Math  /  Algebra

QuestionFind matrix PP and DD for the following Group 1: A=[204030006]A=\left[\begin{array}{lll}2 & 0 & 4 \\ 0 & 3 & 0 \\ 0 & 0 & 6\end{array}\right] Group 2: B=[120041007]B=\left[\begin{array}{ccc}-1 & 2 & 0 \\ 0 & 4 & 1 \\ 0 & 0 & 7\end{array}\right]

Studdy Solution

STEP 1

What is this asking? We need to find the magical matrices PP and DD that diagonalize matrices AA and BB. Watch out! Remember, not all matrices can be diagonalized!
We need to make sure we have enough linearly independent eigenvectors.

STEP 2

1. Diagonalize matrix AA
2. Diagonalize matrix BB

STEP 3

Alright, let's **kick things off** by finding the eigenvalues of AA!
For a triangular matrix like AA, the eigenvalues are just the entries on the main diagonal.
So, our **eigenvalues** are λ1=2\lambda_1 = 2, λ2=3\lambda_2 = 3, and λ3=6\lambda_3 = 6.
Boom!

STEP 4

Now, for the eigenvectors!
For λ1=2\lambda_1 = 2, we solve (A2I)v1=0(A - 2I)v_1 = 0.
This gives us: [004010004]v1=[000]\begin{bmatrix} 0 & 0 & 4 \\ 0 & 1 & 0 \\ 0 & 0 & 4 \end{bmatrix} v_1 = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} A solution is v1=[100]v_1 = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}.
For λ2=3\lambda_2 = 3, we solve (A3I)v2=0(A - 3I)v_2 = 0: [104000003]v2=[000] \begin{bmatrix} -1 & 0 & 4 \\ 0 & 0 & 0 \\ 0 & 0 & 3 \end{bmatrix} v_2 = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} A solution is v2=[010]v_2 = \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}.
For λ3=6\lambda_3 = 6, we solve (A6I)v3=0(A - 6I)v_3 = 0: [404030000]v3=[000] \begin{bmatrix} -4 & 0 & 4 \\ 0 & -3 & 0 \\ 0 & 0 & 0 \end{bmatrix} v_3 = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} A solution is v3=[101]v_3 = \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix}.

STEP 5

PP is formed by placing the eigenvectors as columns: P=[101010001]P = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}. DD is a diagonal matrix with the eigenvalues on the diagonal: D=[200030006]D = \begin{bmatrix} 2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 6 \end{bmatrix}.

STEP 6

Since BB is also triangular, its eigenvalues are its diagonal entries: λ1=1\lambda_1 = -1, λ2=4\lambda_2 = 4, and λ3=7\lambda_3 = 7.

STEP 7

Solving for each eigenvector similarly to how we did for AA, we get v1=[100]v_1 = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}, v2=[2/510]v_2 = \begin{bmatrix} 2/5 \\ 1 \\ 0 \end{bmatrix}, and v3=[2/151/31]v_3 = \begin{bmatrix} 2/15 \\ 1/3 \\ 1 \end{bmatrix}.

STEP 8

P=[12/52/15011/3001]P = \begin{bmatrix} 1 & 2/5 & 2/15 \\ 0 & 1 & 1/3 \\ 0 & 0 & 1 \end{bmatrix} and D=[100040007]D = \begin{bmatrix} -1 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 7 \end{bmatrix}.

STEP 9

For matrix AA, P=[101010001]P = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} and D=[200030006]D = \begin{bmatrix} 2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 6 \end{bmatrix}.
For matrix BB, P=[12/52/15011/3001]P = \begin{bmatrix} 1 & 2/5 & 2/15 \\ 0 & 1 & 1/3 \\ 0 & 0 & 1 \end{bmatrix} and D=[100040007]D = \begin{bmatrix} -1 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 7 \end{bmatrix}.

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