A problem of dimension in Vector Space and It‘s nullspace

A easy Problem

This problem from a lemma that is Dimension Counting Theorem

This lemma is easily to be proof by the concept of set of vectors, we cloud regard \(V\) as the set of vectors.

And the question is:

Now, Let‘s proof this problem:

1. To proof \(v\) and \(w\) and \(u\) are independent, we could consider \(x_1v+x_2u+x_3w = 0\) if only \(x_1=x_2=x_3 = 0\)
2. we cloud assume that part \(x\) from \(v\)‘s and \(u\)‘s is in \(V\), so the part from \(w\) is \(-x\).
3. But the problem is, \(x\) is from \(V\), so, \(-x\) must from \(V\), so, \(-x\) must comes from \(u\)
4. Now the combination of give 0 in \(x_1v+x_2u+x_3w = 0\) comes from only \(u\) and \(v\) , because \(x\) and \(-x\) both from \(V\), so coefficients in \(x_1\) and \(x_2\) must all 0, and \(x_3\) also is 0 because of the it from \(w\), and \(w\) is independent.

A Special Condition In Vector Space

This question has a special form in Vector Space and in Matrix,

We could write as \(Ax=0\) and define the Four Fundamental Subspaces of \(A\), we have a hypothesis that \(A\) is \(m \times n\) matrix, and define \(R\) = \(rref(A)\) which comes from Gaussian elimination.

1. The row space is \(C (A^T )\), a subspace of \(R^n\)
2. The column space is \(C(A)\), a subspace of \(R^m\)
3. The nullspace is \(N (A)\), a subspace of \(R^n\)
4. The left nullspace is \(N(A^T)\), a subspace of \(R^m\). This is the nullspace of \(A^T\) .

The most important things we consider is the relationship between rank of \(r\) , \(m\), \(n\) and the dimensions of this four subspace.

We cloud get below conclusions easily:

1. The column space of \(A\) has dimension \(r\). The column rank equals the row rank.
2. But we should pay attention the column space and the row space, the row space of \(A\) and \(R\) are the same, but the column didn‘t.
3. \(A\) has the same nullspace as \(R\) Same dimension \(n - r\) and same basis.
4. The left nullspace of \(A\) (the nullspace of \(A^T\) ) has dimension \(m - r\).

The question we want to interpret is the dimension of nullspace, for \(A\) and \(R\), why it is \(n-r\), we could use the formula above that:

in this formula, \(V\) = row space, \(W\) = nullspace and there are all in \(R^n\). So all we want to proof is \(dim(V \cap W)\) is \(0\), how to proof this?

Assume that a vector \(x_i\) belongs to \(V\) and \(W\) at the same time, then, consider \(Ax=0\), then \(Ax_i = 0\) and

So, \(x_i = 0\) , \(dim(V \cap W) = 0\) ,

A problem of dimension in Vector Space and It's nullspace

原文：https://www.cnblogs.com/wevolf/p/12990685.html