Basis of linear subspace

In summary: But points can also be vectors. So the vector space R^3 consists of all points in three dimensional space that are not collinear.You can find a basis for a subspace S by solving a system of linear equations. Linear systems are easy to solve and so this is a quick way to find a basis.
  • #1
I have a given point (vector) P in R^3 and a 2-dimensional linear subspace S (a plane) which consists of all elements of R^3 orthogonal to P.
The point P itself is element of S.

So I can write

P' ( x - P ) = 0

to characterize all such points x in R^3 orthogonal to P. P' means the transpose of P.

My problem is to find a basis of S. This basis should depend on point P.

I tried to find such a basis (alpha,beta) using the parameter form of the plane

x = P + alpha u + beta v

but I am unable to find two vectors u and v orthogonal to P.

I expect that this problem should be easy but I am nevertheless unable to solve it :(

Please help me a bit.
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  • #2
Technically a "point" is not a "vector" but, yes, given a point, $p= (x_0, y_0, z_0)$ we can assign to to it the vector from (0, 0 0) to p, $<x_0, y_0, z_0>$. If <x, y, z> is orthogonal to that vector then $xx_0+ yy_0+ zz_0= 0$ so, solving for z, $z= \frac{xx_0+ yy_0}{z_0}$ (this is assuming $z_0\ne 0$. If it is choose to solve for x or y instead. If all three of $x_0$, $y_0$, and $z_0$, this is the zero vector and all vectors are orthogonal to it.)

Letting x= 0, $\left<0, y, \frac{y_0}{z_0}y\right>$ and taking y= 0, $\left<x, 0, \frac{x_0}{z_0}x\right>$.

Those two vectors are orthogonal to p. You should be able to show that they are independent and so a basis for the subspace. Choosing a value for y in the first and a value for x in the second gives vectors of specific length. They can be chosen so that the vectors have length one.
  • #3
Thanks for your reply - it helped me out of my blockade.
I was too fixed at the implicit characterisation of orthogonal points to P by the equation

P' ( x - P ) = 0.

You are right, exept a typo in the sign for z in your formula.

Points or vectors, that is an old discussion. "Vectors" are often used in physics and "points" are often used in math.
As you know in math a "vector space" consists of some set with an additional algebraic structure. His elements (for example real numbers
complex numbers or integrable functions) are often called points.

1. What is a linear subspace?

A linear subspace is a subset of a vector space that is closed under addition and scalar multiplication. This means that if you take any two vectors from the subspace and add them together or multiply them by a scalar, the result will still be in the subspace.

2. How is a linear subspace different from a vector space?

A vector space is a set of vectors that can be added and multiplied by scalars, while a linear subspace is a subset of a vector space that also has these properties. In other words, a linear subspace is a smaller, self-contained version of a vector space.

3. What is the basis of a linear subspace?

The basis of a linear subspace is a set of linearly independent vectors that span the subspace. This means that any vector in the subspace can be written as a linear combination of the basis vectors.

4. How do you determine the basis of a linear subspace?

To determine the basis of a linear subspace, you can use the process of Gaussian elimination to find a set of linearly independent vectors that span the subspace. Alternatively, you can also use the Gram-Schmidt process to find an orthogonal basis for the subspace.

5. Can a linear subspace have more than one basis?

Yes, a linear subspace can have multiple bases. This is because there can be different sets of linearly independent vectors that span the same subspace. However, all bases for a given subspace will have the same number of vectors, known as the dimension of the subspace.

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