If $A$ and $B$ are $n \times n$ matrices that have the same row space, then $A$ and $B$ have the same column space.
This is false of course. I could just come up with examples though. Can one prove this?
[Math] Same row space is equivalent to same column space
linear algebramatricesvector-spaces
Related Solutions
When you row-reduce a matrix, the dimension of the column space stays fixed, so if $A,B$ have the same reduced echolon form then the dimensions of the column spaces are equal, but the column spaces might not be equal:
$$A=\begin{pmatrix}1&2\\1&2\end{pmatrix}\hspace{10pt}B=\begin{pmatrix}1&2\\2&4\end{pmatrix}$$
The have the same reduced echolon form, but different column-spaces.
In general, the only way to make sure that two matrices have the same column space is to column-reduce them (unless both are of full rank).
Lemma 1: Given an $m\times n$ matrix $A,$ the null space of $A^T$ is the orthogonal complement of the column space of $A.$
Proof: Write $A=[c_1\:\cdots\:c_n]$ where the $c_j$ are the columns of $A,$ and note that for any $m$-dimensional vector $x$ we have $$A^Tx=\left[\begin{array}{c}c_1^T\\\vdots\\c_n^T\end{array}\right]x=\left[\begin{array}{c}c_1^Tx\\\vdots\\c_n^Tx\end{array}\right]=\left[\begin{array}{c}c_1\cdot x\\\vdots\\c_n\cdot x\end{array}\right].$$ Since the column space of $A$ is spanned by $c_1,...,c_n$, then $x$ is in the orthogonal complement to the column space of $A$ if and only if $x$ is orthogonal to each $c_j$ if and only if each $c_j\cdot x=0$ if and only if $A^Tx$ is the $n$-dimensional zero vector if and only if $x$ is in the null-space of $A^T.$ $\Box$
Lemma 2: Let $V,W$ be subspaces of some finite-dimensional space $X$. $V$ and $W$ have the same orthogonal complement if and only if $V=W$.
Proof: If $V=W$, then their orthogonal complements are trivially the same.
Suppose $V,W$ have the same orthogonal complement. Take $v\in V$. Since $X$ is the direct sum of $W$ and its orthogonal complement, and since $x\in V\subseteq X$, then there exist unique $w,w'$ such that $v=w+w',$ $w\in W$, and $w'$ in the orthogonal complement of $W$. Since $V,W$ have the same orthogonal complement, then $w'$ is orthogonal to $v,$ and so $$0=v\cdot w'=(w+w')\cdot w'=w\cdot w'+w'\cdot w'.\tag{$\star$}$$ Since $w'$ is in the orthogonal complement of $W$ and $w\in W$, then $w\cdot w'=0$, so it follows by $(\star)$ that $$w'\cdot w'=0.$$ Now, no non-zero vector is self-orthogonal, so $w'$ must be the zero vector, whence $v=w\in W$, and so $V\subseteq W$. By symmetrical arguments, we likewise have $W\subseteq V$, so $V=W$. $\Box$
Proposition: Given matrices $A,B$ of the same dimensions, $A$ and $B$ have the same column space if and only if $A^T$ and $B^T$ have the same reduced row echelon form.
Proof: Let $rref(M)$ indicate the reduced row echelon form of a matrix $M$. Recall that we can obtain $rref(M)$ by Gauss-Jordan elimination, which involves multiplication on the left by some finite collection of elementary matrices--that is, for any $M$, there exist elementary matrices $E_1,\cdots,E_n$ of appropriate dimension such that $rref(M)=E_n\cdots E_1M.$ This collection of elementary matrices is not unique, but that isn't important. Note, though, that elementary matrices are invertible, so it follows that the null spaces of $rref(M)$ and $M$ are the same.
Thus, $A^T$ and $B^T$ have the same reduced row echelon form if and only if they have the same null space. By Lemma 1, $A^T$ and $B^T$ have the same null space if and only if the column spaces of $A$ and $B$ have the same orthogonal complement. By Lemma 2, the column spaces of $A$ and $B$ have the same orthogonal complement if and only if the column spaces of $A$ and $B$ are the same. $\Box$
Upshot: The Proposition lets us get around needing to know what the column spaces of two matrices are, and simply determine whether they have the same column space by converting their transposes to reduced row echelon form.
Best Answer
Theory tells us that if the row space of $A$ equals the row space of $B$, then the ranks of the column spaces of $A$ and $B$ are equal. Let this guide us to a minimal counterexample.
Clearly if the dimensions of the column spaces were zero, then indeed the columns spaces would be trivial (and equal). So the minimal counterexample involves dimension (at least) one.
What do rank one matrices with the same row space as $A$ look like? Do any of them have different columns spaces? (Hint: yes.)
If you know matrix multiplication, here's an easy way to build a pair of matrices with the same one-dimensional row space but different column spaces. Pick a nonzero row vector $v$ of length $n$. Pick two nonzero column vectors $u^T$ and $w^T$, also of length $n$, which are not scalar multiples of one another (so the spaces spanned by $u^T$ and $w^T$ are not equal).
Let $A = u^T v$ and $B = w^T v$. Then the row spaces of $A$ and $B$ are both the one-dimensional space spanned by $\{v\}$, but the column spaces are different. The column space of $A$ is the space spanned by $\{u^T\}$, while the column space of $B$ is the space spanned by $\{w^T\}$, and we just arranged that these would be distinct.
Note that all the rows of $A$ and all the rows of $B$ are scalar multiples of $v$, and in each case there is at least one nonzero row. This proves the claim about the row spaces of $A$ and $B$ being the same. Similarly the column space of $A$ is the span of $\{u^T\}$ and the column space of $B$ is the span of $\{w^T\}$.
Any doubt about reversing the roles of row and column spaces should be dispelled by taking the transposes of $A$ and $B$, or $A^T$ and $B^T$ as we usually denote them. These last two $n\times n$ matrices will have equal column spaces but different row spaces.