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I sat down to write a linear algebra take-home exam problem where I would give a $4\times4$ matrix $A$ and ask for bases for these six spaces: $\mathrm{Col}\,A$, $\mathrm{Row}\,A$, $\mathrm{Nul}\,A$, $\mathrm{Col}\,A^t$, $\mathrm{Row}\,A^t$, $\mathrm{Nul}\,A^t$. Trivially, a basis for $\mathrm{Col}\,A$ will work for $\mathrm{Row}\,A^t$, and likewise with $\mathrm{Row}\,A$ and $\mathrm{Col}\,A^t$; that was meant to be part of the test.

I expected $\mathrm{Nul}\,A$ and $\mathrm{Nul}\,A^t$ to be different. I worked on the first $4\times4$ matrix that came to mind: $\begin{bmatrix}1&2&3&4\\5&6&7&8\\9&10&11&12\\13&14&15&16\end{bmatrix}$. To my surprise, the standard method for finding the basis of the null space led me to see that in this case, $\mathrm{Nul}\,A=\mathrm{Nul}\,A^t$. (And since I do not want students to think this will always be the case, I have to try a different matrix.)

As a consequence of the orthogonality of row and null spaces, $\mathrm{Row}\,A=\mathrm{Row}\,A^t$. So we have a matrix where: $$\mathrm{Col}\,A^t=\mathrm{Row}\,A=\mathrm{Row}\,A^t=\mathrm{Col}\,A$$ $$\mathrm{Nul}\; A=\mathrm{Nul}\;A^t$$ all of which is equivalent to just knowing $$\mathrm{Row}\, A=\mathrm{Col}\,A$$ or to just knowing $$\mathrm{Nul}\, A=\mathrm{Nul}\,A^t$$And these are equivalent to the existence of a matrix $B$ such that $AB=A^t$, since the rows of $A$ (columns of $A^t$) are linear combinations of the columns of $A$.

Now any invertible matrix would automatically have these properties, since the null space of an invertible matrix is $\{\vec0\}$. And any symmetric matrix will automatically have these properties, since $A=A^t$. But it seems like a noninvertible, asymmetric square matrix will rarely have these properties. (In the $2\times 2$ case, there are none.)

With general $n\times n$ matrices again, I would like to know

  1. Is there a name for this kind of matrix ($\mathrm{Row}\,A=\mathrm{Row}\,A^t$)? Perhaps additionally demanding noninvertible asymmetric specimens?
  2. Does this kind of matrix have any interesting applications?
  3. If we fix $\mathrm{rank}\, A$, do matrices of this form make an interesting surface in $M_{n\times n}$?
  4. If so, what is the dimension of this surface?
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    Perhaps it would be useful to view this question in light of the singular value decomposition, $A=U \Sigma V^T$, where $U$ and $V$ are orthonormal and $\Sigma$ is diagonal. Then when the matrix has rank $k$ the question is whether the first $k$ columns of $U$ span the same space as the first $k$ columns of $V$ (rows of $V^T$).2012-02-23
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    To show that the set of these matrices is a manifold, I guess the standard approach would be to find a smooth map from the space of all n-by-n matrices to some other vector space, which is zero for precisely the matrices you are interested in, then apply the Regular Value Theorem.2012-02-23

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