a basic understanding of linear algebra, however I'm a beginner with this. The question provides hints how to prove the above mentioned equation.

Namely that I should use the properties of the dot product and the definition of the norm of a vector. Those being symmetry, scaling and distributivity as the dot product properties and the norm of a vector being the squared root of the dot product between the same vector.

I was thinking about using the fact that if a vector is perpendicular to another vector the dot product between those vectors should be 0. But that is not provided as a hint so I'm not sure. I know the under lying thought behind it is the cosine rule for vectors, that being:

$$x\cdot y = ||x||\,||y|| \cos(\theta)$$

If the angle between the two vectors is perpendicular you should use $\cos(\pi/2)$ which is $0$ and $||x||\cdot 0 = 0$ and $||y||\cdot 0 = 0$ with the vectors not necessarily being $0$. Thus $x\cdot y = 0$. How would I apply this to the equation I first mentioned to prove the Pythagoras theorem?

I have a few more thoughts on how I could prove this but I'm not sure if they're correct. I hope someone could point me in the right direction.

$x,y$ are perpendicular if and only if $x\cdot y=0$. Now, $||x+y||^2=(x+y)\cdot (x+y)=(x\cdot x)+(x\cdot y)+(y\cdot x)+(y\cdot y)$. The middle two terms are zero if and only if $x,y$ are perpendiculat. So, $||x+y||^2=(x\cdot x)+(y\cdot y)=||x||^2+||y||^2$ if and only if $x,y$ are perpendicular.

• Ok now I get it. Using distributivity on the left side I'd get (x.(x+y))+(y.(x+y)) right? Then using distributivity again I'd get (x.x)+(x.y)+(y.x)+(y.y) and because the x.y=0 as I mentioned that results in (x.x)+(y.y) which is the same as the right side of the equation. Can I mark two answers as the answer? Because both helped me realize how to get the answer. Thanks! – Rope Nov 23 '10 at 14:44

The definition of $||x||$ for vectors is: $$||x|| = \sqrt{x\cdot x}.$$

So, you have that \begin{align*} ||x+y||^2 &= (x+y)\cdot(x+y) &\text{(by definition)}\\ &= x\cdot x + x\cdot y + y\cdot x + y\cdot y &\text{(by distributivity)}\\ &= x\cdot x + y\cdot y + 2(x\cdot y) &\text{(by symmetry)}\\ &= ||x||^2 + ||y||^2 + 2(x\cdot y) &\text{(by definition)}\\ &= ||x||^2 + ||y||^2 + 2||x||\,||y||\cos(\theta), \end{align*} where $\theta$ is the angle between $x$ and $y$.

This holds in any case.

So, $||x+y||^2 = ||x||^2 + ||y||^2$ if and only if $2||x||\,||y||\cos(\theta)=0$. One possibility is $||x||=0$; another is $||y||=0$; and the final one is $\cos(\theta)=0$. So the equality holds if and only if one of the following happens:

1. $||x||=0$;
2. $||y||=0$; or
3. $\cos(\theta)=0$ where $\theta$ is the angle between $x$ and $y$, $x\neq 0$, $y\neq 0$.
• "angle", Arturo, "angle". :) – J. M. is a poor mathematician Nov 23 '10 at 15:37
• @J.M. If the devil is in the details, can't the angels be in between the vectors? Thanks. – Arturo Magidin Nov 23 '10 at 15:57

A mistake of a higher order is sometimes committed by more advanced students of mathematics who went beyond trigonometry and ventured into the multidimensional geometry. In multidimensional spaces whose elements are vectors, one often defines what is known as the scalar product and then also an angle between two vectors. Say, for two vectors a and b, if the scalar product is denoted a·b, then the angle γ between the two is defined via the cosine function as in:

cos γ = a·b / ||a|| ||b||, where ||a|| is the norm of vector a: ||a||² = a·a and similarly for ||b||.

For γ = 90°, it follows from the properties of the scalar product that

||a - b||² = ||a||² + ||b||², which, in a 2-dimensional case, is easily seen to express the common Pythagorean theorem.

Critique

The problem with this derivation is that the theory of vector spaces is entirely algebraic (or analytic, if you will). Vectors are defined as n-tuples of real numbers which is followed by the componentwise definition of vector operations. Within this framework, the vector Pythagorean identity above is indeed an easy consequence of the axioms and definitions. However, the relationship between the common geometry and the geometry of vector spaces is that of a model and an abstract theory. The above vector identity does not prove the Pythagorean theorem. It only shows that there is a tight relation between the model and the theory. It confirms this relation, perhaps offers an additional insight into the Pythagorean theorem, but does not prove it by any means.

Credit goes to: http://www.cut-the-knot.org/pythagoras/FalseProofs.shtml

If I understood the question correctly, you have to prove that $$\|x+y\|^2 = \|x\|^2 + \|y\|^2$$ if and only if $x$ and $y$ are perpendicular. That is, as you noted, $x \cdot y = 0$.

Hint: You can use the following formulas: $$\|x\|^2 = x \cdot x$$ $$(x+y) \cdot z = x \cdot z + y \cdot z$$

• Yes I have to prove that equation because, it is stated that it is the Pythagoras theorem expressed in vectors on the symbolic level. I started out with exactly what you hinted about. But how do I use the fact that x and y have to perpendicular? Or do I not use it at all and just go solving the equation by pattern matching the equation? – Rope Nov 23 '10 at 14:17
• There is not much to do after expanding $\|x+y\|^2$ using the dot product and those two rules I gave you. Then you just notice that the result is equal to $\|x\|^2 + \|y\|^2$ if and only if $x \cdot y = 0$. – J. J. Nov 23 '10 at 14:24
• If I were to use the distributivity rule you gave as a hint wouldn't that result in (x.(x+y))+(y.(x+y))? If I use it on the left side of the equation? – Rope Nov 23 '10 at 14:35
• Yes, and then you can use it again (well, strictly speaking you need to know that distributivity works for the second argument as well) to get $x \cdot x + x \cdot y + y \cdot x + y \cdot y = \|x\|^2 + x \cdot y + y \cdot x + \|y\|^2$. – J. J. Nov 23 '10 at 15:57