$A$ is invertible matrix iff $Ax=0$ has the trivial solution only.

Question

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.

If Ax=0 for some x not equal to zero,it will imply that A is not one-one and hence not invertible, since A0=0. — O_huck, Jun 03 '14 at 09:29
You have to prove "$A$ is invertible iff $Ax = 0$ iff $x=0$". Your try established that "$A$ is invertible $\Rightarrow \left(Ax=0 \Rightarrow x=0\right)$". So you still have a bit left to establish. — DanielV, Jun 03 '14 at 11:07

score 9 · Answer 1 · answered Jun 03 '14 at 09:30

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.

score 8 · Accepted Answer · edited Jun 03 '14 at 09:31

8

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.

edited Jun 03 '14 at 09:31

Gamma Function

5,642

answered Jun 03 '14 at 09:27

pointer

1,811

score 1 · Answer 3 · answered Jun 03 '14 at 09:31

1

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).

answered Jun 03 '14 at 09:31

5xum

123,496
6
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204

sorry this is probably a stupid question, what I am not following with regards to all of this is, if x=0 is the only solution, then if say some system of equations has some solution say x = 3, y=2, z=1, does that mean that it will be non invertible? I thought non-invertible is when there are either no solutions or infinitely man, i am confused what happens if there is 1 solution but its not trivial (x=0, y=0, etc)? – Math and ML May 16 '21 at 14:44
@bernando_vialli If the system of equations $Ax=0$ has a solution $[3,2,1]$, then the matrix is not invertible, that's right. That's because in that case, you know that $A$ maps $[3,2,1]$ and $[0,0,0]$ both to the same vector, the zero vector. So the matrix cannot be inverible because it is not injective. – 5xum May 16 '21 at 22:01

$A$ is invertible matrix iff $Ax=0$ has the trivial solution only.

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