(LINALG) : Nullspace of transpose : N(A^T)

[FrogPad]

I'm not sure if I am making a mistake, or my book is wrong, or if both answers are correct. But, it is confusing me, and I would like to know why. We are asked to find the basis of the following subspaces on the matrix A.

Find: $$R(A^T),\,\,N(A),\,\,\,R(A),\,\,N(A^T)$$

I'm having trouble finding N(A^T). Here is how I'm doing it.

$$A = \left[ \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 1 & 1 \\ 0 & 0 & 1 & 1 \\ 1 & 1 & 2 & 2 \end{array} \right]$$


thus:
$$A^T = \left[ \begin{array}{cccc} 1 & 0 & 0 & 1 \\ 0 & 1 & 0 & 1 \\ 0 & 1 & 1 & 2 \\ 0 & 1 & 1 & 2 \end{array} \right]$$

so...

$$rref(A^T) = \left[ \begin{array}{cccc} 1 & 0 & 0 & 1 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 \end{array} \right]$$

then we are left with...

$$\begin{array}{c} x_1+\alpha = 0 \\ x_2 + \alpha = 0 \\ x_3 + \alpha = 0 \\ x_4 = \alpha \end{array}$$

which gives:

$$\alpha\left[ \begin{array}{c} -1\\ -1\\ -1\\ 1 \end{array} \right]$$

The book gives:
$$\left[ \begin{array}{c} 1 \\ 1 \\ 1 \\ -1 \end{array} \right]$$

as the basis for $$N(A^T)$$

Is this the same? And why?

I mean $$\alpha$$ can be anything, so if $$\alpha = -1$$ then I get the same answer as the book. So spanning the set with either "my" vector, or the books accomplishes the same thing. It's just confusing to me why the book would not follow the algorithm to get the answer. Thanks in advance.

 

[Spectre5]

As you noted, using alpha = -1 makes your answer the same. They are the same answer.

 

[thepatientmental]

Your approach to finding the nullspace of A^T is correct. The reason why the book's answer is different is because they have chosen to use a different parameter, let's call it \beta, instead of \alpha. So instead of setting x_4 = \alpha, they set x_4 = \beta. This leads to a different basis vector for the nullspace, but both approaches are correct and will span the same subspace. It is just a matter of personal preference which parameter to use.

Additionally, the book's answer is a more standard form for a basis vector in the nullspace, where the leading coefficient is 1. This is not necessary, but it is often preferred for simplicity and consistency. So while your answer is also correct, it may not be in the most standard form.

In summary, both your answer and the book's answer are correct and will span the same subspace. The only difference is in the choice of parameter and the standard form of the basis vector.