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 October 26th, 2017, 05:51 PM #1 Member   Joined: Nov 2016 From: Kansas Posts: 73 Thanks: 1 Diagonalizable Matrix Let A= $\left(\begin{array}{cc} a & b\\ c & d \end{array}\right)$ be considered as a linear operator on Mat$_2$$_*$$_2$ ($\mathbb{C}$) $\ni$ M where M $\mapsto$ AM (a,b,c,d $\in$ $\mathbb{C}$) Is this matrix diagonalizable for any A. Prove your answer( proof or contradiction by example) Last edited by ZMD; October 26th, 2017 at 06:00 PM.
 October 27th, 2017, 05:08 AM #2 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 894 A matrix is diagonalizable if and only if it has a "complete set" of eigenvectors- that is, a set of eigenvectors that is a basis for the space. Having all distinct eigenvalues is sufficient but not necessary.
 October 31st, 2017, 10:10 AM #3 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 894 Since here A is any 2 by 2 matrix with any complex numbers as entries, you are actually asking "can any 2 by 2 matrix over the complex numbers be diagonalized". The answer to that is "no". There exist matrices that cannot be diagonalized. An example is $\begin{bmatrix}1 & 1 \\ 0 & 1 \end{bmatrix}$. That has the double eigenvalue $\lambda= 1$. The only eigenvectors are multiples of $\begin{bmatrix} 1 \\ 0 \end{bmatrix}$.
 November 4th, 2017, 05:09 AM #4 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 894 Since here A is any 2 by 2 matrix with any complex numbers as entries, you are actually asking "can any 2 by 2 matrix over the complex numbers be diagonalized". The answer to that is "no". There exist matrices that cannot be diagonalized. An example is $\displaystyle \begin{bmatrix}1 & 1 \\ 0 & 1 \end{bmatrix}$. That has the double eigenvalue $\displaystyle \lambda= 1$. The only eigenvectors are multiples of $\displaystyle \begin{bmatrix} 1 \\ 0 \end{bmatrix}$. Thanks from topsquark
 November 4th, 2017, 07:27 AM #5 Math Team     Joined: May 2013 From: The Astral plane Posts: 1,907 Thanks: 773 Math Focus: Wibbly wobbly timey-wimey stuff. @County Boy; Is this a version of on-line stuttering? -Dan

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