If A is a symmetric square matrix. I need to show that it is positive definite only if all eigenvalues are positive.

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I understand that a positive definite matrix by the definition is a symmetric matrix where all eigenvalues are positive. I also know that if $ (x,y) = {x^T}{\cdotp}M{\cdotp}y$ then it is positive definite if ${x^T}{\cdotp}M{\cdotp}x {\geq} 0$ and ${x^T}{\cdotp}M{\cdotp}x = 0$ only if $x=\vec{0}$.

Now I believe I need to go about this by first proving that all eigenvalues of A are positive using $Ax = {\lambda}x$ where x is a real eigenvector and $\lambda$ is a real eigenvalue. After I've done this I believe I need to then answer the question and prove that it is positive definite only if all eigenvalues are positive, but I'm not entirely sure how to do the second part.

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That's not quite right. A symmetric real matrix is said to be positive definite if $x^TMx$ is positive whenever $x\neq\vec 0.$ Nothing is said, here, about eigenvalues.

Now, suppose that $\lambda$ is some eigenvalue of $M,$ meaning that there is some vector $x$ with $x\ne\vec 0$ such that $$Mx=\lambda x.$$ What can you say about $x^TMx$? How can you rewrite $x^TMx$? What does this let you say about $\lambda$?