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If an $n \times n$ symmetric A is positive definite, then all of its eigenvalues are positive, so $0$ is not an eigenvalue of $A$. Therefore, the system of equations $A\mathbf{x}=\mathbf{0}$ has no non-trivial solution, and so A is invertible.

I don't get how knowing that $0$ is not an eigenvalue of $A$ enables us to conclude that $A\mathbf{x}=\mathbf{0}$ has the trivial solution only. In other words, how do we exclude the possibility that for all $\mathbf{x}$ that is not an eigenvector of $A$, $A\mathbf{x}=\mathbf{0}$ will not have a non-trivial solution?

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Note that if $Ax=0=0\cdot x$ for some $x\ne 0$ then by definition of eigenvalues, $x$ is an eigenvector with eigenvalue $\lambda = 0$, contradicting that $0$ is not an eigenvalue of $A$. $$Ax=\lambda x$$

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$$\det A = \prod_{j=1}^n \lambda_j \implies \det A = 0 \Leftrightarrow \exists\ i \in \{1,2,\ldots, n\}:\lambda_i = 0$$

In other words, the determinant is the product of the eigenvalues, and can only be zero if at least one eigenvalue is zero.

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Because if $Ax=0$ for some nonzero $x$, then $0$ would be an eigenvalue.

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  • $\begingroup$ Yes. You have that $0$ is an eigenvalue of $A $ $\iff $ $A $ is not invertible $\iff $ there exists nonzero $x $ with $Ax=0$. $\endgroup$ – Martin Argerami Sep 11 '19 at 11:57

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