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Let $n \in \mathbb{N}$, and $U,V$ two linear subspaces of $\mathbb{R}^n$ of the same dimension. Could one always make a matrix $A \in \mathbb{M}^{n \times n}(\mathbb{R})$ such that $spanA = U$ and $spanA^T = V$?


I find it hard to find a counterexample. I know that for any $n \in \mathbb{N}$, there is such a matrix if $\dim U = \dim V \in \{0,n\}$, we could just take $0,I$ as matrices. Moreover, if $U,V$ are lines, that means $U = \mathbb{R}v$ and $V = \mathbb{R}w$ for some nontrivial vectors $v,w$, then the matrix below works.

$$ \left( \ w_1v \ | \ w_2v \ | \ \cdots \ | \ w_nv \ \right) $$ Here $w_i$ are coordinates of the vector $w$, and each entry represents a column this way.

Then I took two subspaces of dimension two in $\mathbb{R}^3$, namely $$ U \quad = \quad span \left\{ \left( \begin{array}{ccc} 1 \\ 0 \\ 0 \end{array} \right) , \left( \begin{array}{ccc} 0 \\ 1\\ 0 \end{array} \right) \right\} \qquad V \quad = \quad \left( \begin{array}{ccc} 0 \\ 1\\ 0 \end{array} \right)^\perp $$ The matrix we need here is $$ \left( \begin{array}{ccc} 1 & 0 & 0 \\ -1 & 1 & 0 \\ 0 & -1 & 0 \end{array} \right) $$


I find it hard to deal with the general case. Could you give me some help to establish the statement, or give a counterexemple?

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This is indeed always possible. Let $U$ be some $r$-dimensional subspace of $\mathbb{R}^n$. Let $R$ denote any matrix with rowspace $U$. Since $U$ is $r$-dimensional, it follows that the rank of $R$ is $r$ and so the image of $R$ is also $r$-dimensional.

Let $\mathcal{B}'$ denote a basis for the image of $R$ and extend $\mathcal{B}'$ to a basis $\mathcal{B}$ of $\mathbb{R}^n$. Form the matrix $P$ with columns given by $\mathcal{B}$, with the first $r$ columns consisting of the vectors of $\mathcal{B}'$. It follows that $P$ maps the subspace $S$ spanned by $\left\{\mathbf{e}_1,\ \cdots, \mathbf{e}_r\right\}$ to $\mathrm{im}(R)$ and therefore $P^{-1}$ maps $\mathrm{im}(R)$ to $S$.

Now let $\mathcal{C}'$ denote a basis for $V$ and extend $\mathcal{C}'$ to $\mathcal{C}$. Form the matrix $Q$ with the basis as the columns, with the first $r$ columns consisting of the vectors of $\mathcal{C}'$. It follows that $Q$ takes $S$ to $V$.

Then the matrix $QP^{-1}R$ is your desired matrix. Since $QP^{-1}$ is invertible, the rowspace of $R$ is unchanged, therefore $\mathrm{row}(QP^{-1}R) = U$. Also, $P^{-1}$ maps $\mathrm{im}(R)$ to $S$ and $Q$ maps $S$ to $V$. Therefore $\mathrm{im}(QP^{-1}R) = V$, as required.

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  • $\begingroup$ I understand that a map $X \mapsto AX$ does not change the kernel of $X$ if $A$ is invertible, but I don't see how that should work for the rowspace of $X$. Could you explain me? $\endgroup$ Aug 11 '14 at 20:37
  • $\begingroup$ @KoenraadvanDuin There are two immediate ways to see this. First you note that the kernel is invariant under $X\mapsto AX$. The rowspace is the orthogonal complement of the kernel, hence it is also left invariant. Alternatively, if $A$ is invertible then it can be written as a product of elementary matrices, so the overall effect of $A$ is to perform a sequence of elementary row operations on $X$, which does not change the rowspace. $\endgroup$
    – EuYu
    Aug 11 '14 at 23:51
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Let $m$ be the dimension of $U$ and $V$. Let $M_U$ be an $n\times m$ matrix whose columns are a basis of $U$ and $M_V$ be an $n \times m$ matrix whose columns are a basis of $V$. Now $U = \{M_U \vec{x} : \vec{x} \in \mathbb{R}^m \}$ and $V = \{M_V \vec{x} : \vec{x} \in \mathbb{R}^m \}$.

We wish to find a matrix $A$ such that $$ \begin{align*} U &= \{M_U \vec{x} : \vec{x} \in \mathbb{R}^m \} = \{A \vec{x} : \vec{x} \in \mathbb{R}^n \}, \\ V &= \{M_V \vec{x} : \vec{x} \in \mathbb{R}^m \} = \{A^T \vec{x} : \vec{x} \in \mathbb{R}^n \}. \end{align*} $$

We note that $\{ M_V^T \vec{x} : \vec{x} \in \mathbb{R}^n\} = \mathbb{R}^m$, because the rank of $M_V^T$ is $m$ and it has $m$ rows. Thus $$ U = \{M_U \vec{x} : \vec{x} \in \mathbb{R}^m \} = \{M_U M_V^T \vec{x} : \vec{x} \in \mathbb{R}^n \}. $$ But this immediately suggests that $A=M_U M_V^T$, because $\mathrm{span}(M_U M_V^T) = U$. By symmetry, also $\mathrm{span}(M_V M_U^T) = V$, and because $A^T=M_V M_U^T$, we are done.

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  • $\begingroup$ I don't understand what you mean by $M_U^T$, could you please tell me? $\endgroup$ Aug 11 '14 at 21:02
  • $\begingroup$ @KoenraadvanDuin Transpose of the matrix $M_U$. $\endgroup$
    – JiK
    Aug 11 '14 at 21:06
  • $\begingroup$ right, I forgot about that. $\endgroup$ Aug 11 '14 at 21:18

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