91Ó°ÊÓ

To prove that \(T\) is one-to-one if \(UT\) is one-to-one, we will show that if \(T\) is not one-to-one, then \(UT\) cannot be one-to-one. Recall that a transformation is one-to-one if every distinct input maps to a distinct output. Assume that \(T\) is not one-to-one. Then there exists distinct vectors \(v_1, v_2 \in V\) such that \(T(v_1) = T(v_2)\). Now, let's consider the transformation \(UT\). We have: \(UT(v_1) = U(T(v_1)) = U(T(v_2)) = UT(v_2)\) Since \(v_1 \neq v_2\) but \(UT(v_1) = UT(v_2)\), the transformation \(UT\) cannot be one-to-one. Therefore, if \(UT\) is one-to-one, \(T\) must be one-to-one.

No, \(U\) does not have to be one-to-one. Consider the following example: Let \(V=W=Z=\mathbb{R}\) and let \(T, U\) be the linear transformations defined by \(T(x) = x\) and \(U(y) = 2y\). Now, we can see that: \(T\) is one-to-one, as \(T(x_1) = T(x_2) \Rightarrow x_1=x_2\); \(U\) is not one-to-one, as \(U(1) = U(-1) = 2\); \(UT = U(T(x)) = U(x) = 2x\) is one-to-one, as \(UT(x_1) = UT(x_2) \Rightarrow 2x_1=2x_2 \Rightarrow x_1 = x_2\). This example demonstrates that \(U\) does not have to be one-to-one even if both \(T\) and \(UT\) are one-to-one.

To prove that \(U\) is onto if \(UT\) is onto, we will show that if \(U\) is not onto, then \(UT\) cannot be onto. Recall that a transformation is onto if for every element in the codomain, there exists an element in the domain that maps to it. Assume that \(U\) is not onto. Then, there exists a vector \(z \in Z\) such that for all \(w \in W\), \(U(w) \neq z\). Now, let's consider the transformation \(UT\). For every \(v \in V\), we have \(UT(v) = U(T(v))\), which implies that \(UT(v)\) is in the image of \(U\). However, since \(z\) is not in the image of \(U\), it cannot be in the image of \(UT\), and therefore \(UT\) cannot be onto. Thus, if \(UT\) is onto, it must be the case that \(U\) is onto.

No, \(T\) does not have to be onto. Consider the following example: Let \(V=W=Z=\mathbb{R}\) and let \(T, U\) be the linear transformations defined by \(T(x) = 2x\) and \(U(y) = y\). Now, we can see that: \(T\) is not onto, as \(T(x) = 1\) has no solution for \(x\); \(U\) is onto, as \(U(y_1) = U(y_2) \Rightarrow y_1=y_2\); \(UT = U(T(x)) = U(2x) = 2x\) is onto, as for any \(z \in Z\), we can find \(x = \frac{z}{2} \in V\) such that \(UT(x) = 2x = z\). This example demonstrates that \(T\) does not have to be onto even if both \(U\) and \(UT\) are onto.

Let's assume that \(U\) and \(T\) are both one-to-one and onto. To prove that \(UT\) is also one-to-one and onto, we need to show that: 1. For every distinct input \(v_1, v_2 \in V\), \(UT(v_1) \neq UT(v_2)\). 2. For every \(z \in Z\), there exists \(v \in V\) such that \(UT(v) = z\). For the first part, suppose \(UT(v_1) = UT(v_2)\). This means that \(U(T(v_1)) = U(T(v_2))\). Since \(U\) is one-to-one, \(T(v_1) = T(v_2)\). But \(T\) is also one-to-one, so \(v_1 = v_2\). Thus, if \(v_1 \neq v_2\), then \(UT(v_1) \neq UT(v_2)\), and so \(UT\) is one-to-one. For the second part, suppose \(z \in Z\). Since \(U\) is onto, there exists a vector \(w \in W\) such that \(U(w) = z\). Moreover, since \(T\) is onto, there exists a vector \(v \in V\) such that \(T(v) = w\). Now, we have: \(UT(v) = U(T(v)) = U(w) = z\) Thus, for every \(z \in Z\), there exists a vector \(v \in V\) such that \(UT(v) = z\), and so \(UT\) is onto. Therefore, if \(U\) and \(T\) are both one-to-one and onto, then \(UT\) is also one-to-one and onto.