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  1. The determinant of a matrix equals the product of its eigenvalues.
  2. A positive semidefinite matrix is a symmetric matrix with only nonnegative eigenvalues.
  3. A positive definite matrix is a symmetric matrix with only positive eigenvalues.

Combining (1) and (3) yields that a positive definite matrix is always nonsingular since its determinant never becomes zero.

Is is true that for a positive semidefinite matrix at least one of its eigenvalues equals zero and thus its determinant always equals zero => a positive semidefinite matrix is always singular?

You would say that specifically having a positive semidefinite matrix instead of a positive definite matrix implies that at least one of the eigenvalues equals zero.

Is this correct?

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  • $\begingroup$ In a word, yes - if a matrix is only pos semi-definite. Be careful when you specify though, all positive definite matrices are positive semi-definite, but not vice-versa $\endgroup$ – TheMathsGeek Jan 5 '17 at 17:31
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    $\begingroup$ by definition, positive definite(p.d.) matrices are positive semidefinite(p.sd.) matrices, so you can say that the p.d. matrices are the p.sd. matrices that are invertible $\endgroup$ – Exodd Jan 5 '17 at 17:31
  • $\begingroup$ @Exodd That is precisely the confusing element. I would say a positive semidefinite matrix is always singular / a positive definite matrix is always nonsingular. My textbook however states that 'Shows that a positive semidefinite matrix is positive definite if and only if it is nonsingular' (but a psd matrix is automatically singular right?) $\endgroup$ – Anna Jan 5 '17 at 18:03
  • $\begingroup$ @Anna "I would say a positive semidefinite matrix is always singular / a positive definite matrix is always nonsingular." This is incorrect. A positive semidefinite matrix may or may not be singular. If it is not singular, then it is (also) called positive definite. Like Exodd said, positive definite matrices are also positive semidefinite/ $\endgroup$ – ChocolateAndCheese Jan 5 '17 at 19:24
  • $\begingroup$ @ChocolateAndCheese But for any positive semidefinite matrix at least one of its eigenvalues is 0 right? (Otherwise, we would have to call it a positive definite matrix). This would imply that the determinant of the matrix - the product of its eigenvalues - equals 0 and thus that any such matrix is singular. This is what confuses me. $\endgroup$ – Anna Jan 5 '17 at 19:59
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I see there is a bit of confusion, so I'll try to explain better.

A matrix $A$ is positive semidefinite (p.sd.) if it is symmetric and all its eigenvalues are $\ge 0$.

A matrix $A$ is positive definite (p.d.) if it is symmetric and all its eigenvalues are $>0$.

This means that every p.d. matrix is also a p.sd. matrix. The set of positive semidefinite matrices contains the set of positive definite matrices, but there are some p.sd. matrices that are not p.d., like the matrix that has all elements equal to zero.

As an analogy, you can think about the real and integer numbers: all integer numbers are real numbers, because the set of real numbers contains the set of integer numbers, but there are real numbers, such as $\sqrt 2$ that are not integer.

The most common example is the identity matrix $I$: all its eigenvalues are $1>0$, so it is a p.d. matrix, but $1\ge 0$ so it is also a p.sd. matrix. This means that $I$ is a p.sd. matrix and it is invertible.

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If a matrix $M$ is Positive definite then for all non-zero $x$, $x^TMx \gt 0$.

If a matrix $M$ is Positive semidefinite then for all non-zero $x$, $x^TMx \ge 0$.

So, every positive definite matrix is positive semidefinite, but not vice versa.

If there is a matrix $S$ which is positive semidefinite but not positive definite then at least one of its eigen values is zero, hence it is a singular matrix. For proof, you can use schur factorization to diagonalize the matrix $S$ such that $S = QDQ^T$ where $D$ is a real diagonal matrix formed using the eigen values of $S$, and $Q$ is a real (by construction) unitary (hence full-rank) matrix.

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