# Thread: linear algebra quick question

1. ## linear 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?

2. Originally Posted by matt beeson
is a positive definite matrix necessarily symmetric?
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 $\displaystyle \text{Re} \ \bold{x}^*A\bold{x}=\frac{1}{4}((2a+c)^2+(2b+d)^2 + 3(c^2+d^2)) > 0.$

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

3. Originally Posted by NonCommAlg
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 $\displaystyle \text{Re} \ \bold{x}^*A\bold{x}=\frac{1}{4}((2a+c)^2+(2b+d)^2 + 3(c^2+d^2)) > 0.$

right! in fact $\displaystyle A \bold{x}=0,$ for some $\displaystyle \bold{x} \neq \bold{0},$ if and only if $\displaystyle A$ 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

4. 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 $\displaystyle n \times n$ 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 $\displaystyle A.$

5. Originally Posted by NonCommAlg
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 $\displaystyle \text{Re} \ \bold{x}^*A\bold{x}=\frac{1}{4}((2a+c)^2+(2b+d)^2 + 3(c^2+d^2)) > 0.$

right! in fact $\displaystyle A \bold{x}=0,$ for some $\displaystyle \bold{x} \neq \bold{0},$ if and only if $\displaystyle A$ 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.

6. 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 $\displaystyle B=\begin{pmatrix} 1 & -1 \\ 2 & 1 \end{pmatrix}$ is positive definite, but its eigenvalues are not even real!

7. Originally Posted by NonCommAlg
this is true only if M is Hermitian. for example $\displaystyle B=\begin{pmatrix} 1 & -1 \\ 2 & 1 \end{pmatrix}$ 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.