Determinant of a unit columns matrix

In summary: So ##\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-1)##. 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.
  • #1
hedipaldi
210
0
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
 
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  • #2
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.
 
  • #3
Yes,i know that but i am searching for a proof.
 
  • #4
Think about the QR decomposition and see if the property of the unit vectors remains true for the Q and R matrices.
 
  • #5
Office_Shredder said:
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##.
 
  • #6
You mean to decomposite the euclidean space to direct sum ,using torthogonal space of each vector?
If not,can you specify more?
 
  • #7
The volume of a parallelepiped can be found as follows: if the edges are v1, v2,...,vn:

||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,.
 
  • #8
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##.
 

What is the determinant of a unit columns matrix?

The determinant of a unit columns matrix is always equal to 1. This is because a unit columns matrix has only 1's as its diagonal elements and 0's everywhere else, resulting in a determinant value of 1.

How do you calculate the determinant of a unit columns matrix?

To calculate the determinant of a unit columns matrix, you can use the cofactor expansion method or the Gaussian elimination method. In both methods, the determinant value will always be 1.

What is the significance of the determinant of a unit columns matrix?

The determinant of a unit columns matrix is important in linear algebra as it represents the volume scaling factor of the matrix. It also plays a crucial role in solving systems of linear equations and finding the inverse of a matrix.

Can a unit columns matrix have a determinant value other than 1?

No, a unit columns matrix will always have a determinant value of 1. This is because the matrix has linearly independent columns, which means that its determinant cannot be equal to 0, and since it is a square matrix, it cannot be negative.

What happens to the determinant of a unit columns matrix when a row or column is multiplied by a constant?

The determinant of a unit columns matrix will also be multiplied by the same constant when a row or column is multiplied by a constant. This is because the determinant is a linear function of its rows or columns and any constant factor can be pulled out of the determinant calculation.

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