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Determine if each of the following matrices is diagonalizable and explain why.

a)

$$ \begin{bmatrix} 3 & 1 & 4 \\ 0 & 1 & 5 \\ 0 & 0 & 9 \end{bmatrix} $$

b)

$$ \begin{bmatrix} 1 & 2 & 3 & 4 \\ 2 & 5 & 6 & 8 \\ 3 & 6 & 7 & 9 \\ 4 & 8 & 9 & 1 \end{bmatrix} $$

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2  
What, precisely, are you having trouble with? –  Hurkyl May 3 '13 at 23:09
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Fomo Loco: you've been asking lots of questions, and getting many answers, and you keep coming back. Why, then, I wonder, if you're finding the site helpful, haven't you accepted ANY of the answers you've received? –  amWhy May 3 '13 at 23:21
    
Both are diagonalizable because of known facts: you know the eigenvalues of the first one and the second matrix has a very clear property. –  egreg May 3 '13 at 23:22
    
I am sorry amWhy. I wasn't doing it on purpose. How do I accept answers? –  Fomo Loco May 3 '13 at 23:45
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You can also go back and accept earlier answers ;-) You get two reputation points for each answer you accept! –  amWhy May 4 '13 at 3:31

1 Answer 1

The matrix in a) is triangular, and thus it has its eigenvalues on the diagonal. Since the three eigenvalues are distinct, we know that there are three linearly independent eigenvectors, and the matrix is therefore diagonalisable.

The matrix in b) has real elements and is symmetric ($A=A^T$). By the spectral theorem, we know that it is diagonalisable. (We even know that the eigenvalues are real, and that the eigenvectors can be chosen orthogonal.)

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