91Ó°ÊÓ

When n=1, we have only one injective linear map \(S_1\). By definition, this is an injective map. Thus, our base case holds.

Let's also establish the base case for n=2. Given two injective linear maps \(S_1\) and \(S_2\), we want to show that their composition, \(S_1 S_2\), is also injective. Suppose that \(S_1 S_2 (x) = S_1 S_2 (y)\) for some vectors x and y. Then, by the linearity of the maps, we have \(S_1 (S_2 (x)) = S_1 (S_2 (y))\). Since \(S_1\) is injective, we know that \(S_2 (x) = S_2 (y)\). And as \(S_2\) is also injective, it follows that \(x = y\). Thus, the composition \(S_1 S_2\) is injective, and our base case for n=2 holds.

Assume that for some integer \(k \geq 2\), the product of k injective linear maps is injective. That is, if \(S_1, S_2, \ldots , S_k\) are injective linear maps, then their composition, \(S_1 S_2 \cdots S_k\), is injective.

Now, we will show that if the product of k injective linear maps is injective, then the product of \(k+1\) injective linear maps is also injective. Consider the injective linear maps \(S_1, S_2, \ldots , S_k , S_{k+1}\). We want to prove that their composition, \(S_1 S_2 \cdots S_k S_{k+1}\), is injective. Notice that we can rewrite this composition as \((S_1 S_2 \cdots S_k)(S_{k+1})\). Let x and y be vectors such that \((S_1 S_2 \cdots S_k S_{k+1})(x) = (S_1 S_2 \cdots S_k S_{k+1})(y)\). Then, by the linearity of the maps, we get \((S_1 S_2 \cdots S_k)(S_{k+1}(x)) = (S_1 S_2 \cdots S_k)(S_{k+1}(y))\). By our Inductive Hypothesis, the composition \(S_1 S_2 \cdots S_k\) is injective. Thus, we can conclude that \(S_{k+1}(x) = S_{k+1}(y)\). Since \(S_{k+1}\) is also injective, it follows that \(x = y\). Therefore, the composition of \(k+1\) injective linear maps, \(S_1 S_2 \cdots S_k S_{k+1}\), is injective, and our inductive proof is complete.

We have proven that for any positive integer n, the composition of n injective linear maps is also injective.