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Expert-verified Found in: Page 393 ### Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974 # We say that an ${\mathbit{n}}{\mathbf{×}}{\mathbit{n}}$ matrix A is triangulizable if is similar to an upper triangular ${\mathbit{n}}{\mathbf{×}}{\mathbit{n}}$ matrix B. a. Give an example of a matrix with real entries that fails to be triangulizable over R . b. Show that any ${\mathbit{n}}{\mathbf{×}}{\mathbit{n}}$ 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

${\mathrm{\lambda }}_{1}=-\mathrm{i}×\sqrt{6}+1\mathrm{and}{\mathrm{\lambda }}_{1}=\mathrm{i}×\sqrt{6}+1$

b) $⇒{R}^{-1}\left[\begin{array}{cc}\lambda & \stackrel{\to }{v}\\ 0& B\end{array}\right]R={S}^{-1}AS$

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

See the step by step solution

## 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 ${S}^{-1}AS$ 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

$\left[\begin{array}{cc}1& 2\\ -3& 1\end{array}\right]$

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

$det\left(\left[\begin{array}{cc}1& 2\\ -3& 1\end{array}\right]-\left[\begin{array}{cc}\lambda & 0\\ 0& \lambda \end{array}\right]\right)=0\phantom{\rule{0ex}{0ex}}⇒\left|\begin{array}{cc}1-\lambda & 2\\ -3& 1-\lambda \end{array}\right|=0\phantom{\rule{0ex}{0ex}}⇒{\lambda }^{2}-2\lambda +7=0\phantom{\rule{0ex}{0ex}}⇒\left(\lambda +\left(i×\sqrt{6}-1\right)\right)×\left(\lambda -\left(i×\sqrt{6}+1\right)\right)=0\phantom{\rule{0ex}{0ex}}{\lambda }_{1}=-i×\sqrt{6}+1and{\lambda }_{2}=i×\sqrt{6}+1$

The triangulizable matrix over R since it complex eigen values are

${\lambda }_{1}=-i×\sqrt{6}+1and{\lambda }_{2}=i×\sqrt{6}+1$

Consider the given problem for any $n×n$ matrix with complex entries is triangulizable over C .True value for (n-1) .A real eigen value of $\lambda$ for A and eigen vector of length 1 for $\lambda$ .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

${P}^{-1}AP=\left[\begin{array}{cc}\lambda & \stackrel{\to }{V}\\ 0& B\end{array}\right]$

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

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

Consider

$R=\left[\begin{array}{cc}1& 0\\ 0& Q\end{array}\right]$

To evaluate

${R}^{-1}=\left[\begin{array}{cc}\lambda & \stackrel{\to }{V}\\ 0& B\end{array}\right]R=\left[\begin{array}{cc}1& 0\\ 0& {Q}^{-1}\end{array}\right]\left[\begin{array}{cc}\lambda & \stackrel{\to }{V}\\ 0& B\end{array}\right]\left[\begin{array}{cc}1& 0\\ 0& Q\end{array}\right]\phantom{\rule{0ex}{0ex}}=\left[\begin{array}{cc}\lambda & \stackrel{\to }{v}Q\\ 0& {Q}^{-1}BQ\end{array}\right]\phantom{\rule{0ex}{0ex}}⇒{R}^{-1}\left[\begin{array}{cc}\lambda & \stackrel{\to }{V}\\ 0& B\end{array}\right]R=\left[\begin{array}{cc}\lambda & \stackrel{\to }{V}Q\\ 0& T\end{array}\right]$

Hence ${R}^{-1}\left[\begin{array}{cc}\lambda & \stackrel{\to }{V}\\ 0& B\end{array}\right]R$ is upper triangular matrix

Consider as ${R}^{-1}\left[\begin{array}{cc}\lambda & \stackrel{\to }{V}\\ 0& B\end{array}\right]R$

${R}^{-1}\left[\begin{array}{cc}\lambda & \stackrel{\to }{V}\\ 0& B\end{array}\right]R={R}^{-1}{P}^{-1}APR\phantom{\rule{0ex}{0ex}}=\left(PR\right){\right)}^{-1}APR\phantom{\rule{0ex}{0ex}}\left(S=PR\right)\phantom{\rule{0ex}{0ex}}⇒{R}^{-1}\left[\begin{array}{cc}\lambda & \stackrel{\to }{V}\\ 0& B\end{array}\right]R={S}^{-1}AS$

Hence, we can conclude that matrix with complex entries is triangulizable over C ### Want to see more solutions like these? 