# Linear algebra!! HELP?

Find an invertible matrix P and a diagonal matrix D such that (P^−1)AP=D.

A=

4 0 -2

6 -1 -6

4 0 -2

Ahhhhh thank you so much kb! It makes sense now :)

### 1 Answer

- kbLv 77 years agoFavorite Answer
First, find the eigenvalues of A by solving |A - λI| = 0.

|4-λ 0 -2|

|6 -1-λ -6| = 0

|4 0 -2-λ|

Expanding down the second column:

(-1 - λ) [(λ^2 - 2λ - 8) + 8] = 0

==> λ = -1, 0, 2.

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Next, we find corresponding eigenvectors.

For λ = -1, we solve (A - (-1)I)v = 0.

[5 0 -2|0]

[6 0 -6|0]

[4 0 -1|0], which reduces to

[1 0 0|0]

[0 0 1|0]

[0 0 0|0], yielding eigenvector (0, 1, 0)^T.

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For λ = 0, we solve (A - 0I)v = 0.

[4 0 -2|0]

[6 -1 -6|0]

[4 0 -2|0], which reduces to

[1 0 -1/2|0]

[0 1 3|0]

[0 0 0|0], yielding eigenvector (1, -6, 2)^T.

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For λ = 2, we solve (A - 2I)v = 0.

[2 0 -2|0]

[6 -3 -6|0]

[4 0 -4|0], which reduces to

[1 0 -1|0]

[0 1 0|0]

[0 0 0|0], yielding eigenvector (1, 0, 1)^T.

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So, let D be the diagonal matrix of eigenvalues

[-1 0 0]

[0 0 0]

[0 0 2],

and let P be the matrix whose columns are the corresponding eigenvectors

[0 1 1]

[1 -6 0]

[0 2 1].

Then, P^(-1) AP = D, as required.

I hope this helps!

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