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At the demonstration of this, I couldn't understand why the following holds:

"$A$ positive definite $\Rightarrow$ $A$ invertible, because otherwise would exist $X\not=0$ satisfying $AX=0\Rightarrow X^TAX=0$ wich is a contradiction."

I understood the implications, but couldn't get why "$A$ not invertible then exist $X\not=0$ satisfying $AX=0$"

If anyone could explain this to me or give another way of proving the initial problem, I'd be gratefull.

Marcelo
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  • If $Ax=0$ implies $x=0$, then $A$ is invertible. –  Oct 16 '17 at 17:23
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    A matrix is singular (i.e. non-invertible) iff its determinant (i.e. the product of the eigenvalues) equals zero. A positive-definite matrix has a positive determinant, hence it cannot be singular. – Jack D'Aurizio Oct 16 '17 at 17:24

1 Answers1

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Let $n$ be the number of columns, if $A$ is not invertible, its rank is less than $n$ and its nullity is at least $1$.

Hence there exists $x \neq 0$, such that $Ax=0$.

Siong Thye Goh
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