linear algebra quick question

**matt beeson** · June 7th 2008, 11:11 AM

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?

**NonCommAlg** · June 7th 2008, 06:00 PM

Originally Posted by matt beeson

is a positive definite matrix necessarily symmetric?

No! for example the matrix $A=\begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}$ is positive definite but not symmetric. to see why $A$ is positive

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

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

right! in fact $A \bold{x}=0,$ for some $\bold{x} \neq \bold{0},$ if and only if $A$ is not invertible.

**topsquark** · June 7th 2008, 06:18 PM

Originally Posted by NonCommAlg

No! for example the matrix $A=\begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}$ is positive definite but not symmetric. to see why $A$ is positive

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

right! in fact $A \bold{x}=0,$ for some $\bold{x} \neq \bold{0},$ if and only if $A$ is not invertible.

However, doesn't a positive definite matrix have to be at least Hermitian? We can easily see that such a matrix would not need to be symmetric, but if all entries are real then it would have to be.

Perhaps this is a difference in definition?

-Dan

**NonCommAlg** · June 7th 2008, 08:07 PM

Originally Posted by topsquark

However, doesn't a positive definite matrix have to be at least Hermitian? We can easily see that such a matrix would not need to be symmetric, but if all entries are real then it would have to be.

Perhaps this is a difference in definition?

-Dan

No! being Hermitian is not a part of the definition. an $n \times n$ matrix $A$ over $\mathbb{C}$ is positive definite if $\text{Re} (\bold{x}^*A \bold{x}) > 0, \ \forall \ \bold{0} \neq \bold{x} \in \mathbb{C}^n.$

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

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

if $A$ is Hermitian, then $A$ is positive definite if $\bold{x}^* A \bold{x} > 0, \ \forall \ \bold{0} \neq \bold{x} \in \mathbb{C}^n,$ because $\bold{x}^*A \bold{x} \in \mathbb{R},$ for a Hermitian matrix $A.$

**mr fantastic** · June 7th 2008, 08:23 PM

Originally Posted by NonCommAlg

No! for example the matrix $A=\begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}$ is positive definite but not symmetric. to see why $A$ is positive

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

right! in fact $A \bold{x}=0,$ for some $\bold{x} \neq \bold{0},$ if and only if $A$ is not invertible.

Alternatively, A is positive definite because it's eigenvalues are positive:

The matrix M is positive definite if and only if the eigenvalues of M are all positive.

**NonCommAlg** · June 7th 2008, 08:53 PM

Originally Posted by mr fantastic

Alternatively, A is positive definite because it's eigenvalues are positive:

The matrix M is positive definite if and only if the eigenvalues of M are all positive.

this is true only if M is Hermitian. for example $B=\begin{pmatrix} 1 & -1 \\ 2 & 1 \end{pmatrix}$ is positive definite, but its eigenvalues are not even real!

**mr fantastic** · June 7th 2008, 10:36 PM

Originally Posted by NonCommAlg

this is true only if M is Hermitian. for example $B=\begin{pmatrix} 1 & -1 \\ 2 & 1 \end{pmatrix}$ is positive definite, but its eigenvalues are not even real!

Thanks. My mistake. I should have said:

The matrix M is positive definite if and only if all the real eigenvalues of M are all positive.

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