Dimension of vector space (using matrices)

Question

I have to find $\dim W$, where

$$W=\operatorname{span}\bigl\{(3,1,2,1),(1,-1,1,2),(2,3,0,-3),(2,-1,1,2)\bigr\}$$

What is the difference between writing it as:

\begin{bmatrix}3&1&2&1\\ \vdots&\vdots&\vdots&\vdots \\ \vdots&\vdots&\vdots&\vdots \\ \vdots&\vdots&\vdots&\vdots\end{bmatrix}

and

\begin{bmatrix}3&\cdots&\cdots&\cdots\\1&\cdots&\cdots&\cdots\\2&\cdots&\cdots&\cdots\\1&\cdots&\cdots&\cdots\end{bmatrix}

Is there any? Or in finding dimension there is no difference and in other tasks there might be?

There's no difference, as the row rank equals the column rank — GNUSupporter 8964民主女神地下教會, Jan 16 '18 at 11:48
Mhm. But for example when I would like to find a basis, I would have to use row notation, am I right? — James Smith, Jan 16 '18 at 11:51
My habit is to write basis vectors as columns since a certain linear combination of them can solve ([tag:linear-programming]) problems. — GNUSupporter 8964民主女神地下教會, Jan 16 '18 at 11:54

score 1 · Accepted Answer · answered Jan 16 '18 at 13:48

Yes, there are a difference, even both must to give the same result, you have these two algorithms for found a basis given a set (preferable ordered) of vectors:

Row space Algorithm:
- Step 1. Form the matrix A whose rows are the given vectors.
- Step 2. Row reduce A to an echelon form.
- Step 3. Output the nonzero rows of the echelon matrix.
Casting-Out Algorithm:
- Step 1. Form the matrix M whose columns are the given vectors.
- Step 2. Row reduce M to echelon form.
- Step 3. For each column Ck in the echelon matrix without a pivot, delete (cast-out) the vector from the given vectors.
- Step 4. Output the remaining vectors (which correspond to columns with pivots).

user · Answer 2 · 2018-01-16T14:40:52.707

In order to find the dimension there is no difference: you can put the vectors in rows or in columns and operate by rows or columns operation to obtain a Reduced Echelon Form; at the end of the process the number of pivot will give you the dimension of W.

In order to find dimension and basis there are some differences:

Since the row (column) operation don't change the row (colomn) space, if you put the vectors in rows (columns) and reduce to echelon form by row (column) operations, at the end the pivot rows (columns) will give you also a basis.
Otherwise if you put the vectors in columns (rows) and reduce to echelon form by row (column) operations, at the end the pivot columns (rows) corresponding to the original matrix will be the basis vectors for the column (row) space.

score 0 · Answer 3 · answered Jan 16 '18 at 20:23

Since the row rank of a matrix equals its column rank, there’s no difference from the point of view of finding the dimension of the space. However, the bases that you get with the two methods are different, and each one has its uses. When you write the vectors as columns, the pivot columns of the RREF let you pick out a linearly-independent subset of the original set of vectors that spans the same space. When you write them as rows, because row-reduction doesn’t change the row space the non-zero rows of the RREF are themselves a basis for the space spanned by the original set of vectors. These vectors will tend to be sparse, with most of their components either $0$ or $1$, so this basis can be much more convenient that one chosen from the original vectors.

Dimension of vector space (using matrices)

3 Answers3