Book edition 5th

Author(s) Otto Bretscher

Pages 442 pages

ISBN 9780321796974

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Short Answer

We say that an $n \times n$ matrix A is triangulizable if is similar to an upper triangular $n \times n$ matrix B.
a. Give an example of a matrix with real entries that fails to be triangulizable over R .
b. Show that any $n \times n$ matrix with complex entries is triangulizable over C . Hint: Give a proof by induction analogous to the proof of Theorem 8.1.1.

a) The triangulizable matrix over R since it complex eigen values are

$λ_{1} = - i \times \sqrt{6} + 1 and λ_{1} = i \times \sqrt{6} + 1$

b) $\Rightarrow R^{- 1} [\begin{matrix} λ & \vec{v} \\ 0 & B \end{matrix}] R = S^{- 1} A S$

we can conclude that matrix with complex entries is triangulizable over C

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: DefineTriangularizable matric:

Triangularizable matrices are those that are similar to triangular matrices. In a nutshell, this is the same as stabilising a flag: upper triangular matrices are the ones that keep the standard flag, which is determined by the standard ordered basis.

Step 2: Matrix with real entries fails to be triangulizable and complex entries:

Consider an A is triangulizable matrix is identical to and upper triangular matrix . If $S^{- 1} A S$ is an upper triangular matrix then the first column of S is an eigen vector of A. Hence any matrix without real eigen vectors fails to be triangulizable vector over R .Consider the matrix as

$[\begin{matrix} 1 & 2 \\ - 3 & 1 \end{matrix}]$

With real entries. To find the eigen values are as follows

$d e t ([\begin{matrix} 1 & 2 \\ - 3 & 1 \end{matrix}] - [\begin{matrix} λ & 0 \\ 0 & λ \end{matrix}]) = 0 \Rightarrow |\begin{matrix} 1 - λ & 2 \\ - 3 & 1 - λ \end{matrix}| = 0 \Rightarrow λ^{2} - 2 λ + 7 = 0 \Rightarrow (λ + (i \times \sqrt{6} - 1)) \times (λ - (i \times \sqrt{6} + 1)) = 0 λ_{1} = - i \times \sqrt{6} + 1 a n d λ_{2} = i \times \sqrt{6} + 1$

The triangulizable matrix over R since it complex eigen values are

$λ_{1} = - i \times \sqrt{6} + 1 a n d λ_{2} = i \times \sqrt{6} + 1$

Consider the given problem for any $n \times n$ matrix with complex entries is triangulizable over C .True value for (n-1) .A real eigen value of $λ$ for A and eigen vector of length 1 for $λ$ .For every A there exists a complex invertible matrix P whose first column is an eigen vector of A .It can be represented as below

$P^{- 1} A P = [\begin{matrix} λ & \vec{V} \\ 0 & B \end{matrix}]$

Hypothesis of induction if B is triangulizable matrix such that there exists $(n - 1) \times (n - 1)$ matrix Q that satisfies

$Q^{- 1} B Q = T$

Consider

$R = [\begin{matrix} 1 & 0 \\ 0 & Q \end{matrix}]$

To evaluate

$R^{- 1} = [\begin{matrix} λ & \vec{V} \\ 0 & B \end{matrix}] R = [\begin{matrix} 1 & 0 \\ 0 & Q^{- 1} \end{matrix}] [\begin{matrix} λ & \vec{V} \\ 0 & B \end{matrix}] [\begin{matrix} 1 & 0 \\ 0 & Q \end{matrix}] = [\begin{matrix} λ & \vec{v} Q \\ 0 & Q^{- 1} B Q \end{matrix}] \Rightarrow R^{- 1} [\begin{matrix} λ & \vec{V} \\ 0 & B \end{matrix}] R = [\begin{matrix} λ & \vec{V} Q \\ 0 & T \end{matrix}]$

Hence $R^{- 1} [\begin{matrix} λ & \vec{V} \\ 0 & B \end{matrix}] R$ is upper triangular matrix

Consider as $R^{- 1} [\begin{matrix} λ & \vec{V} \\ 0 & B \end{matrix}] R$

$R^{- 1} [\begin{matrix} λ & \vec{V} \\ 0 & B \end{matrix}] R = R^{- 1} P^{- 1} A P R = (P R))^{- 1} A P R (S = P R) \Rightarrow R^{- 1} [\begin{matrix} λ & \vec{V} \\ 0 & B \end{matrix}] R = S^{- 1} A S$

Hence, we can conclude that matrix with complex entries is triangulizable over C

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Linear Algebra With Applications

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Step by Step Solution

Step 1: DefineTriangularizable matric:

Step 2: Matrix with real entries fails to be triangulizable and complex entries:

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Linear Algebra With Applications

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Step 1: DefineTriangularizable matric:

Step 2: Matrix with real entries fails to be triangulizable and complex entries:

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