# I Basis, row and column space

1. Feb 23, 2017

### Lord Anoobis

Hello there. I'm currently trying to come to terms with the aforementioned topics. As I am self studying, a full understanding of these concepts escapes me. There's something I'm not grasping here and I would like to discuss these to clear away the clouds.

As I understand it, a basis for some vector space $R^n$ can be taken to be any set of $n$ linearly independent vectors, in other words an alternate set of axes as opposed to the usual $(1,0,...,n),(0,1,0,...,n)$ , etc. Correct?

So, in a sense it can be thought of as analogous to a Galilean transformation for relative motion?

Last edited by a moderator: Feb 23, 2017
2. Feb 23, 2017

### Lord Anoobis

Heading to work, will continue later or tomorrow.

3. Feb 23, 2017

### mathman

Your notion of basis is correct.

Galilean transformation?

4. Feb 24, 2017

### Lord Anoobis

Yes, the transformation between the coordinates of two different frames of reference. Something that shows up in Newtonian mechanics.

5. Feb 24, 2017

### Lord Anoobis

What I'm not too clear on is basis when it comes to matrices. Take the following 3 x 2 matrix as an example.
$A = \begin{bmatrix} 1 & 3 \\ 1 & 1 \\ 3 & 1 \end{bmatrix}$
Keeping in mind what was stated before, am I correct that:
a) A basis for $A$ is a set of 3 x 2 matrices which are not row equivalent
b) A basis for the row space of $A$ is a set 2 of linearly independent row vectors which is ${[1, 1], [0, 1]}$
c) A basis for the column space of $A$ is the two column vectors since neither is a scalar multiple of the other?

Concerning point (c), does this mean that the column space of $A$ is a plane through the origin since it must be a subspace of $R^3$?

6. Feb 27, 2017

### Staff: Mentor

A better way to say it is a set of 3 x 2 matrices that are linearly independent.
It doesn't have to be those two vectors. Any two linearly independent vectors in $\mathbb{R}^2$ will do.
Those two vectors (the column vectors in the matrix above) are linearly independent, and so would work as a basis for the column space of that matrix. It wouldn't have to be those two, though. Any two vectors that lie in the same plane, and that are linearly independent (point in different directions in this case) would also form a basis for the column space.
Yes. The column space is a two-dimensional subspace of $\mathbb{R}^3$.

7. Mar 1, 2017

### Lord Anoobis

Excellent, that clears up that misunderstanding.

$A = \begin{bmatrix} 1 & 3 & -2 & 0 & 2 & 0 \\ 2 & 6 & -5 & -2 & 4 & -3 \\ 0 & 0 & 5 & 10 & 0 & 15 \\ 2& 6 & 0 & 8 & 4 & 18 \end{bmatrix}$

The matrix A above is example 4 from Anton & Rorres in the chapter on row, column and null spaces. the solution shows the three vectors

$v_1 = \begin{bmatrix} -3\\1\\0\\0\\0\\0 \end{bmatrix}, v_2 = \begin{bmatrix} -4\\0\\-2\\1\\0\\0 \end{bmatrix}, v_3 = \begin{bmatrix} -2\\0\\0\\0\\1\\0 \end{bmatrix}$

forming the basis of the null space of A. What I Don't understand here is how do three vectors form the basis of a space in $R^6$. How do three vectors end up spanning the subspace?