is a positive definite matrix necessarily symmetric?

Also if A is an nxn matrix and is singular (not invertible) then there is an x in Rn such that Ax=0 right?

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- Jun 7th 2008, 11:11 AMmatt beesonlinear algebra quick question
is a positive definite matrix necessarily symmetric?

Also if A is an nxn matrix and is singular (not invertible) then there is an x in Rn such that Ax=0 right? - Jun 7th 2008, 06:00 PMNonCommAlg
No! for example the matrix $\displaystyle A=\begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix} $ is positive definite but not symmetric. to see why $\displaystyle A$ is positive

definite, let $\displaystyle \bold{x}=\begin{pmatrix}a+bi \\ c+di \end{pmatrix} \neq \bold{0} $. then see that $\displaystyle \text{Re} \ \bold{x}^*A\bold{x}=\frac{1}{4}((2a+c)^2+(2b+d)^2 + 3(c^2+d^2)) > 0.$

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Also if A is an nxn matrix and is singular (not invertible) then there is an x in Rn such that Ax=0 right?

- Jun 7th 2008, 06:18 PMtopsquark
- Jun 7th 2008, 08:07 PMNonCommAlg
No! being Hermitian is not a part of the definition. an $\displaystyle n \times n$ matrix $\displaystyle A$ over $\displaystyle \mathbb{C}$ is positive definite if $\displaystyle \text{Re} (\bold{x}^*A \bold{x}) > 0, \ \forall \ \bold{0} \neq \bold{x} \in \mathbb{C}^n.$

if all entries of $\displaystyle A$ are real, then it's easy to see that in order to prove that $\displaystyle A$ is positive definite, we only need to show that

$\displaystyle \bold{x}^T A \bold{x} > 0, \ \forall \ \bold{0} \neq \bold{x} \in \mathbb{R}^n.$ using this fact, we can see easier why the matrix $\displaystyle A$ in my previous post is positive definite.

if $\displaystyle A$ is Hermitian, then $\displaystyle A$ is positive definite if $\displaystyle \bold{x}^* A \bold{x} > 0, \ \forall \ \bold{0} \neq \bold{x} \in \mathbb{C}^n,$ because $\displaystyle \bold{x}^*A \bold{x} \in \mathbb{R},$ for a Hermitian matrix $\displaystyle A.$ - Jun 7th 2008, 08:23 PMmr fantastic
- Jun 7th 2008, 08:53 PMNonCommAlg
- Jun 7th 2008, 10:36 PMmr fantastic