Consider the vector space V Span a b c d where a 1 2 1 3 b

Consider the vector space V = Span {a, b, c, d} where a = (1, 2, -1, 3) b = (-2, -4, 2, -6) c = (1, 3, -4, 5) d = (1, 0, 5, -1) Find a basis set B for V Express the vector x = a + 3b - 2c - d as a linear combination of elements of the basis B Consider the matrix M [2 1 -2 0 -2 3 3 1 4 -2 1 3 2 3 -3 -2 -2 -3 -5 -1] What is the dimension of its row space? Examine the reduced row echelon form of this matrix to find the basis for the row space of M What is the dimension of its column space ? Examine the reduced row echelon form of this matrix to find a basis for the column space of M What is the dimension of the null space of this matrix. Find a basis for the null space of M How many distinct, independent equations does the system m_x = 0 represent, for m as above? How many variables are there? How many parameters are in the general solution.

Solution

Solution : ( 3 )

( a )

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

Divide row1 by 2

Add (-1 * row1) to row2

Add (2 * row1) to row3

Add (2 * row1) to row5

Divide row2 by 3/2

Add (-4 * row2) to row3

Add (-4 * row2) to row4

Add (-1 * row2) to row5

Divide row3 by -11/3

Add (11/3 * row3) to row4

Add (11/3 * row3) to row5

Add (-5/3 * row3) to row2

Add (-1/2 * row3) to row1

Add (-3/2 * row2) to row1

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.  

The rows highlighted below in BOLD comprise a Basis for the Row Space of our matrix:

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( b )

Add (-1/2 * row1) to row2

Add (1 * row1) to row3

Add (1 * row1) to row5

Add (-8/3 * row2) to row3

Add (-8/3 * row2) to row4

Add (-2/3 * row2) to row5

Add (-1 * row3) to row4

Add (-1 * 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, 1 and 2. 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.

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( c )

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

Divide row1 by 2

Add (-1 * row1) to row2

Add (2 * row1) to row3

Add (2 * row1) to row5

Divide row2 by 3/2

Add (-4 * row2) to row3

Add (-4 * row2) to row4

Add (-1 * row2) to row5

Divide row3 by -11/3

Add (11/3 * row3) to row4

Add (11/3 * row3) to row5

Add (-5/3 * row3) to row2

Add (-1/2 * row3) to row1

Add (-3/2 * row2) to row1

The matrix has 3 pivot columns (hilighted in BOLD) 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 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 this column vector :

1	3/2	1/2	-1
1	3	3	-2
-2	1	2	-3
0	4	3	-5
-2	-2	-3	-1