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Chapter 10: Problem 64

Prove that two \(3 \times 3\) matrices with the same minimal and characteristic polynomials are similar.

Short Answer

Expert verified
Since two 3x3 matrices A and B have the same minimal and characteristic polynomials, they share the same eigenvalues. Using the Cayley-Hamilton Theorem, we find that they satisfy their own characteristic equations. Having the same eigenvalues and eigenspaces, an invertible matrix P exists which relates A and B through a similarity transformation. Therefore, any two 3x3 matrices with the same minimal and characteristic polynomials are similar.

Step by step solution

01

Define key concepts

Minimal Polynomial: The minimal polynomial of a square matrix A is the unique monic polynomial m(x) of the least degree such that \(m(A) = 0\), where m(A) is the result of substituting A into the polynomial m(x). Characteristic Polynomial: The characteristic polynomial of a square matrix A is defined as the determinant of the matrix \(A - xI\), where x is a scalar variable and I is the identity matrix of the same order as A. It is denoted as p(x). Similar Matrices: Two matrices A and B are similar if there exists an invertible matrix P such that \(B = P^{-1}AP\).
02

Properties of Minimal and Characteristic Polynomials

1. Both minimal and characteristic polynomials have the matrix eigenvalues as their roots. 2. The minimal polynomial divides the characteristic polynomial.
03

Show that the two matrices have the same eigenvalues

According to properties mentioned above, if two matrices have the same minimal and characteristic polynomials, then they have the same eigenvalues. Now, let's call these two matrices A and B. Since they have the same minimal and characteristic polynomials, we know that they have the same eigenvalues.
04

Use the Cayley-Hamilton Theorem

The Cayley-Hamilton Theorem states that every square matrix satisfies its own characteristic equation. Therefore, \(p(A) = p(B) = 0\), where p(x) is the characteristic polynomial of both matrices.
05

Deduce that the matrices are similar

Since A and B have the same minimal and characteristic polynomials, they have the same roots (eigenvalues) and the same null spaces for each eigenvalue. Thus, there exists an invertible matrix P whose columns are composed of eigenvectors corresponding to the same eigenvalues for both A and B. Additionally, the minimal polynomial condition guarantees that these eigenvectors are linearly independent. Therefore, A and B are diagonalizable and related by a similarity transformation. In conclusion, any two 3x3 matrices with the same minimal and characteristic polynomials are similar because they have the same eigenvalues and eigenspaces, which allows us to find an invertible matrix P relating them through a similarity transformation.

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Most popular questions from this chapter

Let \(V\) be a vector space of odd dimension (greater than 1 ) over the real field \(\mathbf{R}\). Show that any linear operator on \(V\) has an invariant subspace other than \(V\) or \(\\{0\\}\).

Suppose dim \(V=n .\) Show that \(T: V \rightarrow V\) has a triangular matrix representation if and only if there exist \(T\) -invariant subspaces \(W_{1} \subset W_{2} \subset \cdots \subset W_{n}=V\) for which \(\operatorname{dim} W_{k}=k, k=1, \ldots, n\).

Prove the following: The cosets of \(W\) in \(V\) partition \(V\) into mutually disjoint sets. That is, (a) Any two cosets \(u+W\) and \(v+W\) are either identical or disjoint. (b) Each \(v \in V\) belongs to a coset; in fact, \(v \in v+W\) Furthermore, \(u+W=v+W\) if and only if \(u-v \in W,\) and so \((v+w)+W=v+W\) for any \(w \in W\) Let \(v \in V .\) Because \(0 \in W,\) we have \(v=v+0 \in v+W,\) which proves (b). Now suppose the cosets \(u+W\) and \(v+W\) are not disjoint; say, the vector \(x\) belongs to both \(u+W\) and \(v+W\). Then \(u-x \in W\) and \(x-v \in W\). The proof of (a) is complete if we show that \(u+W=v+W\) Let \(u+w_{0}\) be any element in the coset \(u+W\). Because \(u-x, x-v, w_{0}\) belongs to \(W\) \\[ \left(u+w_{0}\right)-v=(u-x)+(x-v)+w_{0} \in W \\] Thus, \(u+w_{0} \in v+W,\) and hence the cost \(u+W\) is contained in the coset \(v+W\). Similarly, \(v+W\) is contained in \(u+W\), and so \(u+W=v+W\) The last statement follows from the fact that \(u+W=v+W\) if and only if \(u \in v+W\), and, by Problem \(10.21,\) this is equivalent to \(u-v \in W\).

Suppose \(E: V \rightarrow V\) is linear and \(E^{2}=E .\) Show that \((\mathrm{a}) E(u)=u\) for any \(u \in \operatorname{Im} E\) (i.e., the restriction of \(E \text { to its image is the identity mapping }) ;\) (b) \(V\) is the direct sum of the image and kernel of \(E: V=\operatorname{Im} E \oplus \operatorname{Ker} E ;\) (c) \(E\) is the projection of \(V\) into \(\operatorname{Im} E,\) its image. Thus, by the preceding problem, a linear mapping \(T: V \rightarrow V\) is a projection if and only if \(T^{2}=T ;\) this characterization of a projection is frequently used as its definition.

Prove Theorem 10.4: Suppose \(W_{1}, W_{2}, \ldots, W_{r}\) are subspaces of \(V\) with respective bases \\[ B_{1}=\left\\{w_{11}, w_{12}, \ldots, w_{1 n_{1}}\right\\}, \quad \ldots, \quad B_{r}=\left\\{w_{r 1}, w_{r 2}, \ldots, w_{r n_{r}}\right\\} \\] Then \(V\) is the direct sum of the \(W_{i}\) if and only if the union \(B=\bigcup_{i} B_{i}\) is a basis of \(V\) Suppose \(B\) is a basis of \(V\). Then, for any \(v \in V\) \\[ v=a_{11} w_{11}+\cdots+a_{1 n_{1}} w_{1 n_{1}}+\cdots+a_{r 1} w_{r 1}+\cdots+a_{m_{r}} w_{m_{r}}=w_{1}+w_{2}+\cdots+w_{r} \\] where \(w_{i}=a_{i 1} w_{i 1}+\cdots+a_{i n_{i}} w_{i n_{i}} \in W_{i} .\) We next show that such a sum is unique. Suppose \\[ v=w_{1}^{\prime}+w_{2}^{\prime}+\cdots+w_{r}^{\prime}, \quad \text { where } \quad w_{i}^{\prime} \in W_{i} \\] Because \(\left\\{w_{i 1}, \ldots, w_{i n}\right\\}\) is a basis of \(W_{i}, w_{i}^{\prime}=b_{i 1} w_{i 1}+\cdots+b_{i n_{i}} w_{i n_{i}},\) and so \\[ v=b_{11} w_{11}+\dots+b_{1 n_{1}} w_{1 n_{1}}+\cdots+b_{r 1} w_{r 1}+\cdots+b_{m_{r}} w_{r m_{r}} \\] Because \(B\) is a basis of \(V, a_{i j}=b_{i j},\) for each \(i\) and each \(j .\) Hence, \(w_{i}=w_{i}^{\prime},\) and so the sum for \(v\) is unique. Accordingly, \(V\) is the direct sum of the \(W_{i}\) Conversely, suppose \(V\) is the direct sum of the \(W_{i}\). Then for any \(v \in V, v=w_{1}+\cdots+w_{r},\) where \(w_{i} \in W_{i} .\) Because \(\left\\{w_{i j_{i}}\right\\}\) is a basis of \(W_{i},\) each \(w_{i}\) is a linear combination of the \(w_{i j},\) and so \(v\) is a linear combination of the elements of \(B\). Thus, \(B\) spans \(V\). We now show that \(B\) is linearly independent. Suppose \\[ a_{11} w_{11}+\dots+a_{1 n_{1}} w_{1 n_{1}}+\cdots+a_{r 1} w_{r 1}+\cdots+a_{r n_{r}} w_{r n_{r}}=0 \\] Note that \(a_{i 1} w_{i 1}+\cdots+a_{i n_{i}} w_{i n_{i}} \in W_{i} .\) We also have that \(0=0+0 \cdots 0 \in W_{i} .\) Because such a sum for 0 is unique, \\[ a_{i 1} w_{i 1}+\cdots+a_{i n_{i}} w_{i n_{i}}=0 \quad \text { for } i=1, \ldots, r \\] The independence of the bases \(\left\\{w_{i j}\right\\}\) implies that all the \(a\) 's are \(0 .\) Thus, \(B\) is linearly independent and is a basis of \(V\)
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