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Let two matrix $A=(a_{ij})_{m\times n}$ and $B=(b_{ij})_{m\times n}$ satisfy $\ker(A)=\ker(B)$ , $\: $($Ax=0\Leftrightarrow Bx=0$)

Prove that exist matrix P invertible then A=PB.

My tried:

$Ax\Leftrightarrow Bx=0\Leftrightarrow PBx=0\to \ker(A)=\ker(PB)$

Come here, I don't know how.

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It's a little bit hard to understand your English. Is the goal to prove that there exists an invertible $P$ such that $A=PB$? In that case you can't just assume the existence of $P$, like it seems you have done. – Svinepels Dec 31 '13 at 12:50

Let $(e_1,\ldots,e_p)$ a basis of $\ker A=\ker B$ which we complete on $(e_1,\ldots,e_n)$. We know that $(Be_{p+1},\ldots,Be_n)$ is a basis of $\operatorname{im}(B)$ which we complete on $(y_1,\ldots, y_p,Be_{p+1},\ldots,Be_n)$ and $(Ae_{p+1},\ldots,Ae_n)$ is a basis of $\operatorname{im}(A)$ which we complete on $(x_1,\ldots, x_p,Ae_{p+1},\ldots,Ae_n)$. Let $P$ the matrix such that $$Py_i=x_i\; i=1\ldots,p\quad\text{and}\quad PBe_i=Ae_i\; i=p+1,\ldots,n$$ hence $P$ is invertible since it tranforms a basis to a basis and we have $A=PB$.

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This is for my اخی Sami. – Babak S. Jan 1 '14 at 6:47

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