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Recently I was asked on my linear algebra course the following question:

Suppose the map $\varphi:L \rightarrow M$ has the matrix $$A = \begin{bmatrix}1&-2&1\\1&-1&0\\2&-2&0\\-2&1&1\end{bmatrix}$$ in some pair of bases($e$, $f$). Can it have the matrix $$A' = \begin{bmatrix}2&-1&2\\0&2&-2\\0&1&-1\\-1&2&2\end{bmatrix}$$ in some other pair of bases?($e'$, $f'$)

I know that the relationship between $A'$ and $A$ is given by $$ A' = T^{-1} A S $$ Where $S$ is the transition matrix $e \rightarrow e'$ and $T$ is the transition matrix $f \rightarrow f'$. But how to find them? Or how to prove that they don't exist? Also I see that $det(A) = det(A^{-1}) = 0$ but I don't think it will help me here.

It really doesn't make any sense.

1 Answers1

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It is not possible. The columns of the matrix $A$ sum to $0$ so it has rank $2$. On the other hand, the rank of the matrix $A'$ is $3$.

mechanodroid
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