00:01
The objective is to find the basis of four fundamental subspaces of 3 by 5 matrix, which is given in the question.
00:12
Let's consider the given matrix as a.
00:15
So to compute the first basis, let's consider the subspace, column space of a.
00:23
So column space of given matrix a is equal to the set 0 -00 -0 -0 -0 -0 -0 -0 -0 -0 -0.
00:40
First, we have to find the set that spans the column space of a.
00:48
As the elements of the column space of a are 0 matrices, what we can see is the set 0 -0 spans this set but this set is not linearly independent so from here what we can say is basis of column space of a is equal to empty set so the dimension of basis of column space of a is equal to zero now let's go to the second basis for second basis let's consider the set column space of a transpose as a is a 3 by 5 matrix a transpose will be a 5 by 3 matrix so the sets that spans the column space of a transpose will be equal to 0 0 0 0 0 and similarly here also will have basis of column space of a transpose to be equal to an empty set and the dimension of column space of a transpose is equal to 0.
02:29
And for the third basis, let's consider the subset null space of a.
02:39
That is, we have to find the elements that satisfies the equation ax is equal to 0.
02:46
As a is a 3 by 5 matrix, the set of all.
02:50
X that satisfies this equation will be of the matrix form 5 by 1.
02:59
Understand that the set of all x belongs to r5 satisfies this equation.
03:08
So the basis of null space of a will be equal to 1 .000000, 0100 ,0100, 01 ,000, 0 ,000, 0 ,000, 0 ,000, 0 ,000, 0, 1 ,00, 0 ,000, 0 ,000 and 0 ,000, 1 .1...