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

Question

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 ?

Answer 1

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 $b\in\mathbb{R}^{10}$ there are numbers $x_1$, $\ldots$, $x_j$, $\ldots$, $x_{12}$ such that $$ \begin{pmatrix} x_1\cdot A_{11}+& \ldots &+x_j\cdot A_{1j}+ & \ldots &+x_{12}\cdot A_{1\,12} \\ \vdots & \ddots & \vdots & \ddots & \vdots \\ x_1\cdot A_{i1}+ & \ldots &+x_j\cdot A_{ij}+ & \ldots &+x_{12}\cdot A_{i\,12} \\ \vdots & \ddots & \vdots & \ddots & \vdots \\ x_1\cdot A_{10\,1}+ & \ldots & +x_j\cdot A_{10\,j} +& \ldots &+x_{12}\cdot A_{10\,12} \end{pmatrix} = \begin{pmatrix} b_1 \\\vdots \\ b_j\\ \vdots \\b_{10} \end{pmatrix} $$ 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 $b\in\mathbb{R}^{10}$ 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} $.

Answer 2

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 $\Bbb{R}^{10}$.

Answer 3

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

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