Let A = [-4 3 -1 -1 -1 1 0 -1 -4 2 -2 2 2 0 1 0] (i) The dimension of the row sp

Question

Let A = [-4 3 -1 -1 -1 1 0 -1 -4 2 -2 2 2 0 1 0] (i) The dimension of the row space of A is 0 1 2 3 4 The dimension of the column space of A is 1 2 3 4 The dimension of the null space of A is 0 1 2 3 4 Make your choices and press the "Submit Answers" button. You'll have to solve (i) correctly before you can attempt part (ii). (ii) Determine a basis for the row space: (You'll have to solve part (i) correctly before you can attempt part (ii)). Determine a basis for the column space: (You'll have to solve part (i) correctly before you can attempt part (ii)) Determine a basis for the null space: (You'll have to solve part (i) correctly before you can attempt part (ii))

Explanation / Answer

Row Space of A

First, we must convert the matrix to reduced row echelon form:

Divide row1 by -4

Add (-3 * row1) to row2

Add (1 * row1) to row3

Add (1 * row1) to row4

Divide row2 by 1/4

Add (-1/4 * row2) to row3

Add (3/4 * row2) to row4

Divide row3 by -1

Add (-4 * row3) to row4

Add (-6 * row3) to row2

Add (1/2 * row3) to row1

Add (-1/4 * row2) to row1

Dimension =3

Column space of A

Add (3/4 * row1) to row2

Add (-1/4 * row1) to row3

Add (-1/4 * row1) to row4

Add (-1 * row2) to row3

Add (3 * row2) to row4

Add (4 * row3) to row4

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 yellow)
Because we have found pivots in columns 0, 1 and 3. 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:

where A, B and C are any real numbers.

Null Pace of A

First, let's put our matrix in Reduced Row Eschelon Form...

Divide row1 by -4

Add (-3 * row1) to row2

Add (1 * row1) to row3

Add (1 * row1) to row4

Divide row2 by 1/4

Add (-1/4 * row2) to row3

Add (3/4 * row2) to row4

Divide row3 by -1

Add (-4 * row3) to row4

Add (-6 * row3) to row2

Add (1/2 * row3) to row1

Add (-1/4 * row2) to row1

The matrix has 3 pivot columns (hilighted in yellow) and 1 free column; 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 yellow)...

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 this column vector:

1 1/4 1 -1/2 3 1 2 0 -1 0 -2 1 -1 -1 2 0

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