Let . Find a basis for the null space of A. I got the answer Is it correct?
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Yes it is correct. By the way the matrix has Rank 3 so the dim of the nullspace must be 1 and you found 1 vector in the null space.
So that, strictly speaking, you have NOT yet solved the problem. The "null space" is a one-dimensional subspace of and you have found one vector in that space. How would you represent any vector in the null space?
any vector in the null space.....do you mean where is a real number?
Yes, or, to phrase it slightly differently, , the set of all real multiples of your original vector= the subspace spanned by that vector.