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  1. #1
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    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?
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    Quote Originally Posted by matt beeson View Post
    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.
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    Quote Originally Posted by NonCommAlg View Post
    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
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    Quote Originally Posted by topsquark View Post
    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.
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    Quote Originally Posted by NonCommAlg View Post
    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.
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    Quote Originally Posted by mr fantastic View Post
    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!
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    Quote Originally Posted by NonCommAlg View Post
    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.
    Last edited by mr fantastic; June 8th 2008 at 12:21 AM.
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