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Suppose $A$ is some matrix

I am trying to prove the following relation:

Im$(A)$ = span{$\begin{bmatrix} 0\\1\\0\\ 1 \end{bmatrix}$, $\begin{bmatrix} 1\\0\\1\\0 \end{bmatrix}$) = $\{x \in \mathbb{R}^4: x_1 - x_3 = 0, x_2 - x_4 = 0\}$

How did the last equality came from? I can't see how they were able to put the image into constraint form.

Attempt:

span{$\begin{bmatrix} 0\\1\\0\\ 1 \end{bmatrix}$, $\begin{bmatrix} 1\\0\\1\\0 \end{bmatrix}$) = $c_1\begin{bmatrix} 0\\1\\0\\ 1 \end{bmatrix}$ + $c_2\begin{bmatrix} 1\\0\\1\\0 \end{bmatrix}$ = $\begin{bmatrix} x_1\\x_2\\x_3\\x_4 \end{bmatrix}$

Then $\begin{bmatrix} c_2\\c_1\\c_2\\c_1 \end{bmatrix}$ = $\begin{bmatrix} x_1\\x_2\\x_3\\x_4 \end{bmatrix}$

So $x_1 = x_3, x_2 = x_4$, then $x_1 - x_3 = 0, x_2-x_4 = 0$

Am I right?

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  • $\begingroup$ Do you know about linear transformations? $\endgroup$ Nov 19, 2015 at 15:38
  • $\begingroup$ @rVitale yet but I don't know the steps to make this mental leap $\endgroup$
    – Fraïssé
    Nov 19, 2015 at 15:39
  • $\begingroup$ Your answer is right, I have another way to think of it that I'll post below. $\endgroup$ Nov 19, 2015 at 15:45
  • $\begingroup$ @rVitale Hmm what if the span is not consisting of the vectors of 0 and 1s but something like $[\alpha, 0, \alpha ,0]$, $[0, \beta, 0, \beta]$. I would like to write the constraint form in terms of $\alpha$ and $\beta$ but the above method will not work $\endgroup$
    – Fraïssé
    Nov 19, 2015 at 15:48
  • $\begingroup$ span($[\alpha,0,\alpha,0],[0,\beta,0,\beta]$)=span([1,0,1,0],[0,1,0,1]) since the spanning vectors are multiples of one another. $\endgroup$ Nov 19, 2015 at 15:51

3 Answers 3

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Let $x=\begin{bmatrix}x_1\\x_2\\x_3\\x_4\end{bmatrix}\in\text{Im}(A)$, there exist real numbers $a$ and $b$ such that

$$a\begin{bmatrix}1\\0\\1\\0\end{bmatrix}+b\begin{bmatrix}0\\1\\0\\1\end{bmatrix}=\begin{bmatrix}x_1\\x_2\\x_3\\x_4\end{bmatrix}\qquad\text{i.e.}\qquad\begin{bmatrix}1&0\\0&1\\1&0\\0&1\end{bmatrix}\begin{bmatrix}a\\b\end{bmatrix}=\begin{bmatrix}x_1\\x_2\\x_3\\x_4\end{bmatrix}$$ By taking the augmented matrix and performing elemental row operations we get $$\left[ \begin{array}{cc|r} 1 & 0 & x_1 \\ 0 & 1 & x_2 \\ 1 & 0 & x_3 \\ 0 & 1 & x_4 \end{array} \right]\sim \left[ \begin{array}{cc|c} 1 & 0 & x_1 \\ 0 & 1 & x_2 \\ 0 & 0 & x_3-x_1 \\ 0 & 0 & x_4-x_2 \end{array} \right] $$ Which has solutions iff $x_3-x_1=x_4-x_2=0$.

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Consider the linear map $T:\mathbb{R}^4 \rightarrow \mathbb{R}^2$ defined by $T(x_1,x_2,x_3,x_4)=(x_1-x_3,x_2-x_4)$. The kernel of this map is given by your constraints, and by the rank-nullity theorem the kernel must be 2-dimensional. Then you just have to check that $[0,1,0,1],[1,0,1,0]$ are in the kernel and are linearly independent to get your equality.

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  • $\begingroup$ You do also need to note that $T$ is surjective to get the kernel to be dimension 2, but looking at the image of $[1,0,0,0]$ and $[0,1,0,0]$ shows this. $\endgroup$ Nov 19, 2015 at 16:08
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Another way to see it, though is pretty much the same as Mario already did, think of the span of this two vectors as the image of the linear transformation given by the matrix $$\begin{bmatrix} 0 & 1 \\ 1 & 0 \\ 0 & 1 \\ 1 & 0 \end{bmatrix}$$ To see what this image is, first note that the matrix is 4x2 so the domain is $\mathbb{R}^2$ and the image is $\mathbb{R}^4$. So start with a vector in $\mathbb{R}^2$, say $\begin{bmatrix} y_1\\ y_2 \end{bmatrix}$ and multiply it by our matrix of interest: $$\begin{bmatrix} 0 & 1 \\ 1 & 0 \\ 0 & 1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} y_1\\ y_2 \end{bmatrix}=\begin{bmatrix} y_2 & y_1 & y_2 & y_1 \end{bmatrix}$$ Here you can see that the image of this linear transformation is the set of those vectors in $\mathbb{R}^4$ whose first and third coordinates are the same and second and fourth coordinates are the same as well. In other words $$\{(x_1, x_2, x_3, x_4): x_1-x_3=0 \text{ and } x_2-x_4=0\}$$

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