Question

Show that to fully parenthesize an expression having \(n\) matrices we need \(n-1\) pairs of parentheses.

Short answer

We have shown through the process of mathematical induction that to fully parenthesize an expression having \(n\) matrices, we need \(n-1\) pairs of parentheses.

Step-by-step solution

Define Full Parenthesization

When we say full parenthesization, it simply means that we place parentheses in the expression in such a way to indicate the order of multiplication. For instance, if we have three matrices \(A\), \(B\), \(C\), then a full parenthesization would look like either \((AB)C\) or \(A(BC)\).

Base Case

To start proving the exercise, the base case is the starting point. This is where we have only one matrix, hence \(n=1\). Since we have just one matrix, there is no need for any set of parentheses, so 0 pair is needed. Therefore, it follows that we need \(n-1 = 1 - 1 = 0\) pairs of parentheses which verifies the base case.

Inductive Hypothesis

Let's assume the statement is true for \(k\) matrices, that is we need \(k-1\) pairs of parentheses for \(k\) numbers. This is our inductive hypothesis.

Inductive Step

We need to verify the statement is true for \(k+1\) matrices assuming it's true for \(k\) matrices. When we have \(k+1\) matrices, we divide these matrices into two groups, the first group having \(i\) matrices (\(1 \leq i \leq k\)) and the second group having \(k+1-i\) matrices. Each group will need to be fully parenthesized which by our inductive hypothesis, will require \(i-1\) and \(k+1-i-1=k-i\) pairs of parentheses respectively. Additionally, one more pair of parentheses is needed for the total product. Therefore, in total, we need \((i-1) + (k-i) + 1 = k-1+1=k\) pairs of parentheses for \(k+1\) matrices, hence the proof.