91影视

We have a \(K\)-vector space \(V\), a linear subspace \(U \subset V\), and an element \(a \in V\). We are looking for the elements \(a\) for which \(a + U:=\{a + u; u \in U\}\) is a linear subspace of \(V\).

A subset \(W\) of a vector space \(V\) is a linear subspace if it satisfies the following properties: 1. The zero vector of \(V\) is in \(W\). 2. If \(w_1, w_2 \in W\), then \(w_1 + w_2 \in W\). (Closed under addition) 3. If \(w \in W\) and \(c \in K\), then \(c \cdot w \in W\). (Closed under scalar multiplication)

Let's verify if the set \(a + U\) satisfies the properties of a linear subspace: 1. Zero vector property: Since \(U\) is a linear subspace, the zero vector, denoted as \(0_V\), is an element of \(U\). Then \(a + 0_V = a\) is an element of \(a + U\) if and only if \(a=0_V\). So, for the set \(a + U\) to be a linear subspace, we must have \(a = 0_V\). 2. Closed under addition: Take any two elements, \(a_1, a_2 \in a+U\). Then, there exist \(u_1, u_2 \in U\) such that \(a_1 = a + u_1\) and \(a_2 = a + u_2\). Their sum is given by: \[a_1 + a_2 = (a + u_1) + (a + u_2) = a + a + u_1 + u_2.\] Now, since \(U\) is a linear subspace and \(u_1, u_2 \in U\), we know that \(u_1 + u_2\in U\). Hence, their sum can be expressed as: \[a_1 + a_2 = a + a + (u_1 + u_2) = a + (a + (u_1 + u_2)).\] For the result to be in \(a + U\), we require \(a + a = 0_V\), i.e., \(2a = 0_V\) or \(a = 0_V\). 3. Closed under scalar multiplication: Take any element \(a_1 \in a + U\) and a scalar \(c \in K\). Then, there exists \(u_1 \in U\) such that \(a_1 = a + u_1\). The product of the scalar and the element is given by: \[c \cdot a_1 = c(a + u_1) = c a + c u_1.\] Now, since \(U\) is a linear subspace and \(u_1 \in U\), we know that \(c \cdot u_1 \in U\). We need the result \(c a + cu_1\) to be in \(a+U\), that means we need \(c a\) to be in \(U\), which implies \(a \in U\). Hence, the set \(a + U\) can be a linear subspace of V if and only if \(a = 0_V\).