Let A 1 3 2 0 0 2 2 6 4 0 0 4 0 0 0 3 3 6 0 0 0 3 1 2 0 0 0

Let A = [1 3 -2 0 0 -2 2 6 -4 0 0 -4 0 0 0 -3 -3 6 0 0 0 -3 1 2 0 0 0 -2 -3 5]. Find a basis for the row space of A, a basis for the column space of A, a basis for the null space of A, the rank of A, and the nullity of A. [1 3 -2 0 0 -2 0 0 0 1 0 -1 0 0 0 0 1 -1 0 0 0 0 0 0 0 0 0 0 0 0].) Row Space basis: Column Space basis: Null Space basis: Rank: 3 Nullity: 2

Solution

Solution :

Row space basis :

Transform the matrix to the reduced row echelon form

Because we have only performed linear operations on rows, the non-zero rows in the reduced row echelon form of the matrix comprise a Basis for the Row Space of the matrix.
(Note that this is not true of the Column Space; the Column Space certainly changes as you perform row operations.)
The rows highlighted below in bold comprise a Basis for the Row Space of our matrix :

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Column space basis :

Add (-2 * row1) to row2

Swapping row3 with row2

Add (-1 * row2) to row4

Add (-2/3 * row2) to row

Swapping row4 with row3

Add (1/4 * row3) to row5

First, we must reduce the matrix so we can calculate the pivots of the matrix (note that we are reducing to row echelon form, not reduced row echelon form):

The matrix has 3 pivots (hilighted above in bold)
Because we have found pivots in columns 0, 3 and 4. We know that these columns in the original matrix define the Column Space of the matrix.
Therefore, the Column Space is given by the following equation :

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Null Space basis :

Transform the matrix to the reduced row echelon form

The matrix has 3 pivot columns (hilighted in bold) and 3 free columns; because the matrix has 3 pivots, the rank of the matrix is 3.

Let\'s take the \'free\' part of the reduced row echelon form matrix (hilighted below in bold)...

and turn it into its own matrix:

Let\'s multiply this matrix by -1 :

Now, we add the Identity Matrix to the rows in our new matrix which correspond to the \'free\' columns in the original matrix, making sure our number of rows equals the number of columns in the original matrix (otherwise, we couldn\'t multiply the original matrix against our new matrix) :

Finally, the Null Space of our matrix is defined by scalar multiples of these column vectors:

The rank of the matrix is 3. It equals the number of leading entries.

The nullity of the matrix is 3. This is the dimension of the null space. It equals the number of columns without leading entries.

1	3	- 2	0	0	-2
0	0	0	1	0	-1
0	0	0	0	1	-1
0	0	0	0	0	0
0	0	0	0	0	0