Hello,Very naive question here...What does it mean for a correlation matrix to be positive definite? I know the mathematical definition but I do not understand the implication. Why is it important that a correlation matrix be positive definite and what is the implicaiton if it is not? Are there any useful links please?

It implies for instance that you can apply Cholesky's decomposition to the matrix.

It means you can easily solve a linear a equation system, but the most important application in finance is:If it is positive definite, you can do Cholesky Decomposition that you have correlation matrix.

The precise answer to your question is:Fact. A matrix M is positive semidefinite iff M is the correlation matrix of some n-dimensional multivariate random variable (i.e. with state space R^n). We can replace semidefinite with definite if we insist that every component of the random variable has nonzero variance.

I have a question about complex positive definite matricies.According this wikipedia article, if x A x* is real for all complex vectors x, then A must be Hermitian. Does anyone know how to prove this?

@repoman: the thing here that matters is, that you cannot restrict to real vectors - complex is right, even for real matrixes. For a proof see for instance, Golub Van Loan.

QuoteOriginally posted by: trippel@repoman: the thing here that matters is, that you cannot restrict to real vectors - complex is right, even for real matrixes. For a proof see for instance, Golub Van Loan.Do you mean "Matrix Computations"? I did not see anything like that in there.

QuoteOriginally posted by: repomanI have a question about complex positive definite matricies... if x A x* is real for all complex vectors x, then A must be Hermitian. Here's my proof: By induction on n where A is n x n. Take n = 2 for the base case. Then taking x_1, x_2 to both be nonzero reals, x A x* is real implies Im(a_12)=-Im(a_21). And x_1 = 1 + i, x_2 = 1 - i gives Re(a_12)=Re(a_21). This proves that A is Hermitian.(There must be more informative way.)The induction step is trivial. For n >= 3, take any indicies 1 <= k,l <= n. Since A has at least 3 row and columns, there exists 1 <= m <= n different from both k and l. Letting B be the (n-1) x (n-1) minor matrix obtained by removing the m-th row and colum from A, by the induction hypothesis B is Hermitian. Since a_kl, a_lk are both elements of B it follows that a_kl is the complex conjugate of a_lk.

QuoteOriginally posted by: repomanQuoteOriginally posted by: trippel@repoman: the thing here that matters is, that you cannot restrict to real vectors - complex is right, even for real matrixes. For a proof see for instance, Golub Van Loan.Do you mean "Matrix Computations"? I did not see anything like that in there.I did indeed mean that book, but I have to admit I didn't read it. I know for sure, that it can be found in the book "Numerical Mathematics" by Quarteroni.

QuoteOriginally posted by: repomanQuoteOriginally posted by: repomanI have a question about complex positive definite matricies... if x A x* is real for all complex vectors x, then A must be Hermitian. Here's my proof: By induction on n where A is n x n. Take n = 2 for the base case. Then taking x_1, x_2 to both be nonzero reals, x A x* is real implies Im(a_12)=-Im(a_21). And x_1 = 1 + i, x_2 = 1 - i gives Re(a_12)=Re(a_21). This proves that A is Hermitian.(There must be more informative way.)The induction step is trivial. For n >= 3, take any indicies 1 <= k,l <= n. Since A has at least 3 row and columns, there exists 1 <= m <= n different from both k and l. Letting B be the (n-1) x (n-1) minor matrix obtained by removing the m-th row and colum from A, by the induction hypothesis B is Hermitian. Since a_kl, a_lk are both elements of B it follows that a_kl is the complex conjugate of a_lk.Base case should be n=1.Despite, there is another simple proof:Since $\forall z\in\mathbb{C}: z^TA\bar z\in\mathbb{R}$ it follows $=\bar{z^TA\bar z} = {\bar z}^T\bar{A} z = z^T\bar{A}^T\bar{z}$ Btw. LaTeX Support sucks in this board!

QuoteOriginally posted by: trippelQuoteOriginally posted by: repomanQuoteOriginally posted by: repomanI have a question about complex positive definite matricies... if x A x* is real for all complex vectors x, then A must be Hermitian. Here's my proof: By induction on n where A is n x n. Take n = 2 for the base case. Then taking x_1, x_2 to both be nonzero reals, x A x* is real implies Im(a_12)=-Im(a_21). And x_1 = 1 + i, x_2 = 1 - i gives Re(a_12)=Re(a_21). This proves that A is Hermitian.(There must be more informative way.)The induction step is trivial. For n >= 3, take any indicies 1 <= k,l <= n. Since A has at least 3 row and columns, there exists 1 <= m <= n different from both k and l. Letting B be the (n-1) x (n-1) minor matrix obtained by removing the m-th row and colum from A, by the induction hypothesis B is Hermitian. Since a_kl, a_lk are both elements of B it follows that a_kl is the complex conjugate of a_lk.Base case should be n=1.No, the induction step does not work to get from n = 1 to n = 2. But of course your simple proof is better.

But why does a correlation matrix have to be positive definite? Thanks,

Because it is invalid in this situation. If it wasn't (semi) positive definite then you could construct a linear combination of the variables in the matrix whose sum has a negative variance.

QuoteOriginally posted by: outrunQuoteOriginally posted by: ACDBecause it is invalid in this situation. If it wasn't (semi) positive definite then you could construct a linear combination of the variables in the matrix whose sum has a negative variance.I think there is subtle difference: its more like a combination of covariances that can only be valid if one could construct a negative variance sequence (which you can;t) I don;t see how you can construct a sequence of real numbers (via linear combination or by hand, or via some other method) with negative variance.If the matrix is not (semi) positive definite then it's possible, hence the matrix is invalid. Assume we have the following correlation matrix:1, 0.9, -0.50.9, 1.0, 0.9-0.5, 0.9, 1.0Let's say each item has unit variance so that the correlation matrix and covariance matrix are identical. If we took the sum of the 3 variables held at these proportions under this matrix (this is just the eigenvector corresponding to the negative eigenvalue):0.546062-0.625320.546062Then the variance of this sum under that matrix is -0.48748 (clearly nonsense caused by the matrix being invalid as it is not positive definite). Your sum above is assuming a diagonal covariance matrix which is always positive definite if the diagonal elements are positive.