# Matrix, Kernel and Span

Q. if $A$ be a $n\times n$ be a matrix. if $A\vec{x}=\vec{0}$ has trivial solution, does it imply columns of $A$ span $\mathbb{R}^n$?

A. No, The equation always has trivial solution.

Is not it that if the $A\vec{x}=\vec{0}$ has trivial solutions equivalent to saying that columns are linearly independent? and hence span $\mathbb{R}$, more precisely, $A\not=0$?

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You seem to be tripping up over language. There's a very big difference between

$Ax = 0$ has the trivial solution

and

$Ax = 0$ has only the trivial solution

The latter is also equivalent to

The solution to $Ax = 0$ is trivial

if by "solution" one implicitly means the "complete solution".

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yes. thanks. merely matter of language. thanks. – user45099 Feb 26 '13 at 12:20
What does your comment mean? Yes the question was "has only trivial solution"? If so then you have two answers already, don't forget to mark one of them as such :) – Sh3ljohn Feb 26 '13 at 13:55

If the only solution is the trivial solution, then the n columns and are linearly independent and span $\mathbb{R}^n$ furthermore the linear operator $T:\mathbb{R}^n \to \mathbb{R}^n$ defined by $T(x) = Ax$ is a bijection, the determinant of the matrix $A$ is nonzero and there are probably 20 more things that are equivalent to this.

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Sorry if I've misunderstood, but if $x$ is the zero vector in $\mathbb{R}^n$ then $Ax = 0$ for all $n \ \text{x} \ n$ matrices $A$. So it isn't true, because if it were, it's essentially implying that all $n \ \text{x} \ n$ matrices $A$ have linearly independent columns (and rows) which isn't true.

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