TITLE: Is it easier to determine that a matrix is singular than it is to determine nonsingular?
QUESTION [4 upvotes]: I came across this line "It is often easier to determine that a matrix is singular than it is to determine that a matrix is nonsingular. The facts below illustrate this.   

Fact 1.10. Let $A\in\mathbb C^{n\times n}$ and $x,b\in\mathbb C^n$.

If $x\ne0$ and $Ax=0$, then $A$ is singular.
If $x\ne0$ and $A$ is nonsingular, then $Ax\ne0$.
If $Ax=b$, where $A$ is nonsingular and $b\ne0$, then $x\ne0$.


I am not sure how that it is illustrated.  I understand the facts and agree with them.  But how they make one to determine if a matrix is singular in a EASIER way, I do not understand.  Any suggestions?

REPLY [2 votes]: I agree with all the other answers that this must be what the book means; however, I rather disagree with the general philosophy of this quotation. 
Indeed, it is the opposite of what is the case in the applied setting: if nature throws a square matrix with real-valued entries at you, you can conclude with sufficiently precise equipment that it's non-singular but you can never conclude (by measurement alone) that it is singular (the point is that a vanishing determinant is a closed condition, so no error can be tolerated in measurements, whereas a non-vanishing determinant is an open condition).
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