1. ## Two basic questions!

Ok so I have two quick questions.

The problem I'm trying to solve is as follows:

Find $a$ so that the null space of $A^{T}$ has dimension 0. Find a basis for the column space of $A^{T}$ once $a$ has been determined.

we know A = $\begin{bmatrix} a &1 &-1 &3 \\ 0 &a &-1 &0 \\ 1 &-1 &0 &a \end{bmatrix}$

so I went ahead and set a = 0 so that the dimension of $A^{T}$ = 0.

With that, I was wondering if I would instantly be able to say:

a basis for the column space of $A^{T}$ = all 3 columns of $A^{T}$ which, when reduced, would be =
$SP{\begin{bmatrix} 1\\0\\0\\0 \end{bmatrix} , \begin{bmatrix} 0\\1\\0\\0 \end{bmatrix}, \begin{bmatrix} 0\\0\\1\\0 \end{bmatrix} }$

For my second question, I was wondering if there's a generalized way to find $a$. For this problem, I pretty much did trial and error and looked at what value I could give $a$ so that, when I put $A^{T}$ into R.E.F. it would be linearly independent. Is that the only way to find a value for a?

2. Since no one has answered this yet, perhaps I need to make it more clear what I was thinking.

I thought that if I can conclude that dim(N(A)) = 0, then the rank = 3. As in an M x N matrix, N = Rank + Nullity. Therefore, since the rank is 3 and n = 3, a basis for the column space would have to be the 3 columns.

Am I correct in thinking this?

Since no one has answered this yet, perhaps I need to make it more clear what I was thinking.

I thought that if I can conclude that dim(N(A)) = 0, then the rank = 3. As in an M x N matrix, N = Rank + Nullity. Therefore, since the rank is 3 and n = 3, a basis for the column space would have to be the 3 columns.

Am I correct in thinking this?
Why don't you just apply $A$ to basis elements of the domain space and pick $a$ intelligently so that the image of those basis elements are linearly independent?

4. Yes. If $a=0$, then the only solution to $A^Tx=0$ is $x=0$, so then $\mathrm{dim} \mathrm{null} A=0$.

In this case, as you said, $\mathrm{rank} A^T=3-0=3$ (I use 3 here because $A^T$ has 3 columns; you have to be careful since $A^T$ is not square.)

Since the 3 columns generate the column space, this forces them to be a basis for it. (The rank of a matrix is equal to its column rank.)