Consider a matrix

This matrix has eigenvalues

So

*A*is a 3-by-3 matrix with 3 different eigenvalues, therefore it is diagonalizable. Note that if there are exactly n distinct eigenvalues in an n×n matrix then this matrix is diagonalizable.
These eigenvalues are the values that will appear in the diagonalized form of matrix , so by finding the eigenvalues of we have diagonalized it. We could stop here, but it is a good check to use the eigenvectors to diagonalize .

The eigenvectors of

*A*are
One can easily check that

Now, let

*P*be the matrix with these eigenvectors as its columns:
Note there is no preferred order of the eigenvectors in

*P*; changing the order of the eigenvectors in*P*just changes the order of the eigenvalues in the diagonalized form of*A*.
Then

*P*diagonalizes*A*, as a simple computation confirms:
Note that the eigenvalues appear in the diagonal matrix.

#### Alternative Method

Starting with: , where , and the Diagonalization matrix is:

Distribute into the column vectors of .

Then can be broken down to its column vectors as follows:

Multiplying on the left side of the equation gives:

Setting each entry of the matrix to its corresponding entry:

Then the equations can be solved as follows, using the same process for both:

and it solves for , which is the first entry in the diagonal matrix, and also the first eigenvalue.

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