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Let $A$ be an $m \times n$ matrix and $\mathbf{x}$ an $n \times 1$ vector. The planes from the $m$ equations $A\mathbf{x} = \mathbf{b}$ are in ?-dimensional space. The combination of the columns of $A$ is in ??-dimensional space.

1st Attempt: Since $A\mathbf{x}$ has $m$ rows, each of which is a plane, thus $\color{ #DC143C}{? = m}$.
Since there are $n$ columns in $A$, thus $\color{#DC143C}{?? = n} \qquad \blacksquare$.

(Too terse) Correct Answers: $? = n \quad \& \quad ?? = m$.

My 1st attempt was wrong so I compassed to work from the definitions of matrix multiplication.

2nd Attempt: By definition of multiplication by rows, for all $\color{ #D555D1}{1 \leq r \leq m}$,
row $\color{ #D555D1}{r}$ of $A_{\color{#D555D1}{m} \times \color{ #43A934}n}\mathbf{x}_{\color{ #43A934}n \times 1} := [\vec{\text{row $\color{ #D555D1}{r}$ of $A_{\color{#D555D1}{m} \times \color{ #43A934}n}$}}]_{1 \times \color{ #43A934}n} \cdot \mathbf{x}_{\color{ #43A934}n \times 1} =$ $\LARGE{\color{red}{¡ ¡ ¡ }}$ scalar (with size $1 \times 1$) $\LARGE{\color{red}{!!!}}$.

But there are $\color{#D555D1}{m}$ rows in $A_{\color{ #D555D1}{m} \times n}\mathbf{x}_{\color{ #43A934}n \times 1}$, so $A_{\color{ #D555D1}{m} \times n}\mathbf{x}$ is of size $\color{#D555D1}{m} \times 1. \quad \Longrightarrow \; ? = \color{ #D555D1}{m}.$

By definition of multiplication by columns, $A_{\color{ #D555D1}{m} \times \color{ #43A934}n}\mathbf{x}_{\color{#43A934}n \times 1} := x_1\color{#43A934}{[\vec{\text{coln 1 of A}}]}_{\color{ #D555D1}{m} \times 1} + ... + x_n\color{#43A934}{[\vec{\text{coln n of A}}]}_{\color{ #D555D1}{m} \times 1} $
$= {\huge[}\sum_{1 \leq k \leq \color{ #43A934}c} x_k\color{#43A934}{[\vec{\text{coln c of A}}]} \, {\huge]}_{\LARGE{\color{#D555D1}{m}} \times 1} \Longrightarrow \;?? = \color{ #D555D1}{m}.$

My second attempt also founders; it gave $? = ?? = \color{ #D555D1}{m}$. What and where are the errors? Please keep answers rudimental; this exercise is from but Chapter $2$.


Hindsight: My work for ?? had been correct. The mistake is signaled with the red exclamation marks. While what I had written was right, it was delusive.
That $1 \times 1$ scalar is $\sum_{1 \leq c \leq n}A(r,c)x_c$, so it's actually one equation of $n$ variables, thus denoting a hyperplane. It contains $n$ variables (ie $x_1, ..., x_n$) which represents but $\color{ #DC143C}{? =}$ the dimension of each of the $m$ rows of $\mathbf{Ax = b}$.

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Actually, having read your question again, I realise you know the answer but you don't agree with the request for a more rigorous proof? If I'm wrong let me know and I'll try another approach.

You can't just claim the rows are planes. You have to spell out that each one of those equations is in the same number of variables as there are columns in $A$.

With the second part, you can't just say it's because of the number of columns, you have to say how. An $m \times n$ matrix $A$ maps from $\mathbb R^n \to \mathbb R^m$ and the column space of $A$, that is, the subspace spanned by its columns, is its range. That subspace is in $\mathbb R^m$ because it is the span of $m$-element vectors - the columns of $A$.

These elementary points are part of linear algebra at this stage and I guess they wanted you to demonstrate that.

Addendum 1:

First an example with a $2 \times 3$ matrix, so $m=2$ and $n=3$: $\begin{bmatrix}a&b&c\\d&e&f\end{bmatrix}\,\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}j\\k\end{bmatrix}$.

\begin{array}{cc} \text{When you multiply this out :} & \begin{align*} ax + by + cz &= j \\ dx + ey + fz &= k \end{align*} \\ \end{array}

From the equivalence with the simultaneous equations, the number of variables is equal to the number of columns, $n=3$, and the number of equations is equal to the number of rows, $m=2$. These equations are planes in $\mathbb R^3$. So the number of columns gives the dimension of the space in which these planes occur. In general they are called hyperplanes and have the general formula:

$$a_{r1}x_1 + a_{r2}x_2 + \cdots + a_{rn}x_n = b_r.$$

Rearrange the previous equations, they are equivalent to:

$$ x\begin{bmatrix}a\\d\end{bmatrix} + y \begin{bmatrix}b\\e\end{bmatrix} + z \begin{bmatrix}c\\f\end{bmatrix} = \begin{bmatrix}j\\k\end{bmatrix} $$

This is a linear combination of the $n=3$ columns to give an $m=2$-dimensional vector, $\begin{bmatrix}j\\k\end{bmatrix}.$

Can you see by your own work in your question that

$A\vec x = x_1\vec c_1 + \cdots + x_n\vec c_n = \vec b$ is giving $\vec b$ as a linear combination of the columns?
There are $m$ rows, so the columns are $m$-dimensional vectors and thus $\vec b$ as a linear combination of $m$-dimensional vectors must be in $\mathbb R^m$.

Addendum 2:

In your rework for the first question you're almost there.

\begin{align*} \text{Row}_r \;\vec x &= \begin{bmatrix}a_{r1}&a_{r2}&\cdots&a_{rn}\end{bmatrix}\begin{bmatrix}x_1\\x_2\\\vdots\\x_n\end{bmatrix}\\ &= a_{r1}x_1 + a_{r2}x_2 + \cdots + a_{rn}x_n \end{align*}

If we set Row$_r\;\vec x = b_r$ then we get

$$a_{r1}x_1 + a_{r2}x_2 + \cdots + a_{rn}x_n = b_r$$

This is the equation of a hyperplane in $n$ variables in $\mathbb R^n$. There are $m$ rows and so we get $m$ equations in $n$ variables. So, the number of rows, $m,$ is the number of equations of hyperplanes. The number of columns, $n$, is the number of variables and also the dimension of the space in which those hyperplanes ($\mathbf{Ax = b}$) exist.

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  • $\begingroup$ Thank you very much. My problem was that although my second attempt begets the right answers, I still can't conceive what's wrong with my 1st attempt. Moreover, I wouldn't have realised that the 1st was wrong had I not checked the answer! $\endgroup$ – Greek - Area 51 Proposal Sep 25 '13 at 10:42
  • $\begingroup$ Also, could you please simlify your answer? This exercise is from a section that precedes "column space", "subspace", "span", etc... $\endgroup$ – Greek - Area 51 Proposal Sep 25 '13 at 10:44
  • $\begingroup$ @LePressentiment See Addendum 1. Hopefully this is now clearer. Let me know how it goes. $\endgroup$ – Geoff Pointer Sep 25 '13 at 12:02
  • $\begingroup$ Many thanks. Upvoted. I'm cottoning on to $?? = m$ now. Did you also explain $? = n$ in your addendum 1? $\endgroup$ – Greek - Area 51 Proposal Sep 29 '13 at 6:48
  • $\begingroup$ @LePressentiment I did. The part about $n=3$ columns and planes in $\mathbb R^3$. I've added another line to show what the row $i$ equation from $A\vec x=\vec b$ would look like in the general $m \times n$ case. Let me know if it's still not clear. $\endgroup$ – Geoff Pointer Sep 29 '13 at 7:34

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