# examining if a matrix is diagonizable

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.

## 3 Answers

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

• 3 different eigen values means it's diagnoalizable, right ? i was going to do that but i wasn't sure about it – Mr.Cat Jun 10 '14 at 17:27
• Yes it does because eigenvectors belonging to distinct eigenvalues are linearly independent. – Timbuc Jun 10 '14 at 17:28
• Corrected Git. Thank you. – Timbuc Jun 10 '14 at 17:31
• what about finding the diagonal matrix :D ? !! – Mr.Cat Jun 10 '14 at 17:31
• @Mr. Cat, having the eigenvalues that already is trivial. – Timbuc Jun 10 '14 at 17:32

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.

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.