Linear Algebra | Special Examples of Basis for Row space / Column space (Part 2)

Hello friends in this video we are going to discuss example of bases for row space and column space. These are the particular examples where the algorithm of finding bases for row space and column space we cannot apply directly. The algorithm associated with the bases of row space and column space we discuss in the part one. If you want to learn the algorithm just click on the suggested video. Now example, find a basis for the row space So first we understand the concept consisting entire row vectors that means row vector should remain same in the basis. If we use the usual algorithm row vectors will be changed as a basis of row space but column vectors will remain same as a basis vector of column space. So we can use the property of basis vector of column space which remains same. And for that we can find the basis of column space of A transpose which is the basis of row space of A. That means the column space of A transpose will give entire column vectors of A transpose which are the row vectors of A for the basis of row space of A. That means here we are going to find the basis for the column space of A transpose. First step, Row echelon form of A transpose. So this is A transpose. Now for the row echelon form in the first column except the first entry other two entries should be zero. For that we apply the row operation multiply first row by minus two and add it to the second row and multiply first row by one and add it to the corresponding entries of third row. Now in the second step we multiply second row by two and add it to the corresponding entries of third row to make this entry zero. This 4, 1, we multiply third row by 1 by 4, then we have row echelon form of A transpose. In the second step, we want to find the basis for the column space of A transpose, therefore we have to identify leading ones in the row echelon form of A transpose, which are contained in first column, second column, third column. Corresponding columns C1, C2, C3 will form a basis for column space of A transpose. And the step 3, as we know the basis for row space of A is equal to the basis for column space of A transpose. Therefore, basis for row space of A is 1, 2, minus 1, 0, 1, minus 2, 1, 4, minus 1. These are the three row vectors of A. And therefore, we are able to get the entire row vectors of A, which form a basis for row space of A. Now, example number two, find a basis for the column space of A does not contain all entire column vectors of A. Now, it is completely opposite to the previous example. In the previous example, we want entire row vectors, here we don't want entire column vectors. That means the column vector should be changed in the basis. The usual algorithm. Then the column vectors will remain same as a basis vector of column space but the row vectors will be changed as a basis of row space. So we will use the property of basis of row space where basis vector will be changed and for that we will take A transpose and row space of A transpose is equal to column space of A. Therefore the basis for row space of A transpose is equal to basis for column space Therefore, the basis for row space of A transpose will change the row vectors in a basis of row space of A transpose which are the column vectors of A for the basis of column space of A. That means in the example, we will find the basis for the row space of A transpose. So, first find row echelon form of A transpose. For that, in the first column, other entry should be 0. So, we will try to make the second entry 0. By multiplying 1st row by minus 2 and add it to the corresponding entries of 2nd row then make this entry 1 in the 2nd row by multiplying minus 1 and simultaneously multiply 3rd row by minus 1 in the 3rd column all the entry below the leading one should be 0 so we will try to make this entry 0 by multiplying 3rd row by minus 1 and add it to the corresponding entries of 4th row then we have row echelon form of A transpose. Now, we are going to find the basis for the row space of A transpose. So, non-zero row will form a basis for the row space of A transpose. So, first, second and third row will form a basis for A transpose. But basis for row space of A transpose is nothing but the basis for the column space of A. Therefore, we have column space of A. The first column 1, 1, minus 1. Second column 0, 1, minus 1, 0. Which is not the second column of 801 and 001 minus 1. Therefore this way we are able to change the basis for column space of. Thanks for watching. For more videos like, share, subscribe, press the bell icon of Next Gen Maths.