What are some relationships between a matrix and its transpose?

Question

All I can think of are that If symmetric, they're equivalent

If A is orthogonal, then its transpose is equivalent to its inverse.

They have the same rank and determinant.

Is there any relationship between their images/kernels or even eigenvalues?

Answer 1

The key to understanding the various relations between a matrix and its transpose is to understand what the transpose of a matrix signifies. For simplicity, let's work over the field $\mathbb R$. Recall that for a vector space $V$, its dual, $V^*$, is the vector space of linear functionals: linear transformations $\varphi :V\to \mathbb R$.

Here is an exercise then to explain what transposing a matrix does:

1) Given a linear transformation $T:V\to W$, one obtains a function $T^*:W^*\to V^*$, given by $T^*(\varphi)(v)=\varphi(T(v))$. Actually, $T^*$ is a linear transformation.

2) Fix bases $B_V=(v_1,\cdots, v_n)$ and $B_W=(w_1, \cdots, w_k)$ for $V$ and $W$ respectively (here each $v_i$ is a vector in $V$, each $w_i$ is a vector in $W$). Let $M$ be the matrix representing $T$ in these bases.

3) Recall the dual bases $B^*_{V^*}$ and $B^*_{W^*}$. Prove that the matrix representing $T^*$ with respect to these bases is precisely the transpose of $M$.

Understanding the transpose via the dual spaces explains a lot.

Answer 2

Fix a ring $R$, and let $A \in M_n(R)$. The characteristic polynomial for $A$ is $$\chi_A(x)=\det (xI-A),$$ so that $\chi_{A^T}(x) = \det (x I -A^T)= \det ((xI-A)^T)=\det(xI-A)$. Since the eigenvalues of $A$ and $A^T$ are the roots of their respective characteristic polynomials, $A$ and $A^T$ have the same eigenvalues. (Moreover, they have the same characteristic polynomial.)

It follows that $A$ and $A^T$ have the same minimal polynomial as well, because the minimal polynomial arises as the radical of the characteristic polynomial.

Answer 3

There is a very strong connection between an $m\times n$ matrix $A$ and its transpose $A^T$, at least if the base field is $\def\R{\mathbb{R}}\R$.

Denote by $C(A)$ the column space, that is, the subspace of $\R^m$ generated by the columns of $A$. Denote by $N(A)$ the null space, that is, the set of vectors $v\in\R^n$ such that $Av=0$.

Associated to a matrix $A$ we have four subspaces:

• $C(A)\subseteq\R^m$
• $N(A)\subseteq\R^n$
• $C(A^T)\subseteq\R^n$
• $N(A^T)\subseteq\R^m$

Now the relations (of course, the second one is equivalent to the first)

$$ C(A)\oplus N(A^T)=\R^m,\qquad C(A^T)\oplus N(A)=\R^n $$

which means that every vector of $\R^m$ can be written in one and only one way as the sum of a vector in $C(A)$ and one in $N(A^T)$.

To see why, consider a vector $v\in C(A)\cap N(A^T)$; then $v=Ax$ for some $x\in\R^n$ because $v\in C(A)$; since also $v\in N(A^T)$ we get $$ 0=A^Tv=A^TAx $$ from which we deduce $$ 0=x^T0=A^TAx=(Ax)^T(Ax), $$ so that $Ax=0$ and therefore $v=Ax=0$. Thus $C(A)\cap N(A^T)=\{0\}$ and we only need to show that $C(A)+N(A^T)=\R^m$.

The relation between the rank of $A$ and of $A^T$ and the "rank-nullity theorem" tell then that $$ \dim N(A^T)=m-\mathrm{rk}\,A^T=m-\mathrm{rk}\,A=m-\dim C(A), $$ so that $\dim C(A)+\dim N(A^T)=m=\dim\R^m$ and hence $C(A)+N(A^T)=\R^m$.

The same holds for matrices over $\def\C{\mathbb{C}}\C$, but using the Hermitian transpose instead of the transpose.

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This doesn't hold however for matrices over an arbitrary field, because the step “$(Ax)^T (Ax)=0$ implies $Ax=0$” cannot be done in general.