# Subspace matrix, dimension and basis

1. Mar 14, 2010

### caljuice

So an example was the matrix:

$$A = \left(\begin{array}{cccc} a&a+b\\ b&0\\ \end{array} \right)$$ is a subspace of M2x2.

and is the linear combination $$a*\left(\begin{array}{cccc} 1&1\\ 0&0 \end{array} \right)$$ + $$b*\left(\begin{array}{cccc} 0&1\\ 1&0 \end{array} \right)$$

Meaning it has dimension 2. But i'm not sure how it comes to this conclusion.

Dimension means # of vectors in a basis. However, I don't know how to translate this matrix addition in terms of vectors. Is the dimension 2 because there are 2 matrices being added? Or because we can break it down into the linear combination of indepedent vectors v1 =(1,0) v2=(0,1)? Or is it completely something else? thanks.

2. Mar 14, 2010

### CompuChip

The "vector space" you are talking about here, has matrices as its elements, as opposed to the columns of numbers you were used to. (If you want, for this purpose, you can ignore the matrix structure and you can just view them as vectors with 4 entries written in a funny way).

The dimension being 2, just means that you need precisely two numbers to specify a unique element from the subspace. So if we agree on a way to interpret these numbers to create an element of the subspace, I can give you any two numbers and you will obtain a unique element. For example, we can agree to map (a, b) to
$$a\cdot\left(\begin{array}{cccc} 1&1\\ 0&0 \end{array} \right) + b\cdot\left(\begin{array}{cccc} 0&1\\ 1&0 \end{array} \right).$$

This is the most obvious way to do it, however, we can devise different systems (for example, create a complicated expression when I give you a + b and a - b). Yet, it doesn't matter how you do it, you will always need two and exactly two numbers to specify the element. (This is the intuitive interpretation, which is made precise by introducing the concept of a basis).

3. Mar 14, 2010

### Landau

What they're saying, is that the set B=$$\left\{\left(\begin{array}{cccc} 1&1\\ 0&0 \end{array}\right),\left(\begin{array}{cccc} 0&1\\ 1&0 \end{array} \right)\right\}$$ is in fact a basis of A. Since B contains two elements, the dimension of A is 2. Of course, you still have to prove that B is indeed a basis. As demonstrated, every element of A kan be expressed as a linear combination of the asserted basis elements. This shows that B is generating ("complete"). Now you still have to prove that these linear combinations are unique. Equivalently, prove that the two elements of B are linearly independent.