Determinant of a unit columns matrix

[hedipaldi]

If all columns of a matrix are unit vectors, the determinant of the matrix is less or equal 1

I am trying to prove this assertion,which i guess to be true.
can anybody help me?
Thank's in advance

 

[Office_Shredder]

The determinant is equal to the volume of the parallelepiped formed by the columns - so the question reduces to if you have a parallelepiped which has all edges of length 1 or smaller, is its volume no larger than 1. Intuitively this should be true because the largest volume would occur when the edges are orthogonal, depending on what you know it may or may not be easy to prove though.

 

[hedipaldi]

Yes,i know that but i am searching for a proof.

 

[R136a1]

Think about the QR decomposition and see if the property of the unit vectors remains true for the Q and R matrices.

 

[AlephZero]

so the question reduces to if you have a parallelepiped which has all edges of length 1 or smaller, is its volume no larger than 1.

The proof of that by Euclidean geometry is easy. In 2D, you have the area of a parallelogram with pairs of edges of length a <= 1 and b <= 1. The area is equal to a rectangle with base a and the height h <= b.

The same argument works to transform an n-dimensional parallepiped into an n-dimensional cuboid, so just translate it into vector algebra. It will depend on the triangle inequality, which is equivalent to the fact that $|\cos\theta| <= 1$.

 

[hedipaldi]

You mean to decomposite the euclidean space to direct sum ,using torthogonal space of each vector?
If not,can you specify more?

 

[Office_Shredder]

The volume of a parallelepiped can be found as follows: if the edges are v₁, v₂,...,v_n:

||v₁||*|| the component of v₂ orthogonal to v₁||* || the component of v₃ orthogonal to span(v₁,v₂||...

the orthogonal component always has a norm smaller than the original vector, so we are multiplying a bunch of numbers that are all no larger than 1,.

 

[AlephZero]

Translating into determinant notation, if the columns of the matrix are $v_1, \dots, v_n$, let $v_2 = av_1 + v'_2$ where $v'_2$ is orthogonal to $v_1$.
Then $\det(v_1,v_2,.\dots,v_n) = a\det(v_1,v_1,\dots,v_n) + \det(v_1,v'_2,\dots,v_n)$, $\det(v_1,v_1,\dots,v_n) = 0$ because two columns are identical, and $||v'_2|| \le ||v_2||$.

Repeat for columns $3,\dots,n$.