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# 1. Transposes and symmetry

Recall that the transpose AT of an n×m matrix A is the m×n matrix whose entries are given by the rule (AT)ij = Aji. A matrix is symmetric if it equals its own transpose.

1. Use the definition of matrix multiplication to show that, for any compatible matrices A and B, (AB)T = BTAT.

2. Show that for any matrix A, both of AAT and ATA exist and are symmetric.

3. Prove or disprove: If AAT = ATA, then A is symmetric.

## 1.1. Solution

1. We want to show that ((AB)T)ij = (AB)ji = (BTAT)ij. From the definition of matrix multiplication we have (AB)ji = ∑k AjkBki and (BTAT)ij = ∑k (BT)ik(AT)kj = ∑k AjkBki = (AB)ji.

2. Existence depends only on having compatible dimensions; if A is an n×m matrix, then AT is an m×n matrix, and so both products AAT (n×m by m×n) and ATA (m×n by n×m) work. To show there are symmetric, use the answer to part 1 to observe that (AAT)T = (AT)TAT = AAT and similarly (ATA)T = AT(AT)T = ATA.

3. Disproof: Let A = . Then an easy computation shows AAT = ATA = I, but A is not symmetric.

How I found the counterexample matrix for the last part: Suppose A is a 2-by-2 matrix and AAT = ATA. Let A = and compute AAT and ATA. Matching up entries gives a2+b2 = a2+c2 from which b2 = c2; d2+b2 = d2+c2; and ac + bd = ab + cd. The second equation follows from the first, and the third can be made true by setting a = d = 0. This leaves b2 = c2, which we can make true without having b=c by setting b = -c.

Other counterexamples probably exist.

# 2. Finite paths

Recall that the adjacency matrix A of a directed graph G = (V,E), where V = {1..n}, is given by the rule Aij = 1 if ij∈E and 0 otherwise.

Use the fact that (Ak)ij counts the number of i-j paths in G of length n to show that if G is acyclic, then the matrix (A-I) is invertible. Hint: Count the total number of i-j paths in G.

## 2.1. Solution

Following the hint, we compute ∑ Ak. No path can have more than n-1 edges without repeating vertices; so if the graph is acyclic, it follows that there are no paths of length k > n-1 and thus that Ak = 0 for all k > n-1, implying the sum converges. But then we have (A-I)(∑ Ak) = A ∑ An - ∑ Ak = -A0 = -I. So (A-I)-1 = -∑ Ak.

# 3. Convex bodies

A set of vectors S is convex if, for any two vectors x and y in S, and any scalar a with 0≤a≤1, the vector ax+(1-a)y is also in S.

1. Prove that if S is convex and T is convex, then S∩T is convex.
2. Give an example that shows even if S and T are convex, S∪T is not necessarily convex.
3. Show that for any vector z and constant b, the half-space H = { x | x⋅z≤b } is convex.

4. Show that for any matrix A and column vector b (of appropriate dimensions), the set of all vectors x such that Ax≤b is convex.1

## 3.1. Solution

1. Let x and y be in S∩T. Then x,y∈S and convexity of S implies z=ax+(1-a)y ∈ S. Similarly x,y∈T implies z∈T. Thus z∈S∩T and S∩T is convex.
2. Consider the sets S = { 0 } and S = { 1 } in ℝ1; then S∪T does not contain (1/2)0 + (1/2)1 = 1/2.

3. Let x⋅z≤b and y⋅z≤b. Then (ax+(1-a)y)⋅z = a(x⋅z) + (1-a)(y⋅z) ≤ ab + (1-a)b = b.
4. Observe that Ax≤b if and only if Ai⋅x ≤ bi for each i, where Ai is the i-th row of A. So the set of points S satisfying the inequality Ax≤b is the intersection of the sets Si = { x | Ai⋅x ≤ bi }. Since each set Si is convex (by the previous part) and the intersection of convex sets is convex (use part 1 plus induction on the number of sets), S is convex.

1. When x and y are vectors, the notation x≤y means that xi≤yi for all indices i. (1)

2014-06-17 11:57