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(b) $A$ and $A^T$ have the same left nullspace.

Soln: False. Any nonsquare matrix gives a counterexample, so do certain square matrices: $B = \begin{bmatrix} k & 0 \\ k & 0 \\ \end{bmatrix}$. Since $EB = REF(B) \implies \begin{bmatrix} 1 & 0 \\ \color{green}{-1} & \color{green}1 \\ \end{bmatrix}B = \begin{bmatrix} k & 0 \\ \color{green}0 & \color{green}0 \\ \end{bmatrix}$
thus left nullspace of $B$ = $= \{\color{green}{(-1, 1)}\}$.
Left nullspace of $\color{olive}{B^T} = N([\color{olive}{B^T}]^T) = \{\mathbf{x} : (B^T)^T \mathbf{x = 0} \}$
whence $\mathbf{Bx = 0}$ yields $ \mathbf{x} = [ 0 \quad c]$ for all scalars $c$.

(c) If the row space equals the column space then $A^T = A$.

Soln: False. Any nonsymmetric invertible matrix is a counterexample, like $N = \begin{bmatrix} 1 & 1 \\ 0 & 1 \\ \end{bmatrix}$.

Only posterior to seeing the answers do I realise that (b) and (c) are false. Anterior to finding counterexamples, what alludes/suggests that (b) and (c) are false? Also, how were the counterexamples devised?

For (b), I started with the definitions. Left nullspace of any matrix $A = N(A^T) = \{\mathbf{y} : A^T\mathbf{y = 0 \iff y^TA = 0} \}$
Left nullspace of any matrix $\color{olive}{A^T} = N([\color{olive}{A^T}]^T) = \{\mathbf{x} : \color{olive}{A^T}^T \mathbf{x = 0 \iff Ax = 0} \}$. Then?

For (c), I see that $C(N) = C(N^T)$ but didn't realise to consider nonsymmetric invertible matrices.

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Only posterior to seeing the answers do I realise that (b) and (c) are false. Anterior to finding counterexamples, what alludes/suggests that (b) and (c) are false?

You've put phrasing like this into many of your recent questions (often with different grandiloquent word choices for "before" and "after"). But it seems a little weird to me to make such a point of asking how one would know how to solve the problem in advance: if you know that a statement is false "anterior" to finding counterexamples, then you don't need to do the problem at all. The whole point of solving problems is that you've learned something "posterior" to solving the problem that you didn't know "anterior" to working on it.

Also, how were the counterexamples devised?

There are many ways of finding counterexamples for questions like these. If you want to know the mental processes of the person who wrote up the solution, ask them.

Maybe you are asking "How would I usefully get started thinking about questions like this?" If you're asked whether $X$ is always true, then (assuming you don't know the answer already...) you should start by looking at examples where $X$ can be either true or false. If in any of your examples you find that $X$ is not true, you've answered the question. If you keep trying examples and $X$ is always true, you need to either look for a common feature of the examples that makes $X$ hold, or you need to switch to a broader class of examples. This process is dynamic: you could do this several times before you finally come up with an answer, and what you write as your answer will probably not record your entire thought process but merely what worked in the end.

Your starting point is a good one as far as it goes: the "left null spaces of $A$ and $A^T$" are just the null spaces of $A^T$ and $A$, so the question is slightly strangely stated: it might as well ask whether the null space of $A$ must be equal to the null space of $A^T$. Is this true? As the answer points out, it doesn't make sense if the matrix is not square, so that's a kind of answer, but now let's restrict to square matrices. We try some examples, startin with the simplest: e.g. the zero matrix. Then both null spaces are all of $\mathbb{R}^n$. Or the identity matrix. Then both null spaces are $0$. How about $A = \left[ \begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array} \right]$. The null space of $A$ is spanned by $e_2$. What about the null space of $A^T$? Oh, of course it's the same as the null space of $A$ because $A^T = A$. So we see that in order to find counterexamples we need to look past symmetric matrices. What's the simplest nonsymmetric matrix? Maybe $\left[ \begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array} \right]$. Here the null space of the matrix is spanned by $e_1$ and the null space of the transpose is spanned by $e_2$. They are not equal, so we've answered the question.

Let's look at part c). The question is whether equality of row and column spaces forces the matrix to be symmetric. Again if I just play around with nonsymmetric matrices then I'll soon find $\left[ \begin{array}{cc} 1 & 0 \\ 1 & 1 \end{array} \right]$. It's a counterexample because it's not symmetric and because its row and column space are equal. In fact, they're equal because they're both $\mathbb{R}^2$. Since the row space of an $n \times n$ matrix is $\mathbb{R}^n$ if and only if the matrix is invertible if and only if the column space is $\mathbb{R}^n$, we see that we've found a much broader class of counterexamples, so perhaps we record that as an answer rather than the special case we found just by exploring the example space.

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  • $\begingroup$ Thank you very much. I really appreciated the more general lessons in your answer, especially because my textbook and instructor don't teach them. I'll try to apply this. $\endgroup$ – Greek - Area 51 Proposal Dec 7 '13 at 6:45
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Note that for (b), $A$ and $A^\mathrm{T}$ having the same left-nullspace is equivalent to saying that $A$ has nullspace equal to left-nullspace (since the left-nullspace of $A^\mathrm{T}$ is just the nullspace of $A$). But this means the orthogonal complements of the nullspace and the left-nullspace are also equal, i.e. the rowspace and columnspace of the matrix must be equal. This fact should help you in finding counter examples, and taking a non-square matrix necessarily forces the rowspace and columnspace, and hence the nullspace and left-nullspace, to be unequal.

More simply, you can simply notice that for an $m\times n$ matrix, the nullspace is a subspace of $\mathbb{R}^n$ while the left-nullspace is a subspace of $\mathbb{R}^m$. For non-square matrices, they are not even subspaces of the same vector space, let alone equal.

For (c), it's useful to keep in mind that being symmetric is a very strong condition for a matrix. Not only do the rowspace need to equal the columnspace, but the columns and rows of the matrix must represent the same ordered basis for the two spaces. The simplest way to produce a counter-example then is for the rows and columns of an $n\times n$ matrix to span all of $\mathbb{R}^n$. That way we don't have to worry about the rows and columns spanning unequal subspaces. But this means the matrix must be invertible, so we look at non-symmetric invertible matrices for a counter-example.

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