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If matrices $A$ and $B$ have the same fundamental subspace than $A=cB$ where $c$ is some scalar.

I think it is true, because I try to disprove with this two matrices, they have the same dimension of fundamental subspace, but I think in this question want that we have the same vector in fundamental space of matrices. So than is true that exist c.

A=$\begin{bmatrix} 1& 0& 0\\ 0& 1& 0\\ 0 & 0& 1\\ 0 & 0& 2 \end{bmatrix}$. B=$\begin{bmatrix} 3& 0& 0\\ 0& 2& 0\\ 0 & 0& 1\\ 0 & 0& 2 \end{bmatrix}$.

But what do you think?

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  • $\begingroup$ What do you mean with fundamental subspace? Also, in your case $A\neq cB$ for any $c$. $\endgroup$ – blub Aug 20 '18 at 13:43
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If by fundamental subspace you mean the space generated by rows and the kernel of $A$ and $A^t$ as stated here then your assertion is false. Consider $$ A = \begin{bmatrix} 0 & 1 \\ 1 & 0 \\ \end{bmatrix} \qquad B = \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix} $$ Those two have the same fundamental spaces but there exists no $c \in \mathbb{R}$ such that $A = cB$

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  • $\begingroup$ I think that everything is the same but $R(A)=(0,1)$ and $R(B)=(1,0)$ so they do not have the same fundamental subspaces everything other is the same but there is different $\endgroup$ – Marko Škorić Sep 14 '18 at 10:31
  • $\begingroup$ what is $R(A)$? $\endgroup$ – JayTuma Sep 14 '18 at 13:29
  • $\begingroup$ Image of linear transformation $\endgroup$ – Marko Škorić Sep 14 '18 at 13:40
  • $\begingroup$ image of what transfomation? The induce one from $\mathbb{R}^2$ to $\mathbb{R}^2$? $\endgroup$ – JayTuma Sep 14 '18 at 13:42
  • $\begingroup$ It does not write a dimension but I understand that you choose a some matrices and you just change one entrie and you still have the same fundamental subspace, but c does not exist since you if you multiply with c you do not change one entries you change every entries so it muss not exist some c such that A=cB $\endgroup$ – Marko Škorić Sep 14 '18 at 13:48

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