Anonymous
Anonymous asked in Science & MathematicsMathematics · 7 years ago

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

Update:

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

1 Answer

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  • kb
    Lv 7
    7 years ago
    Favorite 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.

    -------------

    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.

    --------------

    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.

    --------------

    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.

    --------------

    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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