- #1
xalvyn
- 17
- 0
Hihi
I've been working on this problem for some time: if A is a (m x n) matrix, and A' denotes its transpose, then the null space of A is equal to the null space of A'A. Is this always true? I thought of a proof for the special case where A is given in reduced row echelon form, but fail to see how the general case can be proved or disproved. My proof runs like this:
1) Suppose A is in reduced row echelon form. Then, every column of A either i) contains a single 1 element, while the other entries are 0, or ii) its first i rows consist of elements which are not 1 and not necessarily 0, where i < column number.
2) Now take A' and multiply it by A. Then we have i) every row of A is contained in A'A, and ii) every row of A'A is a linear combination of rows from A.
3) Thus if A'A(X) = 0, then A(X) = 0. Similarly, if A(X) = 0, then A'A(X) = 0. Hence the null spaces of A and of A'A are the same.
There is also the weaker general result: the null space of A is a subset of the null space of A'A, because if A(X) = 0, then A'A(X) = A'(0) = 0.
i think the solution to the problem lies in the column space of A, but I'm not sure exactly how. Will be really grateful to anyone who can share his/her insight on this problem. Thanks. :)
I've been working on this problem for some time: if A is a (m x n) matrix, and A' denotes its transpose, then the null space of A is equal to the null space of A'A. Is this always true? I thought of a proof for the special case where A is given in reduced row echelon form, but fail to see how the general case can be proved or disproved. My proof runs like this:
1) Suppose A is in reduced row echelon form. Then, every column of A either i) contains a single 1 element, while the other entries are 0, or ii) its first i rows consist of elements which are not 1 and not necessarily 0, where i < column number.
2) Now take A' and multiply it by A. Then we have i) every row of A is contained in A'A, and ii) every row of A'A is a linear combination of rows from A.
3) Thus if A'A(X) = 0, then A(X) = 0. Similarly, if A(X) = 0, then A'A(X) = 0. Hence the null spaces of A and of A'A are the same.
There is also the weaker general result: the null space of A is a subset of the null space of A'A, because if A(X) = 0, then A'A(X) = A'(0) = 0.
i think the solution to the problem lies in the column space of A, but I'm not sure exactly how. Will be really grateful to anyone who can share his/her insight on this problem. Thanks. :)