Explain proof that any positive definite matrix is invertible
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?
$\endgroup$4 Answers
$\begingroup$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 $A$. $$Ax=\lambda x$$
$\endgroup$ 1 $\begingroup$$$\det A = \prod_{j=1}^n \lambda_j \implies \det A = 0 \Leftrightarrow \exists\ i \in \{1,2,\ldots, n\}:\lambda_i = 0$$
In other words, the determinant is the product of the eigenvalues, and can only be zero if at least one eigenvalue is zero.
$\endgroup$ 0 $\begingroup$Because if $Ax=0$ for some nonzero $x$, then $x$ can be said to be in the null space of $A, N(A)$, hence it is non-invertible. We can also see that $$Ax=\lambda x$$ by definition of eigenvalue/vector so $0$ would be an eigenvalue.
Therefore by the chain of equivalences, $λ=0\implies A$ Is non-invertible.
$\endgroup$ 1 $\begingroup$Note that $A^TA$ is symmetric and positive definite and we have$$(\det A)^2=\det A\det A^T= \det (A^TA)\neq 0 $$
this reduces the question to the symmetric case.
Now if A is symmetric. Assume that is $0$ is an eigenvalue, then we have $Ax=0$ with $x\neq 0$. By positive definite, we get $$0=(Ax,x)>0$$ which is impossible. So $x=0$ is the only solution to $Ax=0$.
NB: note that by definition, an eigenvector is never zero
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