91影视

We are given that \(\left(-2,-4\right)\) is a generating set of \(U_1 \cap U_2\). We have to check whether this vector is in both \(U_1\) and \(U_2\). For \(U_1\), there must exist a scalar \(k_1\) such that: $$\left(-2,-4\right) = k_1 \cdot \left(\begin{array}{l}1 \\\ 2\end{array}\right)$$ Solving for \(k_1\), we find that indeed \(k_1=-2\). So, the vector \(\left(-2,-4\right)\) is in \(U_1\). For \(U_2\), there must exist scalars \(k_2\) and \(k_3\) such that: $$\left(-2,-4\right) = k_2 \cdot \left(\begin{array}{l}1 \\\ 1\end{array}\right) + k_3 \cdot \left(\begin{array}{c}1 \\\ -2\end{array}\right)$$ Solving the system of linear equations, we find that \(k_2 = -\frac{4}{3}\) and \(k_3 = -\frac{10}{3}\). Therefore, the vector \(\left(-2,-4\right)\) is also in \(U_2\). Hence, \(\left\\{\left(\begin{array}{l}-2 \\\ -4\end{array}\right)\right\\}\) is indeed a generating set of \(U_1 \cap U_2\), so statement (a) is correct.

We have to check if the empty set is a basis for the intersection \(U_1 \cap U_3\). If the intersection is the zero-vector space, this statement is true. Let's see if there exists some scalar \(k\) to satisfy: $$k \cdot \left(\begin{array}{l}1 \\\ 2\end{array}\right) = k' \cdot \left(\begin{array}{c}1 \\\ -3\end{array}\right)$$ From the first components, we get \(k=k'\). For the second components, we get \(2k = -3k'\). But as \(k=k'\), the only solution is \(k = k' = 0\). So, the intersection of \(U_1\) and \(U_3\) is only the zero-vector space. The empty set is indeed a basis for \(U_1 \cap U_3\), making statement (b) correct.

We are given the vector \(\left(\begin{array}{l}1 \\\ 4\end{array}\right)\) and we need to check whether it is linearly independent in \(U_2\). Let's first write this vector as a linear combination of the basis vectors of \(U_2\): $$\left(\begin{array}{l}1 \\\ 4\end{array}\right) = k_4 \cdot \left(\begin{array}{l}1 \\\ 1\end{array}\right) + k_5 \cdot \left(\begin{array}{c}1 \\\ -2\end{array}\right)$$ Solving the system of linear equations, we get \(k_4=\frac{9}{3}, k_5=\frac{2}{3}\). As this vector can be written as a linear combination of the basis vectors, it is not linearly independent in \(U_2\). Therefore, the statement (c) is false.

We are given the union of \(U_1\) and \(U_3\), and we need to check whether their span is \(\mathbb{R}^{2}\). To do this, let's check if the basis vectors for \(U_1\) and \(U_3\) generate \(\mathbb{R}^{2}\). We can start by creating a matrix with the basis vectors as columns: $$\left(\begin{array}{cc}1 & 1 \\ 2 & -3\end{array}\right)$$ Taking the determinant of this matrix, we find that it is nonzero. This means that the column vectors are linearly independent, indicating that their span is \(\mathbb{R}^2\). So, the statement (d) is true. In conclusion, statements (a), (b), and (d) are true while statement (c) is false.