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I tried doing this

$M$ is invertible so $M^{-1}$ exists, therefore $(M^{-1})Mv = (M^{-1})0 = v = 0$

then I am confused on how to prove that this is the only solution. Can I just say...

Assume $M$ is not invertible, then its column vectors are not linearly independent, therefore for column vectors $\left\{v_1, v_2, \ldots , v_n\right\}$in $M$ and constants $\left\{c_1, c_2, \ldots , c_n\right\}$

$$c_1v_1 + c_2v_2 + \ldots + c_nv_n = 0$$ is a non-trivial solution to $Mv = 0$ because some ($c_kv_k$) term is not equal to zero?

sorry for the format

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  • $\begingroup$ $M^{-1} (Mv) = (M^{-1} M) v = I v = v$. Hence if $Mv = 0$, then $(M^{-1} M) v = 0$ from which we must have $v=0$. $\endgroup$ – copper.hat Oct 26 '16 at 3:37
  • $\begingroup$ Thank you very much I see now from your answer that we must have v = 0. $\endgroup$ – BadAtMath Oct 26 '16 at 3:58
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For your second part.

No. The condition is if $M$ is invertible. The moment you assume the case that $M$ is not invertible, it is irrelevant.

As for the first part. We started from the equation $Mv=0$ and we are interested to figure out what is $v$. Multiplying by $M^{-1}$ shows that $v$ has to be $0$, and hence it has a unique solution.

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  • $\begingroup$ Thank you very much. I upvoted you but my karma is too low for it to show up. $\endgroup$ – BadAtMath Oct 26 '16 at 3:58

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