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Suppose $CA=I_n$ (the $n \times n$ identity matrix. Show that the equation $A\textbf{x}=\textbf{0}$ has only the trivial solution.

The textbook solution is as follows:

If x satisfies $A\textbf{x} = \textbf{0}$, then $CA\textbf{x} = C\textbf{0} = \textbf{0}$ and so $I_n\textbf{x} = \textbf{0}$ and $\textbf{x} = \textbf{0}$. This shows that the equation $A\textbf{x} = \textbf{0}$ has no free variables. So every variable is a basic variable and every column of A is a pivot column.

Now I get that $I_n\textbf{x} = \textbf{0}$ would mean $\textbf{x}$ must be the trivial solution since there are no free variables and you have a pivot position at every column. However, I don't get how that implies that $A\textbf{x}=\textbf{0}$ has the trivial solution since $A \neq I_n$. I don't quite get how $C$ comes into play for solving this problem.

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  • $\begingroup$ If $A$ is not invertible, then $A$ cannot be full rank, thus having infinite solutions. I don't know if you know the equivalent formulation of invertiability though. $\endgroup$
    – macton
    Jan 27, 2021 at 8:30
  • $\begingroup$ So this problem is supposed to be solved without invertibility since it hasn't been covered yet. $\endgroup$
    – jlcv
    Jan 27, 2021 at 8:35
  • $\begingroup$ $I_nx=x$ by definition because $I_n$ is the identity matrix. Thus, $I_nx=0$ and $x=0$ is the same thing. No need to mention free variables or anything else. $\endgroup$
    – A.Γ.
    Jan 27, 2021 at 8:39
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    $\begingroup$ Does this answer your question? Linear Algebra - Suppose $CA=I_n$. Show that the equation $Ax = 0$ has only the trivial solution. $\endgroup$
    – Ugo
    Jul 14, 2021 at 9:28

2 Answers 2

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$$ A\mathbf{x} = \mathbf{0} $$

$$ \Rightarrow CA\mathbf{x} = C\mathbf{0} $$

$$ \Rightarrow I_n\mathbf{x} = C\mathbf{0} $$

$$ \Rightarrow \mathbf{x} = C\mathbf{0} $$

$$ \Rightarrow \mathbf{x} = \mathbf{0} $$

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The second sentence in the texbook solution is incomplete. What they have proved is $Ax=0$ implies $x=0$ so no vector other than $0$ is a solution.

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  • $\begingroup$ Sorry I had missed one of the solution sentences, I updated my question to include the missing sentence. $\endgroup$
    – jlcv
    Jan 27, 2021 at 8:34

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