What's your working definition of "positive semidefinite" or "positive definite"? In floating point arithmetic, you'll have to specify some kind of tolerance for this.

You could define this in terms of the computed eigenvalues of the matrix. However, you should first notice that the computed eigenvalues of a matrix scale linearly with the matrix, so that for example the matrix I get by multiplying $A$ by a factor of one million has its eigenvalues multiplied by a million. Is $\lambda=-1.0$ negative eigenvalue? If all of the other eigenvalues of your matrix are positive and on the order of $10^{30}$, then $\lambda=-1.0$ is effectively 0 and shouldn't be treated as a negative eigenvalue. Thus it's important to take scaling into account.

A reasonable approach is to compute the eigenvalues of your matrix, and declare that the matrix is numerically positive semidefinite if all eigenvalues are larger than $-\epsilon \left| \lambda_{\max} \right|$, where $ \lambda_{\max}$ is the largest eigenvalue.

Unfortunately, computing all of the eigenvalues of a matrix is rather time consuming. Another commonly used approach is that a symmetric matrix is considered to be positive definite if the matrix has a Cholesky factorization in floating point arithmetic. Computing the Cholesky factorization is an order of magnitude faster than computing the eigenvalues. You can extend this to positive semidefiniteness by adding a small multiple of the identity to the matrix. Again, there are scaling issues. One fast approach is to do a symmetric scaling of the matrix so that the diagonal elements are 1.0 and add $\epsilon$ to the diagonal before computing the Cholesky factorization.

You should be careful with this though, because there are some problems with the approach. For example, there are circumstances where the $A$ and $B$ are postive definite in the sense that they have floating point Cholesky factorizations, but $(A+B)/2$ does not have a Cholesky factorization. Thus the set of "floating point Cholesky factorizable positive definite matrices" isn't convex!

What's your working definition of "positive semidefinite" or "positive definite"? In floating point arithmetic, you'll have to specify some kind of tolerance for this.

You could define this in terms of the computed eigenvalues of the matrix. However, you should first notice that the computed eigenvalues of a matrix scale linearly with the matrix, so that for example, the matrix I get by multiplying $A$ by a factor of one million has its eigenvalues multiplied by a million. Is $\lambda=-1.0$ a negative eigenvalue? If all of the other eigenvalues of your matrix are positive and on the order of $10^{30}$, then $\lambda=-1.0$ is effectively 0 and shouldn't be treated as a negative eigenvalue. Thus it's important to take scaling into account.

A reasonable approach is to compute the eigenvalues of your matrix, and declare that the matrix is numerically positive semidefinite if all eigenvalues are larger than $-\epsilon \left| \lambda_{\max} \right|$, where $ \lambda_{\max}$ is the largest eigenvalue.

Unfortunately, computing all of the eigenvalues of a matrix is rather time consuming. Another commonly used approach is that a symmetric matrix is considered to be positive definite if the matrix has a Cholesky factorization in floating point arithmetic. Computing the Cholesky factorization is an order of magnitude faster than computing the eigenvalues. You can extend this to positive semidefiniteness by adding a small multiple of the identity to the matrix. Again, there are scaling issues. One fast approach is to do a symmetric scaling of the matrix so that the diagonal elements are 1.0 and add $\epsilon$ to the diagonal before computing the Cholesky factorization.

You should be careful with this though, because there are some problems with the approach. For example, there are circumstances where the $A$ and $B$ are postive definite in the sense that they have floating point Cholesky factorizations, but $(A+B)/2$ does not have a Cholesky factorization. Thus the set of "floating point Cholesky factorizable positive definite matrices" isn't convex!

What's your working definition of "positive semidefinite" or "positive definite"? In floating point arithmetic, you'll have to specify some kind of tolerance for this.

You could define this in terms of the computed eigenvalues of the matrix. However, you should first notice that the computed eigenvalues of a matrix scale linearly with the matrix, so that for example the matrix I get by multiplying $A$ by a factor of one million has its eigenvalues multiplied by a million. Is $\lambda=-1.0$ is negative eigenvalue? If all of the other eigenvalues of your matrix are positive and on the order of $10^{30}$, then $\lambda=-1.0$ is effectively 0 and shouldn't be treated as a negative eigenvalue. Thus it's important to take scaling into account.

A reasonable approach is to compute the eigenvalues of your matrix, and declare that the matrix is numerically positive semidefinite if all eigenvalues are larger than $-\epsilon \left| \lambda_{\mbox{max}} \right|$, where $ \lambda_{max}$ is the largest eigenvalue.

Unfortunately, computing all of the eigenvalues of a matrix is rather time consuming. Another commonly used approach is that a symmetric matrix is considered to be positive definite if the matrix has a Cholesky factorization in floating point arithmetic. Computing the Cholesky factorization is an order of magnitude faster than computing the eigenvalues. You can extend this to positive semidefiniteness by adding a small multiple of the identity to the matrix. Again, there are scaling issues. One fast approach is to do a symmetric scaling of the matrix so that the diagional elements are 1.0 and add $\epsilon$ to the diagonal before computing the Cholesky factorization.

You should be careful with this though, because there are some problems with the approach. For example, there are circumstances where the $A$ and $B$ are postive definite in the sense that they have floating point Cholesky factorizations, but $(A+B)/2$ does not have a Cholesky factorization. Thus the set of "floating point Cholesky factorizable positive definite matrices" isn't convex!

What's your working definition of "positive semidefinite" or "positive definite"? In floating point arithmetic, you'll have to specify some kind of tolerance for this.

You could define this in terms of the computed eigenvalues of the matrix. However, you should first notice that the computed eigenvalues of a matrix scale linearly with the matrix, so that for example the matrix I get by multiplying $A$ by a factor of one million has its eigenvalues multiplied by a million. Is $\lambda=-1.0$ negative eigenvalue? If all of the other eigenvalues of your matrix are positive and on the order of $10^{30}$, then $\lambda=-1.0$ is effectively 0 and shouldn't be treated as a negative eigenvalue. Thus it's important to take scaling into account.

A reasonable approach is to compute the eigenvalues of your matrix, and declare that the matrix is numerically positive semidefinite if all eigenvalues are larger than $-\epsilon \left| \lambda_{\max} \right|$, where $ \lambda_{\max}$ is the largest eigenvalue.

Unfortunately, computing all of the eigenvalues of a matrix is rather time consuming. Another commonly used approach is that a symmetric matrix is considered to be positive definite if the matrix has a Cholesky factorization in floating point arithmetic. Computing the Cholesky factorization is an order of magnitude faster than computing the eigenvalues. You can extend this to positive semidefiniteness by adding a small multiple of the identity to the matrix. Again, there are scaling issues. One fast approach is to do a symmetric scaling of the matrix so that the diagonal elements are 1.0 and add $\epsilon$ to the diagonal before computing the Cholesky factorization.

You should be careful with this though, because there are some problems with the approach. For example, there are circumstances where the $A$ and $B$ are postive definite in the sense that they have floating point Cholesky factorizations, but $(A+B)/2$ does not have a Cholesky factorization. Thus the set of "floating point Cholesky factorizable positive definite matrices" isn't convex!