Writing the span of two vectors in constraint form

Question

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?

Answer 1

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 $x_3-x_1=x_4-x_2=0$.

Answer 2

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.

Answer 3

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 $$\{(x_1, x_2, x_3, x_4): x_1-x_3=0 \text{ and } x_2-x_4=0\}$$