Why does the following statemnet true?
$A$ is invertible matrix iff $Ax=0$ has only the trivial solution.
My try:
Let $x$ a solution of $Ax = 0$.
Then, because $A$ is invertible there is $A^{-1}$. Hence,
$$AA^{-1}x = 0$$ $$I_nx = 0$$ $$x=0$$
I used the associative property of matrix multiplication.
$\endgroup$ 43 Answers
$\begingroup$If $\operatorname{rank}(A)=r$ the solutions of $Ax=0$ is $n-r$ dimensional space, so if $Ax=0$ has only trivial solutions it means that $\operatorname{rank}(A)=\operatorname{size}(A)$, so $A$ is invertible.
$\endgroup$ $\begingroup$If $A$ is invertible just multiply by $A^{-1}$ we get $x=0$. Now assume the other way round that $x=0$ is the only solution. For the sake of contradiction assume that $A$ is not invertible. Then columns are not independent. Therfore $\Sigma \lambda_n C_n=0:all $ $\lambda_n$ are not identically $0$. So now $(\lambda_1,...,\lambda_n)$ is a solution. Contradicting the hypothesis.
$\endgroup$ $\begingroup$The proof in one direction is trivial, that is, if $A$ is invertible, then obviously, since $A0=0$ and $A$ is injective, $0$ is the only solution to $Ax=0$.
If $Ax=0$ has only one solution, then it is also simple to show that $A$ is injective (try it!). Then, use the fact that the image of $A$ is a vector space. Since $A$ is injective, the dimension of the image of $A$ is the same as the dimension of the domain of $A$, that is $n$. This means that $\text{im}(A)$ is a $n$-dimensional subspace of a $n$-dimensional space, meaning it is the whole space (therefore, $A$ is surjective).
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