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2 Proof. ( ) Let A be an m n matrix with Hadamard inverse Â. Then we know A Â J mn. That is, [A Â] ij [A] ij [Â] ij 1. Multiplying by inverses in C we know that [Â] ij (1)(([A] ij ) 1 ) ([A] ij ) 1, which is only possible when all entries of A are invertible (in C). In other words, [A] ij 0 for all 1 i m, 1 j n. ( ) Take any m n matrix A with entries in C such that [A] ij 0 for all 1 i m, 1 j n. Then there exists ([A] ij ) 1 for all i,j. This implies [A] ij ([A] ij ) 1 ([A] ij ) 1 [A] ij 1, and so A has an inverse Â defined by [Â] ij ([A] ij ) 1 for all i,j. Note: again we have denoted the Hadamard inverse as Â as to avoid confusion with the usual matrix inverse, A 1. The Hadamard identity matrix and the Hadamard inverse are both more limiting than helpful, so we will not explore their use further. One last fun fact: the set of m n matrices with nonzero entries form an abelian (commutative) group under the Hadamard product (Prove this!). Theorem 1.5. The Hadamard Product is Linear. Suppose α C, and A, B and C are m n matrices. Then C (A + B) C A + C B. Furthermore, α(a B) (αa) B A (αb). Proof. Part 1. Part 2. [C (A + B)] ij [C] ij [A + B] ij [C] ij ([A] ij + [B] ij ) [C] ij [A] ij + [C] ij [B] ij [C A] ij + [C B] ij [C A + C B] ij [α(a B)] ij α[a B] ij α[a] ij [B] ij [αa] ij [B] ij [αa B] ij First Equality α[a] ij [B] ij [A] ij α[b] ij [A] ij [αb] ij [A αb] ij Second Equality 2 Diagonal Matrices and the Hadamard Product We can relate the Hadamard product with matrix multiplication via considering diagonal matrices, since A B AB if and only if both A and B are diagonal. For example, a simple calculation reveals that the Hadamard product relates the diagonal values of a diagonalizable matrix A with its eigenvalues: Theorem 2.1. Let A be a diagonalizable matrix of size n with eigenvalues λ 1, λ 2,..., λ n and diagonalization A SDS 1 where D is a diagonal matrix such that [D] ii λ i for all 1 i n. Also, let [A] ij a ij for all i,j. Then a 11 a 22. a nn [ S (S 1 ) ] T 2 λ 1 λ 2. λ n (304, [4]).

3 Proof. Let d C n such that [d] i λ i for all 1 i n. We will show a ii [S (S 1 ) T d] i. a ii [A] ii [SDS 1 ] ii [S] ik [D] kj [S 1 ] ji j1 k1 [S] ik [D] kk [S 1 ] ki k1 [S] ik λ k [S 1 ] ki k1 [S] ik [S 1 ] ki λ k k1 [S] ik [(S 1 ) T ] ik λ k k1 [S (S 1 ) T ] ik λ k k1 [S (S 1 ) T ] ik [d] k k1 [S (S 1 ) T d] i We obtain a similar result when we look at the singular value decomposition of a square matrices: Theorem 2.2. Let A be a matrix of size n with singular values σ 1, σ 2,..., σ n and singular value decomposition A UDV, where D is a diagonal matrix with [D] ii σ i for all 1 i n. Then, a 11 σ 1 a 22. [ U V ] σ 2. a nn σ n Note that this relation only makes sense with square matrices, since otherwise we cannot compute U V. Proof. Note that (V ) T V. The proof is similar to our proof of (2.1). We have started to see that the Hadamard product behaves nicely with respect to diagonal matrices and normal matrix multiplication. The following are some more general properties that expand on this idea. The proofs all involve straightforward sum manipulation and are left as exercises for the reader. Theorem 2.3. Suppose A, B are m n matrices, and D and E are diagonal matrices of size m and n, respectively. Then, D(A B)E (DAE) B (DA) (BE) (AE) (DB) A (DBE) (304, [4]) 3

4 We will need a quick definition before our next theorem. Definition 2.4. Define the diagonal matrix, D x, of size n with entries from a vector x C n by { [x]i if i j, [D x ] ij 0 otherwise. Theorem 2.5. Suppose A, B are m n matrices and x C n. Then the ith diagonal entry of the matrix AD x B T coincides with the ith entry of the vector (A B)x for all i 1, 2,..., m (305, [4]). That is, [AD x B T ] ii [(A B)x] i for all 1 i m. Theorem 2.6. Let A, B and C be m n matrices. Then the ith diagonal entry of the matrix (A B)C T coincides with the ith diagonal entry of the matrix (A C)B T. That is, [(A B)C T ] ii [(A C)B T ] ii for all 1 i m (305, [4]). 3 Schur Product Theorem The purpose of this section is to set up and prove the Schur Product Theorem. This theorem relates positive definite matrices to the Hadamard product, which is important when analyzing the determinants of matrices since we want real, nonnegative numbers to compare. Note that if A is positive semidefinite of size n then clearly A n i1 λ i 0 since λ i 0 for all 1 i n. Lemma 3.1. Any rank one positive semidefinite matrix A of size n can be written as the product A xx T where x is some vector in C n. Proof. This proof follows the constructive proof of Theorem ROD [3]. Let A be a rank one positive semidefinite matrix of size n. Since A is positive semidefinite, we know it is Hermitian and therefore normal. This allows us to write the orthonormal diagonalization A UDU where D is the matrix of eigenvalues of A and U is a unitary matrix made up of orthonormalized eigenvectors of A (Note that U has real entries, so U U T ). Also since A is positive semidefinite, we know all the diagonal values of D are nonnegative numbers. In the proof of Theorem ROD [3] we now let X {u 1,..., u n } be the columns of U and let Y {u 1,..., u n } be the rows of U (which are the rows of U T ) converted to column vectors (which are the columns of U). Next we define A k λ k u k u T k and we see that A A A n. But A is rank one, so (ordered properly) A A 1 + O O A 1 λ 1 u 1 u T 1. We also know λ 1 > 0, so we can define x λ 1 u 1. So now we have xx T ( λ 1 u 1 )( λ 1 u 1 ) T λ 1 u 1 u T 1 λ1 λ1 λ1 u 1 u T 1 λ 1 u 1 u T 1 A. Therefore any positive semidefinite matrix A of rank one and size n can be written xx T, where x C n. Lemma 3.2. Let B be a positive semidefinite matrix of size n and rank r. Then the matrices B 1,..., B r from the rank one decomposition of B are all also positive semidefinite. Proof. Let B be a positive semidefinite matrix of size n and rank r with rank one decomposition B B B r. Then from the constructive proof of Theorem ROD [3] together with the orthonormal diagonalization of B we know that B k UD k U where U is a unitary matrix and D k is the diagonal matrix with [D k ] kk λ k and [D k ] ij 0 for all other entries. Since B is positive 4

5 semidefinite, then we know that all eigenvalues are positive (as ordered for the decomposition). That is, λ k > 0 for all 1 k r. Clearly then D k is positive semidefinite for all 1 k r. Now consider < B k x, x > for any x C n. Then < B k x, x >< UD k U x, x >< D k U x, U x > 0. This is true for all 1 k r and so B k is positive semidefinite for all k. Lemma 3.3. Let B be a matrix of size n and rank r. Furthermore, suppose the matrices B 1,..., B r from the rank one decomposition of B are all positive semidefinite. Then B itself is positive semidefinite. Proof. Let B B B r as described above and let x C n. Then Therefore B is positive semidefinite. < Bx, x > < (B B r )x, x > < B 1 x B r x, x > < B 1 x, x > < B r x, x > 0. Theorem 3.4. Schur Product Theorem. Suppose A and B are positive semidefinite matrices of size n. Then A B is also positive semidefinite. Proof. (141, [2]). Let A and B be positive semidefinite matrices of size n. Suppose B is of rank zero. This is true if and only if B O, and therefore A B O, which is clearly positive semidefinite. Now suppose B is of rank one. Then we can write B xx T for some vector x C n (Lemma 3.1). Then [A B] ij [A] ij [B] ij [A] ij [x] i [x T ] j [D x AD x ] ij. Now, since A is positive semidefinite, then so is D x AD x. Take any vector v C n. Then < D x AD x v, v > < AD x v, (D x ) v > < AD x v, D x v > < A(D x v), (D x v) > 0. Now suppose B is of rank r, 1 < r n. Then we can decompose B into a sum of rank one matrices B 1,..., B r (Theorem ROD, [3]) where each matrix B i is also positive semidefinite (Lemma 3.2). Then A B A (B B r ) A B A B r (Theorem 1.5). We know that each A B i is positive semidefinite for each i, and so A B is positive semidefinite (Lemma 3.3). Therefore for any two positive semidefinite matrices A and B, A B is also positive semidefinite. 4 Some Inequalities Now for some matrix analysis. In mathematics, the term analysis means there are tons of inequalities (I have no proof for this). Recall the importance of the Schur Product Theorem is that if two matrices A and B are positive semidefinite, then so is A B. This allows us to compare determinants of these matrices, since they are always nonnegative, real numbers. Theorem 4.1. Oppenheim s Inequality. Let A and B be positive semidefinite matrices of size n. Then A B [A] 11 [A] nn B. 5

6 I am not including a proof of Oppenheim s Inequality because it is somewhat long and requires a bit of setup. However, it is not hard to follow and so instead I will simply refer you to page 144 of Bapat [2]. Theorem 4.2. Hadamard s Inequality. Suppose A is positive semidefinite of size n. Then A [A] 11 [A] nn. Proof. Let A be any positive semidefinite matrix of size n. Note that I n is a postive semidefinite matrix of size n. Now we have the following: A [I n ] 11 [I n ] nn A I n A (Oppenheim s Inequality) [A] 11 [A] nn Corollary 4.3. Let A and B be positive semidefinite matrices of size n. Then A B AB. Proof. A B [A] 11 [A] nn B (Oppenheim s Inequality) A B (Hadamard s Inequality) AB Theorem 4.4. Let A and B be positive semidefinite matrices of size n. Let λ 1,..., λ n be the eigenvalues of A B and let λ 1,..., λ n be the eigenvalues of AB. Then n λ i ik n λ i, k 1, 2,..., n. (144, [2]) ik Note that if k 1, then this is true by Corollary 4.3. This theorem dates back to Oppenheim in 1930, but has only recently been proven by Ando [1] and Visick [6] in 1995 (144, [2]). This last inequality does not require A and B to be positive semidefinite, although the statement can be made stronger if they are (107 [5]). Theorem 4.5. Let A and B be square matrices of size n. Then rank (A B) (rank A)(rank B). Proof. We will write the rank one decompositions of A and B. Suppose A has rank ρ 1 with eigenvalues λ k, 1 k n and B has rank ρ 2 with eigenvalues λ l, 1 l n. Then by the constructive proof of Theorem ROD [3] ρ 1 A λ k x k yk T, and B k1 ρ 2 l1 λ l v l w T l. 6

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