Matrix equation : $\;AB=A^t\;$ implies the column space equals the row space

Question

I have the following exercise:

Let $\;A,B\;$ be $\;n\times n\;$ matrices s.t. $\;AB=A^t\;$ . Prove that the row space of $\;A\;$ equals its column space.

Now, this is what I've succeeded so far: I know that the dimensions are equal, and if we denote by $\;R_1,...,R_n\,,\,C_1,...,C_n\;$ the rows and columns of $\;A\;$ resp., I know that it is then enough to prove either

$$\forall\,i\;,\;\;\;C_i\in Sp\{R_1,...,R_2\}\;\;,\;\;or\;\;\;\forall\,i\;,\;\;R_i\in Sp\{C_1,...,C_n\}$$

, because of the dimensions' equality.

I also know that the rows of $\;A^t=$ the columns of $\;A\;$, but I am unable to give the last step and prove what I want: if $\;A=(a_{ij})\;,\;\;B=(b_{ij})\;$ , then we're given that

$$AB=\left(\sum_{k=1}^na_{ik}b_{kj}\right)=(a_{ji})=A^t$$

and it seems that from the above it should follows that the columns of $\;A^t(=$ the rows of $\;A\;$) are a linear combination of the columns of $\;A\;$, as $\;A\;$ is being multiplied from the right by $\;B\;$....and here I am ridiculously stuck!

Answer 1

Since you have that the dimensions are equal, perhaps this will help:

$$ \operatorname{Col}(A^t) = \{ A^t x : x \in \Bbb{R}^n\} = \{AB x : x \in \Bbb{R}^n\} = \{A x : x\in \operatorname{Ran}(B)\}\subset\{Ax : x \in \Bbb{R}^n\} = \operatorname{Col}(A). $$

Answer 2

Let the column $n\times 1$ matrix $e_k$, be defined as follows $e_k=[0,\ldots,0,\underbrace{1}_k,0,\ldots,0]^t$. Then the $j^{\rm th}$ row of $A^t$ is given by $A^te_j$ $$ A^te_j=ABe_j=A\sum_{i=1}^nb_{ij}e_i= \sum_{i=1}^nb_{ij}A e_i \in\hbox{Range} A $$ that is $\hbox{Range} A^t\subset \hbox{Range} A$. and the conclusion follows as you remarked.