Finding the Rank and Basis for a matrix

[Logan Land]

Find rank and the bases for column and row spaces of the matrices

1 0 1
2 -1 3
3 -1 4

Now I can see instantly that row 3 is just row 1 + row 2 so it must be dependent. So that means that row 3 will turn to a row of zeros and thus the rank(A)=2

if I reduced matrix A to row echelon it becomes
1 0 1
0 1 -1
0 0 0

so the rank must be 2 correct?

now what I am a little confused on is the bases
since I have reduced row echelon of
1 0 1
0 1 -1
0 0 0
would the bases of column space be (1,2,3),(0,-1,-1)?
and bases of row space be (1,0,1),(0,1,-1)?

 

[Evgeny.Makarov]

Now I can see instantly that row 3 is just row 1 + row 2 so it must be dependent. So that means that row 3 will turn to a row of zeros and thus the rank(A)=2

Yes, since the first two rows are linearly independent.

if I reduced matrix A to row echelon it becomes
1 0 1
0 1 -1
0 0 0

so the rank must be 2 correct?

Correct.

would the bases of column space be (1,2,3),(0,-1,-1)?

Yes, because row operations preserve (in)dependence of columns. Since in the resulting matrix the first two columns are independent, so are the first two columns in the original matrix.

and bases of row space be (1,0,1),(0,1,-1)?

Yes, or the first two rows of the original matrix.

 

[Logan Land]

Yes, or the first two rows of the original matrix.

oh so either can be the basis? the original matrix rows or the row echelon rows?

(1,0,1),(2,-1,3) or (1,0,1),(0,1,-1)

both are row basis?

 

[Evgeny.Makarov]

Yes. Each vector space (over $\Bbb R$ and of positive dimension) has infinitely many bases. Since the first two rows of the original matrix and independent and the third row is expressible through them, the first two rows form a basis.

In general, it is probably better to take independent rows of the echelon form rather than of the original matrix. Even though row operations do not change the row space, they may change which rows are independent. For example, if after some row operations it turns out that 1st, 3rd and 10th rows are independent, it does not follow that 1st, 3rd and 10th rows of the original matrix were independent.