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For the given equation $Ax=0$ where $A$ is a square matrix and $x$ is a column vector, why $A$ must be a singular matrix (determinant $0$) in order to have $x$ different from null column vector?.

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  • $\begingroup$ Think of regular multiplication, for example, the equation $xy=0$. For $y\not=0$, the only solution is $x=0$. Similarly, for a non-singular matrix, the only vector that will satisfy $Ax=0$ will be the null column vector. $\endgroup$
    – Alex D
    Dec 16, 2017 at 23:05
  • $\begingroup$ While I'm not directly answering your question I think looking at the YouTube series "Essence of Linear Algebra" is a must if you are studying linear algebra and wish to deepen your understanding , it's a series of 10 minute videos . $\endgroup$
    – Matthew
    Dec 16, 2017 at 23:05
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    $\begingroup$ Invertible matrices represent bijective linear maps $\mathbb{R}^n \to \mathbb{R}^n$. In particular, such maps are injective and hence have trivial kernels. $\endgroup$ Dec 16, 2017 at 23:11
  • $\begingroup$ @Nuwanda If you are ok, you can set as solved. Thanks! $\endgroup$
    – user
    Dec 17, 2017 at 8:06

2 Answers 2

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That's simply a matter of linear dependence of the columns of A.

Indeed note that the multiplication $Ax$ correspond to a combination of the columns vectors of A by the coefficient of the vector $x$.

Thus:

  1. if the columns of A are linearly independent (i.e. A not singular) the only way to have Ax=0 is that $x=0$

  2. if the columns of A are linearly dependent (i.e. A singular) you may have $Ax=0$ also for $x\neq0$

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Because if $A$ is non-singular then its inverse $A^{-1}$ exists, and you have

$$0=A^{-1}0=A^{-1}\left(Ax\right)=\left(A^{-1}A\right)x=Ix=x$$

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