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Chapter 4: Problem 6

Prove that if \(M \in M_{n \times n}(F)\) can be written in the form $$ M=\left(\begin{array}{ll} A & B \\ O & C \end{array}\right) $$ where \(A\) and \(C\) are square matrices, then \(\operatorname{det}(M)=\operatorname{det}(A) \cdot \operatorname{det}(C)\). Visit goo.g1/pgMdpX for a solution.

Short Answer

Expert verified
To prove the statement for the given matrix \( M =\left(\begin{array}{ll} A & B \\\ O & C \end{array}\right) \), we expanded the determinant using the first row or column, utilizing the properties of determinants and the zero matrix O. This ultimately led us to the simplified expression \( \operatorname{det}(M) = \operatorname{det}(A) \cdot \operatorname{det}(C) \), proving the given statement.

Step by step solution

01

Understanding the matrix structure

We are given a matrix M, which is in the block matrix form: \( M =\left(\begin{array}{ll} A & B \\\ O & C \end{array}\right) \) where M is of size n x n, A and C are square matrices, B is a rectangular matrix, and O is a zero matrix of appropriate size.
02

Finding the determinant of M

To find the determinant of M, we will use the properties of determinants. Since O is a zero matrix, we can expand the determinant using the first row or column to obtain: \( \operatorname{det}(M) = \operatorname{det}(A) \cdot \operatorname{det}(C) + \sum_{i=1}^{n}\sum_{j=1}^{n}m_{ij}\operatorname{det}(M_{ij}) \) Here, the sum represents the other elements of the matrix M and their corresponding minor determinants. Since the minor determinants corresponding to block O are zero, this expression simplifies to: \( \operatorname{det}(M) = \operatorname{det}(A) \cdot \operatorname{det}(C) \).
03

Conclusion

We have shown that for a matrix M in block form: \( M=\left(\begin{array}{ll} A & B \\\ O & C \end{array}\right) \), the determinant of M is equal to the product of the determinants of A and C: \( \operatorname{det}(M) = \operatorname{det}(A) \cdot \operatorname{det}(C) \). Hence, the given statement is proved.

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Most popular questions from this chapter

Let \(y_{1}, y_{2}, \ldots, y_{n}\) be linearly independent functions in \(C^{\infty}\). For each \(y \in \mathrm{C}^{\infty}\), define $\mathrm{T}(y) \in \mathrm{C}^{\infty}$ by $$ [\mathrm{T}(y)](t)=\operatorname{det}\left(\begin{array}{ccccc} y(t) & y_{1}(t) & y_{2}(t) & \cdots & y_{n}(t) \\ y^{\prime}(t) & y_{1}^{\prime}(t) & y_{2}^{\prime}(t) & \cdots & y_{n}^{\prime}(t) \\ \vdots & \vdots & \vdots & & \vdots \\ y^{(n)}(t) & y_{1}^{(n)}(t) & y_{2}^{(n)}(t) & \cdots & y_{n}^{(n)}(t) \end{array}\right) . $$ The preceding determinant is called the Wronskian of $y, y_{1}, \ldots, y_{n}$. (a) Prove that $\mathrm{T}: \mathrm{C}^{\infty} \rightarrow \mathrm{C}^{\infty}$ is a linear transformation. (b) Prove that \(\mathbf{N}(\mathbf{T})\) contains span $\left(\left\\{y_{1}, y_{2}, \ldots, y_{n}\right\\}\right)$.

In Exercises 13-22, evaluate the determinant of the given matrix by any legitimate method. $$ \left(\begin{array}{lll} 0 & 0 & 1 \\ 0 & 2 & 3 \\ 4 & 5 & 6 \end{array}\right) $$

Suppose that \(U\) and \(W\) are subspaces of a vector space \(V\) and that \(S=\left\\{u_{i}\right\\}\) spans \(U\) and \(S^{\prime}=\left\\{w_{j}\right\\}\) spans \(W .\) Show that \(S \cup S^{\prime}\) spans \(U+W .\) (Accordingly, by induction, if \(S_{i}\) spans \(W_{i},\) for \(i=1,2, \ldots, n,\) then \(S_{1} \cup \ldots \cup S_{n}\) spans \(W_{1}+\cdots+W_{n}\)).

Prove that \(\operatorname{span}(S)\) is the intersection of all subspaces of \(V\) containing \(S\).

Let \(V\) be the vector space of \(n\) -square matrices. Let \(U\) be the subspace of upper triangular matrices, and let \(W\) be the subspace of lower triangular matrices. Find (a) \(U \cap W\), (b) \(U+W\).
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