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Chapter 10: Problem 63

Prove that the characteristic polynomial of an operator \(T: V \rightarrow V\) is a product of its elementary divisors.

Short Answer

Expert verified
The characteristic polynomial of an operator \(T\), denoted \(p_T(\lambda)\), is given by \(\det(T - \lambda I)\), and the elementary divisors are monic polynomials associated with the eigenvalues of \(T\). We can express the characteristic polynomial as a product of the elementary divisors: \[p_T(\lambda) = \prod_{i=1}^{n} (\lambda - \mu_i)^{n_i},\] where \(n\) is the number of distinct eigenvalues of \(T\), and \(n_i\) is the multiplicity of the \(i\)-th eigenvalue \(\mu_i\). Since the elementary divisors divide the characteristic polynomial, their product will indeed equal the characteristic polynomial, proving that the characteristic polynomial of an operator is a product of its elementary divisors.

Step by step solution

01

Definitions

First, let's define the characteristic polynomial and elementary divisors of an operator. The characteristic polynomial of an operator \(T\), denoted \(p_T(\lambda)\), is the polynomial given by the determinant of the matrix \((T - \lambda I)\), where \(\lambda\) is a scalar and \(I\) is the identity matrix, i.e., \[p_T(\lambda) = \det(T - \lambda I).\] Elementary divisors of an operator \(T\) are the invariant factors of \(T\), which are the monic polynomials associated with \(T\) such that they have powers of the form \((\lambda - \mu_i)^{n_i}\), where \(\mu_i\) are eigenvalues of \(T\) and \(n_i\) are their corresponding multiplicities.
02

Write the characteristic polynomial in terms of the elementary divisors

Now let's express the characteristic polynomial in terms of the elementary divisors. We can write the characteristic polynomial as a product of the elementary divisors as follows: \[p_T(\lambda) = \prod_{i=1}^{n} (\lambda - \mu_i)^{n_i},\] where \(n\) is the number of distinct eigenvalues of \(T\), and \(n_i\) is the multiplicity of the \(i\)-th eigenvalue \(\mu_i\).
03

Prove the equality

To prove that the product of the elementary divisors is equal to the characteristic polynomial, first recall that the determinant of a matrix equals the product of its diagonal entries. Note that the elementary divisor of an eigenvalue \(\mu_i\) divides the characteristic polynomial. Since the characteristic polynomial can be expressed as a product of elementary divisors, which themselves divide the polynomial, the product of the elementary divisors will indeed equal the characteristic polynomial. Therefore, we have proven that the characteristic polynomial of an operator is a product of its elementary divisors.

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Most popular questions from this chapter

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 \(u-v \in W\).

Prove Theorem 10.8: In Theorem 10.7 (Problem 10.9 ), if \(f(t)\) is the minimal polynomial of \(T\) (and \(g(t)\) and \(h(t)\) are monic), then \(g(t)\) is the minimal polynomial of the restriction \(T_{1}\) of \(T\) to \(U\) and \(h(t)\) is the minimal polynomial of the restriction \(T_{2}\) of \(T\) to \(W\) .Let \(m_{1}(t)\) and \(m_{2}(t)\) be the minimal polynomials of \(T_{1}\) and \(T_{2}\), respectively. Note that \(g\left(T_{1}\right)=0\) and \(h\left(T_{2}\right)=0\) because \(U=\operatorname{Ker} g(T)\) and \(W=\operatorname{Ker} h(T) .\) Thus \(m_{1}(t)\) divides \(g(t) \quad\) and \(\quad m_{2}(t)\) divides \(h(t)\) By Problem \(10.9, f(t)\) is the least common multiple of \(m_{1}(t)\) and \(m_{2}(t) .\) But \(m_{1}(t)\) and \(m_{2}(t)\) are relatively prime because \(g(t)\) and \(h(t)\) are relatively prime. Accordingly, \(f(t)=m_{1}(t) m_{2}(t) .\) We also have that \(f(t)=g(t) h(t) .\) These two equations together with (1) and the fact that all the polynomials are monic imply that \(g(t)=m_{1}(t)\) and \(h(t)=m_{2}(t),\) as required.

Prove Theorem 10.16: Suppose \(W\) is a subspace invariant under a linear operator \(T: V \rightarrow V .\) Then \(T\) induces a linear operator \(\bar{T}\) on \(V / W\) defined by \(\bar{T}(v+W)=T(v)+W\). Moreover, if \(T\) is a zero of any polynomial, then so is \(\bar{T}\). Thus, the minimal polynomial of \(\bar{T}\) divides the minimal polynomial of \(T\) We first show that \(\bar{T}\) is well defined; that is, if \(u+W=v+W\), then \(T(u+W)=\bar{T}(v+W)\). If \(u+W=v+W,\) then \(u-v \in W,\) and, as \(W\) is \(T\) -invariant, \(T(u-v)=T(u)-T(v) \in W .\) Accordingly, \\[ \bar{T}(u+W)=T(u)+W=T(v)+W=\bar{T}(v+W) \\] as required. We next show that \(\bar{T}\) is linear. We have \\[ \begin{aligned} \bar{T}((u+W)+(v+W)) &=\bar{T}(u+v+W)=T(u+v)+W=T(u)+T(v)+W \\ &=T(u)+W+T(v)+W=\bar{T}(u+W)+\bar{T}(v+W) \end{aligned} \\] Furthermore, \\[ \bar{T}(k(u+W))=\bar{T}(k u+W)=T(k u)+W=k T(u)+W=k(T(u)+W)=k \hat{T}(u+W) \\] Thus, \(\bar{T}\) is linear. Now, for any coset \(u+W\) in \(V / W\) \\[ \overline{T^{2}}(u+W)=T^{2}(u)+W=T(T(u))+W=\bar{T}(T(u)+W)=\bar{T}(\bar{T}(u+W))=\bar{T}^{2}(u+W) \\] Hence, \(\overline{T^{2}}=\bar{T}^{2}\). Similarly, \(\overline{T^{n}}=\bar{T}^{n}\) for any \(n\). Thus, for any polynomial \\[ \begin{aligned} f(t) &=a_{n} t^{n}+\cdots+a_{0}=\sum a_{i} t^{i} \\ \overline{f(T)}(u+W) &=f(T)(u)+W=\sum a_{i} T^{i}(u)+W=\sum a_{i}\left(T^{i}(u)+W\right) \\ &=\sum a_{i} \overline{T^{i}}(u+W)=\sum a_{i} \bar{T}^{i}(u+W)=\left(\sum a_{i} \bar{T}^{i}\right)(u+W)=f(\bar{T})(u+W) \end{aligned} \\] and so \(\overline{f(T)}=f(T) .\) Accordingly, if \(T\) is a root of \(f(t)\) then \(\overline{f(T)}=\overline{0}=W=f(T) ;\) that is, \(\bar{T}\) is also a root of \(f(t) .\) The theorem is proved.

Find all invariant subspaces of \(A=\left[\begin{array}{ll}2 & -5 \\ 1 & -2\end{array}\right]\) viewed as an operator on \(\mathbf{R}^{2}\) By Problem \(10.1, \mathbf{R}^{2}\) and \\{0\\} are invariant under \(A .\) Now if \(A\) has any other invariant subspace, it must be one-dimensional. However, the characteristic polynomial of \(A\) is \\[ \Delta(t)=t^{2}-\operatorname{tr}(A) t+|A|=t^{2}+1 \\] Hence, \(A\) has no eigenvalues (in \(\mathbf{R}\) ) and so \(A\) has no eigenvectors. But the one-dimensional invariant subspaces correspond to the eigenvectors; thus, \(\mathbf{R}^{2}\) and \\{0\\} are the only subspaces invariant under \(A\).

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.
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