/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 47 Suppose \(T_{1}\) and \(T_{2}\) ... [FREE SOLUTION] | 91影视

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Chapter 10: Problem 47

Suppose \(T_{1}\) and \(T_{2}\) are nilpotent operators that commute (i.e., \(T_{1} T_{2}=T_{2} T_{1}\) ). Show that \(T_{1}+T_{2}\) and \(T_{1} T_{2}\) are also nilpotent.

Short Answer

Expert verified
We show that if \(T_1\) and \(T_2\) are nilpotent operators that commute, then their sum \(T_1 + T_2\) and their product \(T_1T_2\) are also nilpotent. Let \(n_1\) and \(n_2\) be the integers such that \(T_1^{n_1} = 0\) and \(T_2^{n_2} = 0\). We consider \((T_1 + T_2)^{n_1n_2}\) and expand it using the binomial theorem. By dividing the sum into two parts and using the commutativity property of \(T_1\) and \(T_2\), we can show that the entire sum is equal to zero, which means that \(T_1 + T_2\) is nilpotent. Similarly, we consider the product of the operators, \(T_3 = T_1T_2\), and show that raising it to an appropriate power yields a zero operator, meaning that \(T_1T_2\) is also nilpotent.

Step by step solution

01

Using binomial theorem to expand the sum of operators

Let \(n_1\) and \(n_2\) be the integers such that \(T_1^{n_1} = 0\) and \(T_2^{n_2} = 0\). Consider \((T_1 + T_2)^{n_1n_2}\) and expand using the binomial theorem: \((T_1 + T_2)^{n_1n_2} = \sum_{k=0}^{n_1n_2} {n_1n_2 \choose k} T_1^{n_1n_2-k} T_2^k\)
02

Dividing the sum into two parts and using commutativity

We can divide the sum into two parts: one where \(k \geq n_2\) and one where \(k < n_2\). Using the commutativity property of \(T_1\) and \(T_2\), we can rewrite the sum as: \((T_1 + T_2)^{n_1n_2} = \sum_{k=0}^{n_1n_2} {n_1n_2 \choose k} T_1^{n_1n_2-k} T_2^k = \sum_{k=0}^{n_1(n_2-1)} {n_1n_2 \choose k} T_1^{n_1n_2-k} T_2^k + \sum_{k=n_1(n_2-1)+1}^{n_1n_2} {n_1n_2 \choose k} T_1^{n_1n_2-k} T_2^k\)
03

Proving that the sum is zero

Since \(T_1^{n_1} = 0\) and \(T_2^{n_2} = 0\), in the first sum, for \(k \geq n_2\), the term \(T_1^{n_1n_2-k} T_2^k = T_2^k T_1^{n_1(n_2-1)} = 0\). Similarly, in the second sum, for \(n_1n_2 - k \geq n_1\), the term \(T_1^{n_1n_2-k} T_2^k = T_1^{n_1(n_2-1)} T_2^k = 0\). Therefore, the entire sum is equal to zero, and \((T_1 + T_2)^{n_1n_2} = 0\), which means that \(T_1 + T_2\) is nilpotent.
04

Proving that the product of the operators is nilpotent

Now, consider the product of the operators, \(T_3 = T_1T_2\). We need to find an appropriate power such that \(T_3^m = (T_1T_2)^m = 0\). Let's start by raising \(T_3\) to the power \(n_1n_2 - 1\): \[(T_1T_2)^{n_1n_2 - 1} = T_1^{n_1n_2 - 1} T_2^{n_1n_2 - 1}\] We can notice that \((T_1^{n_1n_2 - 1})^{n_1} = T_1^{n_1n_2(n_1 - 1)} = (T_1^{n_1})^{n_1(n_2-1)} = 0^{n_1(n_2-1)} = 0\). Similarly, \((T_2^{n_1n_2 - 1})^{n_2} = 0\). Thus, the product of these nilpotent operators is also nilpotent. In conclusion, we have shown that both the sum and the product of the given nilpotent operators \(T_1\) and \(T_2\) are nilpotent, as required.

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Most popular questions from this chapter

Show that two nilpotent matrices of order 3 are similar if and only if they have the same index of nilpotency. Show by example that the statement is not true for nilpotent matrices of order 4.

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

Prove Theorem 10.12: Let \(Z(v, T)\) be a \(T\) -cyclic subspace, \(T_{v}\) the restriction of \(T\) to \(Z(v, T),\) and \(m_{v}(t)=t^{k}+a_{k-1} t^{k-1}+\cdots+a_{0}\) the \(T\) -annihilator of \(v .\) Then (i) The set \(\left\\{v, T(v), \ldots, T^{k-1}(v)\right\\}\) is a basis of \(Z(v, T) ;\) hence, \(\operatorname{dim} Z(v, T)=k\) (ii) The minimal polynomial of \(T_{v}\) is \(m_{v}(t)\) (iii) The matrix of \(T_{v}\) in the above basis is the companion matrix \(C=C\left(m_{v}\right)\) of \(m_{v}(t)\) [which has 1 's below the diagonal, the negative of the coefficients \(a_{0}, a_{1}, \ldots, a_{k-1}\) of \(m_{v}(t)\) in the last column, and \(0 \text { 's elsewhere }]\). (i) \(\quad\) By definition of \(m_{v}(t), T^{k}(v)\) is the first vector in the sequence \(v, T(v), T^{2}(v), \ldots\) that, is a linear combination of those vectors that precede it in the sequence; hence, the set \(B=\left\\{v, T(v), \ldots, T^{k-1}(v)\right\\}\) is linearly independent. We now only have to show that \(Z(v, T)=L(B)\), the linear span of \(B\). By the above, \(T^{k}(v) \in L(B) .\) We prove by induction that \(T^{n}(v) \in L(B)\) for every \(n .\) Suppose \(n>k\) and \(T^{n-1}(v) \in L(B)-\) that \(\quad\) is, \(\quad T^{n-1}(v) \quad\) is \(\quad\) a \(\quad\) linear \(\quad\) combination \(\quad\) of \(\quad v, \ldots, T^{k-1}(v) . \quad\) Then \(T^{n}(v)=T\left(T^{n-1}(v)\right)\) is a linear combination of \(T(v), \ldots, T^{k}(v) .\) But \(T^{k}(v) \in L(B) ;\) hence, \(T^{n}(v) \in L(B)\) for every \(n .\) Consequently, \(f(T)(v) \in L(B)\) for any polynomial \(f(t) .\) Thus, \(Z(v, T)=L(B),\) and \(\operatorname{so~} B\) is a basis, as claimed. (ii) Suppose \(m(t)=t^{s}+b_{s-1} t^{s-1}+\cdots+b_{0}\) is the minimal polynomial of \(T_{v} .\) Then, because \(v \in Z(v, T)\) \\[ 0=m\left(T_{v}\right)(v)=m(T)(v)=T^{s}(v)+b_{s-1} T^{s-1}(v)+\cdots+b_{0} v \\] Thus, \(T^{s}(v)\) is a linear combination of \(v, T(v), \ldots, T^{s-1}(v),\) and therefore \(k \leq s\). However, \(m_{v}(T)=\mathbf{0}\) and so \(m_{v}\left(T_{v}\right)=\mathbf{0} .\) Then \(m(t)\) divides \(m_{v}(t),\) and so \(s \leq k .\) Accordingly, \(k=s\) and hence \(m_{v}(t)=m(t)\) (iii) \\[ \begin{array}{ll} T_{v}(v) & =\quad T(v) \\ T_{v}(T(v)) & = \\ & = \\ T_{v}\left(T^{k-2}(v)\right) & = \\ T_{v}\left(T^{k-1}(v)\right) & =T^{k}(v)=-a_{0} v-a_{1} T(v)-a_{2} T^{2}(v)-\cdots-a_{k-1} T^{k-1}(v) \end{array} \\] By definition, the matrix of \(T_{v}\) in this basis is the tranpose of the matrix of coefficients of the above system of equations; hence, it is \(C,\) as required.

Show that any Jordan nilpotent block matrix \(N\) is similar to its transpose \(N^{T}\) (the matrix with 1 's below the diagonal and \(0^{\prime}\) s elsewhere)

Let \(\hat{T}\) denote the restriction of an operator \(T\) to an invariant subspace \(W\). Prove (a) For any polynomial \(f(t), f(\hat{T})(w)=f(T)(w)\) (b) The minimal polynomial of \(\hat{T}\) divides the minimal polynomial of \(T\) (a) If \(f(t)=0\) or if \(f(t)\) is a constant (i.e., of degree 1 ), then the result clearly holds. Assume deg \(f=n>1\) and that the result holds for polynomials of degree less than \(n\). Suppose that Then $$\begin{aligned} & f(t)=a_{n} t^{n}+a_{n-1} t^{n-1}+\cdots+a_{1} t+a_{0} \\ f(\hat{T})(w) &=\left(a_{n} \hat{T}^{n}+a_{n-1} \hat{T}^{n-1}+\cdots+a_{0} I\right)(w) \\ &=\left(a_{n} \hat{T}^{n-1}\right)(\hat{T}(w))+\left(a_{n-1} \hat{T}^{n-1}+\cdots+a_{0} I\right)(w) \\ &=\left(a_{n} T^{n-1}\right)(T(w))+\left(a_{n-1} T^{n-1}+\cdots+a_{0} I\right)(w)=f(T)(w) \end{aligned}$$ (b) Let \(m(t)\) denote the minimal polynomial of \(T .\) Then by (a), \(m(\hat{T})(w)=m(T)(w)=\mathbf{0}(w)=0\) for every \(w \in W ;\) that is, \(\hat{T}\) is a zero of the polynomial \(m(t) .\) Hence, the minimal polynomial of \(T\) divides \(m(t)\)
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