91影视

Let's first check if \(U_{1}\) contains the zero vector. The zero vector in 鈩澛� is \(\begin{pmatrix} 0\\ 0\\ 0 \end{pmatrix}\). Since \(v_{1}+v_{2}=0+0=2\) does not hold, the zero vector is not in \(U_{1}\). Therefore, \(U_1\) is not an 鈩澛� subspace. #Problem b#

Now let's check if \(U_{2}\) contains the zero vector. From the given condition, \(v_{1}+v_{2}=v_{3}\). The zero vector makes this equation true: \(0+0=0\). So the zero vector is in \(U_{2}\).

Now we need to check if \(U_{2}\) is closed under addition. Suppose \(\vec{u}\) and \(\vec{v}\) are two vectors in \(U_{2}\), both fulfilling the condition \(v_{1}+v_{2}=v_{3}\). We need to prove that the sum of these two vectors, i.e. \(\vec{u}+\vec{v}\), also satisfies this condition. Consider: \((u_{1}+v_{1})+(u_{2}+v_{2})=u_{3}+v_{3}\) Since both \(u\) and \(v\) belong to \(U_{2}\), we can substitute the condition: \((u_{1}+u_{2})+(v_{1}+v_{2})=u_{3}+v_{3}\) This condition holds true, so \(U_{2}\) is closed under addition.

Now let's check if \(U_{2}\) is closed under scalar multiplication. Take any scalar 伪 and a vector \(\vec{u}=\begin{pmatrix} u_{1}\\ u_{2}\\ u_{3} \end{pmatrix} \in U_{2}\). The scalar multiplication of \(\vec{u}\) and 伪 is: \(\alpha\vec{u}=\begin{pmatrix} \alpha u_{1}\\ \alpha u_{2}\\ \alpha u_{3} \end{pmatrix}\). The transformed vector is in \(U_{2}\) if the condition holds: \(\alpha u_{1}+\alpha u_{2}=\alpha u_{3}\) The original vector \(\vec{u}\) belongs to \(U_{2}\), so the condition \(u_{1}+u_{2}=u_{3}\) is true. Multiply both sides by 伪: \(\alpha(u_{1}+u_{2})=\alpha u_{3}\) This matches the condition for the scalar multiplication, so \(U_{2}\) is closed under scalar multiplication. Since all three necessary properties hold for \(U_{2}\), it is a subspace of 鈩澛�. #Problem c#

In the case of \(U_{3}\), according to the given condition, \(v_{1} v_{2}=v_{3}\). The zero vector makes the equation true: \(0*0=0\), so it is in \(U_{3}\).

Now we need to check if \(U_{3}\) closed under addition. Suppose \(\vec{u}\) and \(\vec{v}\) are two vectors in \(U_{3}\), both fulfilling the condition \(u_{1} u_{2}=u_{3}\) and \(v_{1} v_{2}=v_{3}\). We need to prove that the sum of these vectors, i.e. \(\vec{u}+\vec{v}\), also satisfies this condition. However, consider the following example vectors: \(\vec{u}=\begin{pmatrix} 1\\ 1\\ 1 \end{pmatrix}, \vec{v}=\begin{pmatrix} 2\\ 1\\ 2 \end{pmatrix}\) Both vectors fulfill the condition of \(U_{3}\). After adding these two vectors, we have: \(\vec{u}+\vec{v}=\begin{pmatrix} 3\\ 2\\ 3 \end{pmatrix}\) But this vector does not satisfy the condition: \(3*2=6 \neq 3\) Therefore, \(U_{3}\) is **not** closed under addition and thus cannot be a subspace of 鈩澛�. #Problem d#

For the subset \(U_{4}\), check if the zero vector is in \(U_{4}\). Here, we have two conditions: either \(v_{1}=v_{2}\) or \(v_{1}=v_{3}\). The zero vector satisfies both the conditions: \(0=0\), so it is in \(U_{4}\).

Now we need to check if \(U_{4}\) is closed under addition. Suppose that \(\vec{u}\) and \(\vec{v}\) are two vectors in \(U_{4}\), both fulfilling one of the conditions given above. We need to prove that adding them results in another vector in \(U_{4}\). If \(u_{1}=u_{2}\) and \(v_{1}=v_{3}\), their sum must either equal the other component or equal something else. However, consider the following example vectors: \(\vec{u}=\begin{pmatrix} 1\\ 1\\ 3 \end{pmatrix}, \vec{v}=\begin{pmatrix} 2\\ 4\\ 2 \end{pmatrix}\) Both vectors fulfill one of the conditions of \(U_{4}\). After adding these two vectors, we get: \(\vec{u}+\vec{v}=\begin{pmatrix} 3\\ 5\\ 5 \end{pmatrix}\) But this vector does not satisfy any of the conditions (i.e. \(3 \neq 5\) and \(3 \neq 5\)). Hence, \(U_{4}\) is **not** closed under addition and thus cannot be a subspace of 鈩澛�. In conclusion, among the given subsets, only \(U_{2}\) is a subspace of 鈩澛�.