2
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if $\begin{bmatrix} 1\\ 1\\ 0 \end{bmatrix}$, $\begin{bmatrix} 0\\ 0\\ 1 \end{bmatrix}$ $\in R(A)$, and $\begin{bmatrix} 2\\ 5 \end{bmatrix}$, $\begin{bmatrix} 1\\ 2 \end{bmatrix}$ $\in R(A^{T})$

I think that is not exist because, from here we can see that this matrices must be $3x2$ but when I find base of $N(A)$ and $N(A^{T})$ they have dimension 1, so than we need matrice $3x3$ which is not what we expect.

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0
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Let consider

$$A=\begin{bmatrix}1&0\\1&0\\0&1\end{bmatrix}$$

and note that $\operatorname{rank}(A)=2$ therefore $R(A^T)$ = $\mathbb{R^2}$.

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  • $\begingroup$ But from this matrices you can not get the same base for $R(A^{T})$ as I write $\endgroup$ – Marko Škorić Aug 20 '18 at 14:18
  • $\begingroup$ @MarkoŠkorić The condition you gave is that $\in R(A)$, and $\begin{bmatrix} 2\\ 5 \end{bmatrix}$, $\begin{bmatrix} 1\\ 2 \end{bmatrix}$ $\in R(A^{T})$ and of course that condition is fulfilled by A. $\endgroup$ – gimusi Aug 20 '18 at 14:19
  • $\begingroup$ Yes but do you know how to get base of fundamental subspace when you have some matrices? $\endgroup$ – Marko Škorić Aug 20 '18 at 14:21
  • $\begingroup$ @MarkoŠkorić Yes of course a basis for $R(A)$ is $(1,1,0), (0,0,1)$ and for $R(A^T)$ is $(1,0), (0,1)$ or $(2,5), (1,2)$. $\endgroup$ – gimusi Aug 20 '18 at 14:24
  • $\begingroup$ Sorry you are right, I hate this when I watch this and do not see that is true, thank you! $\endgroup$ – Marko Škorić Aug 20 '18 at 14:26

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