Csdrhrt

If $A$ is a matrix $10$ by $12$ and $A x=b$ is solvable for every $b$, then th column space of $A$ is?

The column space of $A$ is the whole of $R^m$ which is $12$ is this answer the right answer ?

According to the hypothesis of your problem, for all fixed vector
$
b=begin{pmatrix} b_1 \vdots \ b_j\ vdots \b_{10} end{pmatrix}
$
in $mathbb{R}^{10}$ exists a vector
$
x=begin{pmatrix} x_1 \vdots \ x_j\ vdots \x_{12} end{pmatrix}
$
in $mathbb{R}^{12}$ such that the equality below is true
$$
begin{pmatrix}
A_{11} & ldots & A_{1j} & ldots & A_{1,12}
\
vdots & ddots & vdots & ddots & vdots
\
A_{i1} & ldots & A_{ij} & ldots & A_{i,12}
\
vdots & ddots & vdots & ddots & vdots
\
A_{10,1} & ldots & A_{10,j} & ldots & A_{10,12}
end{pmatrix}
begin{pmatrix} x_1 \vdots \ x_j\ vdots \x_{12} end{pmatrix}
=
begin{pmatrix} b_1 \vdots \ b_j\ vdots \b_{10} end{pmatrix}
$$
By performing the multiplication of the matrix $A$ by the column matrix $x$ we have that the following statement is valid: for all fixed $binmathbb{R}^{10}$ there are numbers $x_1$, $ldots$, $x_j$, $ldots$, $x_{12}$ such that
$$
begin{pmatrix}
x_1cdot A_{11}+& ldots &+x_jcdot A_{1j}+ & ldots &+x_{12}cdot A_{1,12}
\
vdots & ddots & vdots & ddots & vdots
\
x_1cdot A_{i1}+ & ldots &+x_jcdot A_{ij}+ & ldots &+x_{12}cdot A_{i,12}
\
vdots & ddots & vdots & ddots & vdots
\
x_1cdot A_{10,1}+ & ldots & +x_jcdot A_{10,j} +& ldots &+x_{12}cdot A_{10,12}
end{pmatrix}
=
begin{pmatrix} b_1 \vdots \ b_j\ vdots \b_{10} end{pmatrix}
$$
Let the vector $A_j$ be the j-th column of matrix $A$.The last statement can then be rewritten as follows. For all vector $binmathbb{R}^{10}$ there are real numbers $x_1$, $ldots$, $x_j$, $ldots$, $x_{12}$ such that
$$
x_1 begin{pmatrix} A_{11} \vdots \ A_{i1}\ vdots \A_{10 ,1} end{pmatrix}
+ldots +
x_j begin{pmatrix} A_{1j} \vdots \ A_{ij}\ vdots \A_{10 ,j} end{pmatrix}
+ldots +
x_{12} begin{pmatrix} A_{1,12} \vdots \ A_{i,12}\ vdots \A_{10 ,12} end{pmatrix}
=
begin{pmatrix} b_{1} \vdots \ b_{j}\ vdots \b_{10} end{pmatrix}
$$
Since the vector $ b $ is arbitrary, it follows that the columns $A_1,ldots, A_{12} $ of $ A $ are vectors that generate the space $mathbb {R}^{10} $.

If $A$ is a $10 times 12$ matrix then $A$ is the matrix associated with a linear transformation $T:mathbb{R}^{12}tomathbb{R}^{10}$. Since $Ax=b$ is solvable so it has full rank.

Hence, Column Space ($A) = $ Range($T$) = $mathbb{R}^{10}$.

Your answer is not correct. The maximum rank of $A$ can be only $10$. Since $AX=b$ has a solution for every $b$, this means the rank of $A$ must be $10$. Thus two ($12-10=2$) columns must be free. This means the column space is $Bbb{R}^{10}$.