Question

Find the bases for Col A and Nul A, and then state the dimension of these subspaces for the matrix A and an echelon form of A below.

[1 3 6 -1 -12] [1 3 6 -1 -12]
A= [2 7 16 0 -20] - [0 1 4 2 4]
[-3-12 -30 2 44] [0 0 0 1 4]
[3 13 34 0 -40] [0 0 0 0 0]

Answer 1

The dimension of Col A is 3, and the dimension of Nul A is 2.

The matrix A and its reduced row echelon form are as follows:

`A= \left[\begin{array}{ccc}1&3&6&-1&-12\\2&7&16&0&-20\\-3&-12&-30&2&44\\3&13&34&0&-40\end{array}\right]`

`rref(A)= \left[\begin{array}{ccc}1&3&6&-1&-12\\0&1&4&-1&-12\\0&0&0&1&4\\0&0&0&0&0\end{array}\right]`

Now, let's find the bases for Col A (column space) and Nul A (null space).

The columns corresponding to pivot columns in the reduced row echelon form form the basis for Col A.

`Col A= Span = \left[\begin{array}{ccc}1&3&6\\2&7&16\\-3&-12&-30\\3&13&34\end{array}\right]`

To find the basis for Nul A, we solve Ax=0 for the special solutions corresponding to the free variables (columns without pivots). In this case, the fourth and fifth columns don't have pivots.

Let
`x_4 =s` and
`x_5 =t`. Then the equations from the augmented matrix are:

`x_1 +3x_2+6x_3-x_4-12x_5= 0\\x_2+4x_3+2x_4+4x_5=0\\x_4+4x_5=0`

​Solving these equations gives
`x_1 =−3s-6t, x_2 =-4s-8t, x_3 =s, x_4 =s,` and
`x_5 =-t`. Thus, the null space basis is:

`Nul A=Span = \left[\begin{array}{ccc}-3&-6\\-4&-8\\1&0\\1&1\\0&-1\end{array}\right]`

The dimension of Col A is 3, and the dimension of Nul A is 2.