Let $A$ be an $m × n$ matrix with entries in $\mathbb Q,$ and let $T: \mathbb Q^n → \mathbb Q^m$ be the linear transformation $T(x) = Ax.$ We may also regard the matrix $A$ as having entries in $\mathbb R$ or $\mathbb C.$ Let $T':\mathbb R^n → \mathbb R^m$ and $T'': \mathbb C^n → \mathbb C^m$ be the corresponding linear transformations $T'(x) = Ax,$ and $T''(x) = Ax.$ Show that rank($T$) = rank($T'$) = rank($T''$).
The rank of $T$ is dim(im($T$)). As we have that $\dim(\mathbb Q^m)=\dim(\mathbb R^m)=\dim(\mathbb C^m)= m$, can we directly conclude? I'm unsure whether or not we have that im($T$) $= \mathbb Q^m.$ Can the image be smaller just like with functions? Therefore, if $A$ stays the same, then the im($T$)=im($T'$)=im($T''$) no matter the vector spaces where you apply the transformation. I am just really confused and don't know how to start this proof. I can sense why this is true but don't really know how to prove it. Thank you for your help.
You're correct in that the rank is the dimension of the range, but it is also the maximal number of linearly independent columns of $A.$ Let's say the rank is $r$. That means that there is a set of columns $X_1, X_2, \cdots, X_r$ such that the only solution to $$a_1X_1+a_2X_2+\cdots+a_rX_2=0\tag 1$$ is $a_1=a_2=\cdots =a_r=0.$ Now the question is, if we allow the $a_i$ to be elements of $\mathbb R$ or $\mathbb C$ might we not be able to find a nonzero solution? After all, we have a lot more numbers to choose from.
Think about solving the system of equations (1). The $a_i$ are the unknowns, and the coefficients are elements of the $X_i$ so they're in $\mathbb Q.$ As you go through your favorite algorithm for solving a system of linear equations, you never use any operations other than addition, subtraction, multiplication, and division, so the coefficients always remain in $\mathbb Q,$ and you'll never get any solution with elements that are not in $\mathbb Q.$
So your guess is right. The rank is the same.
EDIT Reading this over, I see that the argument only proves that the rank can't get any smaller. I leave it to you to prove that the rank can't get any bigger. (Hint: This part is really easy.)