Suppose that the matrix $M$ is invertible. Then the equation $Mv=0$ has exactly one solution, namely $v = 0$.

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

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

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.