1. Outline
Eigenvalue
Trace
Rank
Determinant

2. Eigenvalue of A Matrix The definition of eigenvalues
Important properties on the eigenvalues
eig(AB) = eig(BA)
[A]n*n at most n different eigenvalues
rank(A) = r, then A at most r nonzero eigenvalues
eig(A-1) = 1/eig(A)
eig(I + cA) = 1 + c*eig(A), eig(A - cI) = eig(A)-c

3. Eigenvalue of A Matrix eig(AB) = eig(BA)
rank(A) = r, then A at most r nonzero eigenvalues
The eigenvectors corresponding to different eigenvalues for linearly independent
eig(A-1) = 1/eig(A)
eig(I + cA) = 1 + c*eig(A), eig(A - cI) = eig(A)-c

4. Eigenvalue of A Matrix eig(AB) = eig(BA)
rank(A) = r, then A at most r different nonzero eigenvalues
The eigenvectors corresponding to different eigenvalues for linearly independent

5. Positiveness and Negativeness
Hermitian Matrix
Semi-positiveness and eigenvalues
Positive definite: all positive eigenvalues
Positive semidefinite: all non-negative eigenvalues
Negative definite: all negative eigenvalues
Negative semidefinite: all non-positive eigenvalues

6. Trace of A Matrix The summation of diagonal elements
Important properties on the trace
tr(AB) = tr(BA), very important
tr(A) = summation of eigenvalues
Eigenvalue polynomial:
tr(A2) ≤ tr(ATA)

7. Rank of A Matrix Maximum number of linearly independent rows/columns
Important properties on the rank
rank(A+B) ≤ rank([A B]) ≤ rank(A) + rank(B)
[A]m*n, [B]n*p: rank(A) + rank(B) - n ≤ rank(AB)
rank(A) = rank(AAH) = rank(AHA)
[A]m*m, rank(A) = m ↔ det(A) ≠ 0 ↔ A non-singular

8. Rank of A Matrix rank(A+B) ≤ rank([A B]) ≤ rank(A) + rank(B)
A+B = [A B]*[I I]T
[A B] has linearly dependent rows for any rank(A) + rank(B) columns
[A]m*n, [B]n*p: rank(A) + rank(B) - n ≤ rank(AB)
Consider the orthogonal space of A, denoted as A┴
rank(B) = rank([A A┴]B) ≤ rank(AB) + rank(A┴B)
≤ rank(AB) + rank(A┴) = rank(AB) + n – rank(A)

9. Rank of A Matrix rank(A) = rank(AAH) = rank(AHA)
It is easy to prove that rank(A) ≥ rank(AAH);
On the other hand, we have the following
rank(A) = rank([AH (A┴)H]A) ≤ rank(AHA) + rank(A┴HA)
= rank(AHA) + 0 = rank(AHA)

10. Determinant of A Matrix
Determinant and Volume
|det(A)| = volume of parallelepiped spanned by columns of A;
Hardmard Determinant Bound
V = [V1 V2 … VN], |det(V)| ≤ ∏||Vn||2
If A is not full rank, then volume is zero;
and det(A) = 0

11. Determinant of A Matrix
Important properties on the determinant
Hermitian matrix A: det(A) real
det(A) = det(AH) = conj(det(AT))=conj(det(A))
det(A-1) = det(A)-1
det(AB) = det(A)det(B)

12. Determinant of A Matrix
Important properties on the determinant
Hermitian matrix A: det(A) real
|det(AHB)|2 ≤ det(AHA)det(BHB)
Positive definition A: det(A) > 0
Positive semi-definite A: det(A) ≥ 0
A and B positive semi-definite: det(A+B) ≥ det(A) + det(B)
A positive definite, B positive semi-definite: det(A+B) ≥ det(A)

13. Determinant of A Matrix
|det(AHB)|2 ≤ det(AHA)det(BHB)
det(A) = volume of parallelepiped spanned by columns of A;
V = [V1 V2 … VN], |det(V)| ≤ ∏||Vn||2 ≤ 1
SVD of A and B, the final step is to prove that the determinant of a submatrix of unitary matrix is less than or equal to one;

14. Determinant of A Matrix
A and B positive semi-definite: det(A+B) ≥ det(A) + det(B)
We can prove: det(A+I) ≥ det(A) + 1, via eigenvalue decomposition of A
det(B-1/2AB-1/2+I) ≥ det(B-1/2AB-1/2) + 1;
which is equivalent to
det(A+I) ≥ det(A) + 1;
Then perform EVD for A, we need to prove that the above is satisfied for diagonal matrix A

15. Determinant of A Matrix
A positive definite, B positive semi-definite : det(A+B) ≥ det(A)
Proof from information theory. Consider Gaussian random variables X and Y, with covariance matrix A and B. We have the differential entropy
h(X+Y) ≥ h(X) = 1/2log2{2πedet(A)}
Note that we have h(X+Y) ≤ 1/2log2{2πedet(A+B)}
Thus we have 1/2log2{2πedet(A+B)} ≥ 1/2log2{2πedet(A)}