Suppose the first 2 columns b1 and b2 of B are equal. What can be said about the

Question

Suppose the first 2 columns b1 and b2 of B are equal. What can be said about the columns of AB? Why? Suppose that the first 2 columns of B are linearly independent. What can be said about the first 2 columns of AB? Why? Prove or disprove: If the columns of B Rn times P are linearly dependent then so are the columns of AB (for any A Rm times n). Prove or disprove: If the columns of B Rn times p are linearly independent then so are the columns of AB (for any A Rm times n). Prove or disprove: If the columns of B Rn times p are linearly independent as well as those of A. then so are the columns of AB (for A Rm times n).

Explanation / Answer

a) the first 2 columns of AB would also be equal AB(,2) = A*B(,2) =A*(B(,1)) = AB(,1)
b) let them be denoted by x1 and x2 , if they are independent
A*x1 and A*x2 are the 2 columns of AB
we know that k1*(A*x1) + k2*(A*x2) = 0 ==> A*((k1*x1) + k2*x2) = 0
if i can find k1 and k2, scalars s.t (k1*x1+ k2*x2) belongs to the null space of A, then they are not linearly independent
c) True, similar to (a) , consider one column of B can be written as a linear combination of other columns of B, we can always see that corresondingly , the respective column of AB can also be written in terms of the respective columns of B
d) disprove
consider a 2x2 matrix A whose 2 columns are (1,0) and (1,0)
then B has two linearly independent columns (1,0)and (0,1)
but AB = A and hence it does not have columns independent
e)if columns of A and B are linearly independent , then the columns of AB are also linearly independent
we can see that consider a linear combination of columns of AB =0 ==> A(linear combination of columns of B) =0 since A has full column rank
==> linear combination of columns of B = 0 but since B is also full column rank
==> all the coefficients are equal to 0 ==> columns of AB are linearly independent

for any doubts you can comment back

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