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.

 

[blue_raver22]

I would like to first clarify that the process of finding the rank and basis for a matrix is a fundamental concept in linear algebra and is used in many fields of science, including physics, engineering, and computer science.

Now, let's address the content provided. The rank of a matrix is defined as the maximum number of linearly independent rows or columns in a matrix. In this case, since the third row is a linear combination of the first two rows, the rank is 2.

To find the basis for the column space, we can look at the columns that correspond to the pivot positions in the reduced row echelon form of the matrix. In this case, the first and second columns correspond to the pivot positions, so the basis for the column space would be the vectors (1,2,3) and (0,-1,-1).

Similarly, to find the basis for the row space, we can look at the rows that correspond to the pivot positions in the reduced row echelon form. In this case, the first and second rows correspond to the pivot positions, so the basis for the row space would be the vectors (1,0,1) and (0,1,-1).

It is important to note that the basis for the column space and row space are not unique, as there can be multiple sets of vectors that span the same space. However, the number of vectors in the basis will always be equal to the rank of the matrix.

In conclusion, the rank of the given matrix is 2 and the bases for the column and row spaces are (1,2,3),(0,-1,-1) and (1,0,1),(0,1,-1) respectively. I hope this helps clarify the concept of finding the rank and basis for a matrix.