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

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.

• Sorry i don't know the vector notation in latex, also i'll edit the question a bit sorry – L G Jan 10 at 12:56
• The definition is not what you say it is! (If that were the definition then there'd be nothing to prove - what you say you want to prove would reduce to $A$ is positive definite if and only if $A$ is positive definite.) – David C. Ullrich Jan 10 at 14:32
• Obligaroty remark: for beginner questions about positive definiteness, always remember to state the definition of positive definite matrix, because different authors define positive definite matrix differently. E.g. in one old textbook that I've read, a real symmetric matrix $A$ is said to be positive definite if $A=P^TDP$ for some nonsingular matrix $P$ and some positive diagonal matrix $D$. – user1551 Jan 11 at 13:25

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 $$\lambda$$?
• Would i be right in introducing a diagonal matrix (let's call it D) and an orthogonal matrix (let's say B) and then saying how $M = BD{B^{-1}}$ – L G Jan 10 at 13:06
• @LG You certainly could, but there's no need. We can immediately use the definitions to conclude that $x^TMx$ is positive and that $x^TMX=x^T\lambda x,$ and since $\lambda$ is a scalar, then $x^TMx=\lambda x^Tx.$ Can you take it from there? – Cameron Buie Jan 10 at 13:32
• I think i understand, since ${x^T}Mx$ is positive then ${\lambda}{x^T}x$ is also positive and since ${x^T}x$ is positive (as ${||x||^2}$) then $\lambda$ must also be positive – L G Jan 10 at 13:32