October 26th, 2017, 05:51 PM  #1 
Member Joined: Nov 2016 From: Kansas Posts: 68 Thanks: 0  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: 2,944 Thanks: 797 
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: 2,944 Thanks: 797 
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 . That has the double eigenvalue . The only eigenvectors are multiples of .

November 4th, 2017, 05:09 AM  #4 
Math Team Joined: Jan 2015 From: Alabama Posts: 2,944 Thanks: 797 
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 $\displaystyle \begin{bmatrix} 1 \\ 0 \end{bmatrix}$.

November 4th, 2017, 07:27 AM  #5 
Math Team Joined: May 2013 From: The Astral plane Posts: 1,664 Thanks: 653 Math Focus: Wibbly wobbly timeywimey stuff. 
@County Boy; Is this a version of online stuttering? Dan 

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