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Chapter 7: Problem 5

Describe all linear operators_ \(T\) on \(R^{2}\) such that \(T\) is diagonalizable and \(\mathrm{T}^{3}-2 \mathrm{~T}^{2}+\mathrm{T}=\mathrm{T}_{0}\).

Short Answer

Expert verified
The linear operators T on \(R^2\) such that T is diagonalizable and \(T^3 - 2T^2 + T = T_0\) are represented by the following diagonal matrices: 1. \(D_1 = \left[ \begin{array}{cc} 0 & 0 \\ 0 & 0 \end{array} \right]\) - The zero operator (all eigenvalues and eigenvectors are zero). 2. \(D_2 = \left[ \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right]\) - The identity operator (eigenvalues equal one and eigenvectors are the standard basis vectors).

Step by step solution

01

Start with diagonalizable operator representation

Since T is diagonalizable, there exists a basis such that the matrix representation of T with respect to that basis is a diagonal matrix. Let's denote this diagonal matrix by capital D. Then, we have: \[D = \left[ \begin{array}{cc} \lambda_1 & 0 \\ 0 & \lambda_2 \end{array} \right]\] where \(\lambda_1\) and \(\lambda_2\) are the eigenvalues of T.
02

Apply the given equation to the matrix D

Now, we can apply the given equation \(T^3 - 2T^2 + T = T_0\) to the matrix representation D: \[(D^3 - 2D^2 + D) = 0\] \[D^3 - 2D^2 + D = \left[ \begin{array}{cc} 0 & 0 \\ 0 & 0 \end{array} \right]\]
03

Calculate D^n for n = 1,2,3

Let's calculate \(D^2\) and \(D^3\): \(D^2 = \left[ \begin{array}{cc} \lambda_1 & 0 \\ 0 & \lambda_2 \end{array} \right]^2 = \left[ \begin{array}{cc} \lambda_1^2 & 0 \\ 0 & \lambda_2^2 \end{array} \right]\) \(D^3 = \left[ \begin{array}{cc} \lambda_1 & 0 \\ 0 & \lambda_2 \end{array} \right]^3 = \left[ \begin{array}{cc} \lambda_1^3 & 0 \\ 0 & \lambda_2^3 \end{array} \right]\)
04

Substitute the calculated values into the equation

Now substitute the calculated values into the equation: \[\left[ \begin{array}{cc} \lambda_1^3 & 0 \\ 0 & \lambda_2^3 \end{array} \right] - 2 \left[ \begin{array}{cc} \lambda_1^2 & 0 \\ 0 & \lambda_2^2 \end{array} \right] + \left[ \begin{array}{cc} \lambda_1 & 0 \\ 0 & \lambda_2 \end{array} \right] = \left[ \begin{array}{cc} 0 & 0 \\ 0 & 0 \end{array} \right]\]
05

Simplify and solve for eigenvalues

We can simplify the equation: \[\left[ \begin{array}{cc} \lambda_1^3-2\lambda_1^2+\lambda_1 & 0 \\ 0 & \lambda_2^3-2\lambda_2^2+\lambda_2 \end{array} \right] = \left[ \begin{array}{cc} 0 & 0 \\ 0 & 0 \end{array} \right]\] Solve for eigenvalues: \(\lambda_1(\lambda_1^2 - 2\lambda_1 + 1) = 0\) \(\lambda_1(\lambda_1 - 1)^2 = 0\) Similarly, \(\lambda_2(\lambda_2 - 1)^2 = 0\)
06

Determine the possible eigenvalues and their matrices

From the equations, we see that there are two possible diagonal representations for T: 1. \(D_1 = \left[ \begin{array}{cc} 0 & 0 \\ 0 & 0 \end{array} \right]\) when \(\lambda_1 = 0\) and \(\lambda_2 = 0\), 2. \(D_2 = \left[ \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right]\) when \(\lambda_1 = 1\) and \(\lambda_2 = 1\).
07

Conclusion

The linear operators T on \(R^2\) such that T is diagonalizable and \(T^3 - 2T^2 + T = T_0\) are represented by the following diagonal matrices: 1. \(D_1 = \left[ \begin{array}{cc} 0 & 0 \\ 0 & 0 \end{array} \right]\) - The zero operator (all eigenvalues and eigenvectors are zero). 2. \(D_2 = \left[ \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right]\) - The identity operator (eigenvalues equal one and eigenvectors are the standard basis vectors).

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Most popular questions from this chapter

Consider the vector space \(\mathbf{P}(t)\) with inner product \(\langle f, g\rangle=\int_{0}^{1} f(t) g(t) d t .\) Apply the GramSchmidt algorithm to the set \(\left\\{1, t, t^{2}\right\\}\) to obtain an orthogonal set \(\left\\{f_{0}, f_{1}, f_{2}\right\\}\) with integer coefficients. First set \(f_{0}=1 .\) Then find \\[t-\frac{\langle t, 1\rangle}{\langle 1,1\rangle} \cdot 1=t-\frac{\frac{1}{2}}{1} \cdot 1=t-\frac{1}{2}\\] Clear fractions to obtain \(f_{1}=2 t-1 .\) Then find \\[t^{2}-\frac{\left\langle t^{2}, 1\right\rangle}{\langle 1,1\rangle}(1)-\frac{\left\langle t^{2}, 2 t-1\right\rangle}{\langle 2 t-1,2 t-1\rangle}(2 t-1)=t^{2}-\frac{\frac{1}{3}}{1}(1)-\frac{\frac{1}{6}}{\frac{1}{3}}(2 t-1)=t^{2}-t+\frac{1}{6}\\] Clear fractions to obtain \(f_{2}=6 t^{2}-6 t+1 .\) Thus, \(\left\\{1,2 t-1,6 t^{2}-6 t+1\right\\}\) is the required orthogonal set.

Prove Theorem 7.24 : Let \(V\) be a normed vector space. Then the function \(d(u, v)=\|u-v\|\) satisfies the following three axioms of a metric space:

For each linear operator \(T\) on \(V\), find the minimal polynomial of \(T\). (a) \(\mathbf{V}=\mathbf{R}^{2}\) and \(\mathrm{T}(a, b)=(a+b, a-b)\) (b) \(\mathrm{V}=\mathrm{P}_{2}(R)\) and \(\mathrm{T}(g(x))=g^{\prime}(x)+2 g(x)\) (c) \(\mathrm{V}=\mathrm{P}_{2}(R)\) and $\mathrm{T}(f(x))=-x f^{\prime \prime}(x)+f^{\prime}(x)+2 f(x)$ (d) \(\mathrm{V}=\mathrm{M}_{n \times n}(R)\) and \(\mathrm{T}(A)=A^{t}\). Hint: Note that \(\mathrm{T}^{2}=1\).

Suppose \(w_{1}\) and \(w_{2}\) are nonzero orthogonal vectors. Let \(v\) be any vector in \(V\). Find \(c_{1}\) and \(c_{2}\) so that \(v^{\prime}\) is orthogonal to \(w_{1}\) and \(w_{2},\) where \(v^{\prime}=v-c_{1} w_{1}-c_{2} w_{2}\) If \(v^{\prime}\) is orthogonal to \(w_{1},\) then \\[\begin{aligned} 0 &=\left\langle v-c_{1} w_{1}-c_{2} w_{2}, w_{1}\right\rangle=\left\langle v, w_{1}\right\rangle-c_{1}\left\langle w_{1}, w_{1}\right\rangle- c_{2}\left\langle w_{2}, w_{1}\right\rangle \\ &=\left\langle v, w_{1}\right\rangle-c_{1}\left\langle w_{1}, w_{1}\right\rangle-c_{2} 0=\left\langle v, w_{1}\right\rangle- c_{1}\left\langle w_{1}, w_{1}\right\rangle \end{aligned}\\] Thus, \(c_{1}=\left\langle v, w_{1}\right\rangle /\left\langle w_{1}, w_{1}\right\rangle .\) (That is, \(c_{1}\) is the component of \(v\) along \(w_{1}\).) Similarly, if \(v^{\prime}\) is orthogonal to \(w_{2}\) then \\[0=\left\langle v-c_{1} w_{1}-c_{2} w_{2}, \quad w_{2}\right\rangle=\left\langle v, w_{2}\right\rangle-c_{2}\left\langle w_{2}, w_{2}\right\rangle\\] Thus, \(\left.c_{2}=\left\langle v, w_{2}\right\rangle /\left\langle w_{2}, w_{2}\right\rangle . \text { (That is, } c_{2} \text { is the component of } v \text { along } w_{2} .\right)\)

Find \(\cos \theta\) where \(\theta\) is the angle between: (a) \(u=(1,3,-5,4)\) and \(v=(2,-3,4,1)\) in \(\mathbf{R}^{4}\) (b) \(A=\left[\begin{array}{lll}9 & 8 & 7 \\ 6 & 5 & 4\end{array}\right]\) and \(B=\left[\begin{array}{lll}1 & 2 & 3 \\ 4 & 5 & 6\end{array}\right],\) where \(\langle A, B\rangle=\operatorname{tr}\left(B^{T} A\right)\) Use \(\cos \theta=\frac{\langle u, v\rangle}{\|u\|\|v\|}\) (a) Compute: \\[ \langle u, v\rangle=2-9-20+4=-23, \quad\|u\|^{2}=1+9+25+16=51, \quad\|v\|^{2}=4+9+16+1=30 \\] Thus, \\[ \cos \theta=\frac{-23}{\sqrt{51} \sqrt{30}}=\frac{-23}{3 \sqrt{170}} \\] (b) \(\operatorname{Use}\langle A, B\rangle=\operatorname{tr}\left(B^{T} A\right)=\sum_{i=1}^{m} \sum_{j=1}^{n} a_{i j} b_{i j},\) the sum of the products of corresponding entries. \\[\langle A, B\rangle=9+16+21+24+25+24=119\\] Use \(\|A\|^{2}=\langle A, A\rangle=\sum_{i=1}^{m} \sum_{j=1}^{n} a_{i j}^{2},\) the sum of the squares of all the elements of \(A\) \\[\begin{array}{lll}\|A\|^{2}=\langle A, A\rangle=9^{2}+8^{2}+7^{2}+6^{2}+5^{2}+4^{2}=271, & \text { and so } & \|A\|=\sqrt{271} \\ \|B\|^{2}=\langle B, B\rangle=1^{2}+2^{2}+3^{2}+4^{2}+5^{2}+6^{2}=91, & \text { and so } & \|B\|=\sqrt{91} \end{array}\\] Thus, \\[\cos \theta=\frac{119}{\sqrt{271} \sqrt{91}}\\]
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