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I was practicing some linear algebra problems and I stopped at this one:

Without calculating the eigenvectors, show that the following matrix is diagonalizable and find the diagonal matrix to which it is similar.

$$ \begin{bmatrix} 3 & 1 & 4 \\ 0 & 2 & 6 \\ 0 & 0 & 5 \\ \end{bmatrix} $$

How could I know the if its diagonalizable or not without calculating the eigenvectors?
Please help, I'm studying for tomorrow's exam.

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Hint - It is a $\;3\times 3\;$ matrix with $\;3\;$ different eigenvalues

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    $\begingroup$ 3 different eigen values means it's diagnoalizable, right ? i was going to do that but i wasn't sure about it $\endgroup$ – Mr.Cat Jun 10 '14 at 17:27
  • $\begingroup$ Yes it does because eigenvectors belonging to distinct eigenvalues are linearly independent. $\endgroup$ – Timbuc Jun 10 '14 at 17:28
  • $\begingroup$ Corrected Git. Thank you. $\endgroup$ – Timbuc Jun 10 '14 at 17:31
  • $\begingroup$ what about finding the diagonal matrix :D ? !! $\endgroup$ – Mr.Cat Jun 10 '14 at 17:31
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    $\begingroup$ @Mr. Cat, having the eigenvalues that already is trivial. $\endgroup$ – Timbuc Jun 10 '14 at 17:32
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The characteristic polynomial is $(x-3)(x-2)(x-5)$ as you can see directly from the matrix. This is also the minimal polynomial. A matrix is diagonalizabe iff its minimal polynomial is a product of distinct linear factors. The diagonalization has the same diagonal as the original matrix.

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The given matrix is an upper triangular.We know that determinant of an upper triangular matrix is the product of its diagonals.Thus clearly characteristic polynomial is $(x-3)(x-2)(x-5)$ and so it has $3$ distinct eigen values.Now if a $n\times n$ matrix has $n$ distinct eigen values then it is diagonalisable.

Hence given matrix is diagonalisable.

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