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If given the matrix $$\begin{bmatrix} 1 & -1 & 0 \\ 0 & 0 & 0 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \\ \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} b_1 \\ b_2 \\ b_3 \\ b_4 \end{bmatrix}$$ The row echelon form of the matrix is $$\begin{bmatrix} 1 & -1 & 0 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$$ and clearly the row space basis is $\begin{bmatrix} 1 & -1 & 0\end{bmatrix}$ and $\begin{bmatrix}0 & 1 & -1\end{bmatrix}$ and the column space basis is $$\begin{bmatrix} 1 & 0 & 0 & 0 \\ -1 & 0 & 1 & 0 \end{bmatrix}$$

Based on the row space basis, the vector $\begin{bmatrix} 8 & 9 & 6\end{bmatrix}$ would not be in the row space, as its not a linear combination of the row space basis vectors. However, using $x_1=8$, $x_2=9$, $x_3=6$ in the above equation, the vector $b_1=-1$, $b_2=0$, $b_3=3$, $b_4=0$. The vector $\begin{bmatrix} -1 & 0 & 3 & 0\end{bmatrix}^\top$ is in the column space as it can be obtained from a linear combination of the basis vectors for the column space. How is it possible that a vector NOT in the row space produce a vector that IS in the column space?

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    $\begingroup$ Unreadable, sorry. Please use MathJax. $\endgroup$ – amsmath Oct 15 '19 at 3:06
  • $\begingroup$ I don't know how to $\endgroup$ – I.A. Oct 15 '19 at 3:48
  • $\begingroup$ @I.A. I've formatted the question as best as I could. Does this look OK? You can always roll back the edit if it conflicts with what you were trying to write. $\endgroup$ – Theo Bendit Oct 15 '19 at 3:56
  • $\begingroup$ @I.A. If you wish to learn more about MathJax, check out our tutorial. For information about how to format matrices, check out the first answer. Also, please feel free to click "edit" to see how I've formatted your question, and play around with it. $\endgroup$ – Theo Bendit Oct 15 '19 at 3:57
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The matrix maps all vectors in $\mathbb R^3$ to a subset of $\mathbb R^4$

vectors that are multiples of $\begin {bmatrix} 1\\1\\1 \end {bmatrix}$ are in the kernel of the transformation.

And the image is spanned by $\begin {bmatrix} 1\\0\\0\\0 \end {bmatrix},\begin {bmatrix} 0\\0\\1\\0 \end {bmatrix}$

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