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October 26th, 2017, 05:51 PM   #1
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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.
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October 27th, 2017, 05:08 AM   #2
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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.
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October 31st, 2017, 10:10 AM   #3
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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 .
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November 4th, 2017, 05:09 AM   #4
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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 $\displaystyle \begin{bmatrix} 1 \\ 0 \end{bmatrix}$.
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November 4th, 2017, 07:27 AM   #5
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