Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $\{\vec0\}$. But it seems like a noninvertible square matrix will rarely have these properties. In the $2\times 2$ case, such a matrix would have to be of the form $\begin{bmatrix}a^2&ab\\ab&b^2\end{bmatrix}$. All such matrices would form a 2-dimensional surface within the 3-dimensional surface of singular matrices; they are rare, even within the singular matrices.

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

  1. Is there a name for this kind of matrix?
  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?
share|improve this question
1  
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 $V^T$). –  Nick Alger Feb 23 '12 at 6:09
    
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. –  Nick Alger Feb 23 '12 at 6:16
add comment

3 Answers

A geometric characterization of these matrices would be the following:

An $(n\times n)$-matrix $A$ is the same as a linear transformation $A:\ {\mathbb R}^n\to{\mathbb R}^n$. Provide ${\mathbb R}^n$ with the standard scalar product.

Claim: The condition ${\rm Col}\,A={\rm Row}\,A$ is equivalent to ${\rm ker}\,A=({\rm im}\, A)^\perp$. In particular any orthogonal projection would have this property.

Proof. The condition ${\rm Col}\,A={\rm Row}\,A$ means that ${\rm im}\,A={\rm im}\,A^t$. Take an $x\in{\ker}\,A$. Then $$0=\langle A x,y\rangle=\langle x,A^t y\rangle\qquad\forall y\ ;$$ therefore $x$ is orthogonal to ${\rm im}\,A^t={\rm im}\,A$. It follows that ${\rm ker}\,A\subset({\rm im}\,A)^\perp$, whence ${\rm ker}\,A=({\rm im}\, A)^\perp$ by counting dimensions.

Conversely, assume that ${\rm ker}\,A=({\rm im}\, A)^\perp$. Take an $x\in{\rm im}\,A^t$. Then $x=Az$ for a $z\in{\mathbb R}^n$, and we have $$\langle u,x\rangle=\langle u,A^t z\rangle=\langle Au,z\rangle =0\qquad\forall u\in{\rm ker}\,A\ .$$ It follows that ${\rm im}\,A^t\subset({\rm ker}\,A)^\perp={\rm im}\,A$ and therefore ${\rm im}\,A^t={\rm im}\,A$, as $A$ and $A^t$ have the same rank.

share|improve this answer
add comment

Perhaps these should be called scaled $r$-dimensional cosets of the unitary group $U(n)$.

By theorem 14 on page 6 of Travis Schedler's lecture 18, every $A$, when viewed as a linear transformation, has a polar decomposition $A=S\sqrt{A^*A}$, where $S$ is an isometry (unitary over $\mathbb{C}$ or orthogonal over $\mathbb{R}$), $A^*A$ is positive-definite (for $A\ne0$) by a previous theorem, and its square root is defined as the unitary/orthogonal conjugate $U\Lambda'U^*$ of the diagonal matrix $\Lambda'=\sqrt{\Lambda}$ of nonnegative square roots of its eigenvalues $(\Lambda_{ii}=\lambda_i)$ from its (necessarily diagonal) Schur decomposition $A^*A=U\Lambda U^*$. Similarly, $A^*=S\,'\sqrt{A^*A}$ for another isometry $S\,'$, since $(A^*A)^*=A^*A$. Thus $A^T=S\,'S^*A$, so I think your $B$ would have to also be an isometry, and over $\mathbb{R}$, special orthogonal (with determinant $1$). If we place the $\lambda_i$ in nonincreasing order (by permuting the columns of $U$), we can easily see that $\Lambda'$ lives in an embedding (injection) of $r$-dimensional diagonal matrices with strictly positive eigenvalues (scale matrices) from $\mathbb{R}^{r\times r}$ into $\mathbb{R}^{n\times n}$. Thus, $A=S\,U\Lambda'\,U^*$ is an isometry ($S$) or "permuted rotation" of an $r$-dimensional inhomogeneous dilation ($\Lambda'$) about the axes of some orthonomal basis ($U$) of $\mathbb{R}^n$, giving us a pretty good geometric picture of the matrix space $\{A\in\mathbb{R}^{n\times n}\mid A\mathbb{R}^n=A^T\mathbb{R}^n\}$.

Note, also, that unitary matrices (isometries) are of the form $U=e^{iH}$ for $H$ self-adjoint commuting with $U$. In fact, we can define a $C^\infty$ (exponential) map $U:\mathbb{R}\rightarrow\mathbb{R}^{n\times n}$ by $U(t)=e^{itH}$ with derivatives $U^{(n)}(t)=t\,U^*H^n$. The space (a Lie algebra) of self-adjoint matrices $H$ (or anti-self-adjoint if we drop the $i$ above), therefore, gives us the tangent space at each point of the space $U(n)$ of unitary matrices, which is a Lie Group.

I have not found a better necessary condition than Chrisitan's, ${\rm ker}\,A=({\rm im}\, A)^\perp$, or those given in the OP, e.g., $\exists B:A^TB=A$, for $A\mathbb{F}^n=A^T\mathbb{F}^n$ to hold, i.e. for an $n\times n$ matrix $A$ over a field $\mathbb{F}$ to share the same image or column space as its transpose $A^T$ (or conjugate transpose $A^*$). The best way to characterize these matrices is therefore probably with a list of equivalent conditions.

A sufficient condition is that the matrix $A\in\mathbb{F}^{n \times n}$ over the field $\mathbb{F}$ (e.g. $\mathbb{F}=\mathbb{R} \text{ or }\mathbb{C}$) is diagonalizable, meaning (TFAE):

  • $\exists\,G:A=G\Lambda G^{-1}$ with $G$ invertible and $\Lambda$ diagonal in $\mathbb{F}^{n\times n}$ (spectral/eigen- decomposition)
  • the geometric multiplicity of each eigenvalue equals its algebraic multiplicity
  • the sum of the dimensions of its eigenspaces is equal to $n$
  • there exists a basis of $\mathbb{F}^n$ consisting of eigenvectors of $A$
  • the number of linearly independent eigenvectors for each eigenvalue $\lambda$ equals the algebraic degree of $(x-\lambda)$ as a factor of the characteristic polynomial of $A$
  • its minimal polynomial is a product of distinct linear factors (splits and is squarefree) over $\mathbb{F}$
  • ($\mathbb{F}=\mathbb{R}$): each Jordan block in the real Jordan decomposition of $A$ contains no $2\times2$ identity matrix superdiagonal blocks
  • $\exists\,U,T:A=UTU^{-1}$ has a "complete"/complex Schur decomposition over $F$ with $U$ unitary and $T$ upper triangular, which is only true if all eigenvalues are in $F$ (and always true for $F=\mathbb{C}$)
  • ($\mathbb{F}=\mathbb{R}$): the Schur decomposition over the reals $A=QSQ^{-1}$ with $Q$ orthogonal and $S$ block upper triangular has only $1\times 1$ (and no $2\times 2$) diagonal blocks within $S$

A stronger sufficient condition (implying the above) is that $A$ is a normal matrix, meaning (TFAE):

  • $AA^*=A^*A$ ($A$ commutes with its conjugate transpose $A^*$)
  • $A$ is unitarily diagonalizable ($A=U\Lambda U^{-1}$ for $U^*=U^{-1}$)
  • $\exists U:A^*=AU$.

which (linguistically at least) completes the analogy with normal subgroups in group theory, for which left and right cosets are the same.

In these cases, the geometric characterization would be that $A$ is an anisotropic scaling or inhomogeneous dilation (which necessarily includes projection when $\text{rank}\;A\lt n$) after an orthogonal (or unitary for $\mathbb{F}=\mathbb{C}$) change of basis (with no isometry $S$). In this way, it is analogous to a self-adjoint operator.

When $A$ is positive definite, its Schur, spectral, and singular value decompositions all coincide. When $A$ is normal, its polar factors commute: $A=UP=PU$ (for $P$ positive semi-definite). In general there exists a polar factorization $A=UP$, but $U$ and $P$ are not guaranteed to commute.

The list of equivalences of normal (and diagonalizable) matrices is quite long, surely longer than in the wikipedia link, so there are probably many applications. One nice such application is in the classification of quadratic forms. For the dimensions of subspaces of rank $k$, have a look at Grassmanian and Stieffel manifolds.

An earlier version of this post had errors, based on a misunderstanding of this (in fact, an upper triangular, rather than block upper triangular, Schur form only exists for normal matrices whose characteristic or minimal polynomials have all their roots in $\mathbb{F}$).

share|improve this answer
    
This cannot be right. It is not necessary that $AA^*=A^*A$. Let $A=\begin{bmatrix}1&1&0\\0&1&0\\0&0&0\end{bmatrix}$. Then $\mathrm{Row}\,A=\mathrm{Col}\,A=\mathrm{Span}\{\vec{e}_1,\vec{e}_2\}$. So $A$ is the kind of matrix I am asking about. Just for the sake of generality, this matrix also has less than full rank. But $AA^*=\begin{bmatrix}2&1&0\\1&1&0\\0&0&0\end{bmatrix}$ while $A^*A=\begin{bmatrix}1&1&0\\1&2&0\\0&0&0\end{bmatrix}$. –  alex.jordan Feb 23 '12 at 21:14
    
You're absolutely right. It's not necessary at all. Thanks. –  bgins Feb 24 '12 at 0:50
add comment

Let us study this question in a general setting of $n \times n$ matrices.
For a given $n \times n$ matrix $A$, we denote its nullspace by $\mathcal{N}(A)$, and its column space by $\mathcal{C}(A)$.

Recall that the orthogonal complement of a vector subspace $V$ is $$V^\perp := \{\vec{x} : \forall \vec{v} \in V.~(\vec{x}^T \vec{v} = 0) \}.$$ It is well known that for any $n\times n$ matrix $A$ it holds that $(\mathcal{N}(A))^\perp = \mathcal{R}(A)$, where $\mathcal{R}(A)$ is the row space of $A$. In particular, $\mathcal{C}(A) = \mathcal{R}(A^T)$ implies that $(\mathcal{C}(A))^\perp = \mathcal{N}(A^T)$.

Now let $A$ be any $n \times n$ matrix with $\mathcal{N}(A) = \mathcal{C}(A)$. From our preceding discussion, it is thus necessary and sufficient that $\mathcal{N}(A) = \mathcal{N}(A^T)$. We shall first show that such an $n \times n$ matrix $A$ must be of the form $$Q \left(\begin{array}{cc} X & 0 \\ 0 & 0 \end{array}\right) Q^T,$$ where $X$ is an $r \times r$ matrix whose rank, $r$, is equal to that of $A$, and some bordering $0$'s may be absent if $A$ is of full rank $n$.

$Proof$. Extend any given basis $\{e_{r+1},\ldots,e_n\}$ for $\mathcal{N}(A)$ to a basis $\{e_1,\ldots,e_r,e_{r+1},\ldots,e_n\}$ for $\mathbb{R}^n$. Let $P = (\vec{e}_i^T)$ and $Q = P^{-1}$. Clearly, both $P$ (as well as $Q$) is invertible since its column vectors $\{e_1,\ldots,e_n\}$ are linearly independent. Since $\mathcal{N}(A) = \mathcal{N}(A^T)$, it follows that $PAP^T = \left(\begin{array}{cc} X & 0 \\ 0 & 0 \end{array}\right),$ where $r(X) = r(PAP^T) = r(A) = r$. Hence $A$ is of the desired form. The proof is thus complete.

Now we establish that any $n \times n$ matrix $A$ of the above form has to satisfy the equation $$\mathcal{N}(A) = \mathcal{N}(A^T),$$ and hence the condition that $\mathcal{N}(A) = \mathcal{R}(A)$.

$Proof.$ Suppose $A = Q \left(\begin{array}{cc} X & 0 \\ 0 & 0 \end{array}\right) Q^T$, where $X$ is an invertible matrix with $r(X) = r(A)$. Writing the matrix $\left(\begin{array}{cc} X & 0 \\ 0 & 0 \end{array}\right)$ as $Y$, we have that $\vec{x} \in \mathcal{N}(A)$ iff $QYQ^T \vec{x} = 0$ iff $YQ^T \vec{x} = 0$. Now since $X$ is an invertible matrix of size $r$, the last condition is equivalent to $Y^TQ^T \vec{x} = 0$, which in turn is equivalent to $QY^TQ^T \vec{x} = 0$ iff $x \in \mathcal{N}(A^T)$. This shows that $\mathcal{N}(A) = \mathcal{N}(A^T)$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.