University Linear Algebra 2: Have I done this proof correctly?

Question

Proof: Show that columns x and y are orthogonal in Rn if and only if ||x+y|| = ||x-y||

Answer: 
Let x=(x1, x2,...,xn) and y = (y1, y2,...,yn).
columns x and y are orthogonal in Rn when x(dot product)y = 0
when x(dot)y=0, ||x+y|| = ||x||+||y|| and ||x-y|| = ||x||+||-y||=||x||+||y||

Therefore, when x and y are orthogonal in Rn, ||x+y||=||x-y||

Have I explained everything I need to? Is there a way to prove that ||x+y||=||x||+||y|| and ||x-y||=||x||+||-y|| when x(dot)y=0?

? · Accepted Answer

The proof is flawed, because ||x+y|| = ||x||+||y|| is in general not true when x and y are orthogonal (perpendicular).  For example, if x=(1,0) and y=(0,1), then x dot y = 0, but ||x+y|| = ||(1,1)|| = sqrt(2) whereas ||x||+||y|| = ||(1,0)||+||(0,1)|| = 1+1 = 2.

Instead, use the fact that the square of the magnitude of a vector is the dot product of a vector with itself, and use the fact that dot product is distributive over addition (so that taking a dot product of a sum or difference with another sum or difference is just like multiplying out polynomials).

||x+y|| = ||x-y|| 
if and only if
||x+y||^2 = ||x-y||^2, since magnitudes of vectors are never negative
if and only if
(x+y) dot (x+y) = (x-y) dot (x-y)
if and only if
(x dot x) + 2(x dot y) + (y dot y) = (x dot x) - 2(x dot y) + (y dot y)
if and only if
4(x dot y) = 0, since rearranging terms is a reversible step
if and only if
x dot y = 0, since dividing by a nonzero number is a reversible step
if and only if
x and y are orthogonal.

Lord bless you today!

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University Linear Algebra 2: Have I done this proof correctly?

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