Answer to Question #282371 in Linear Algebra for JOY

Question #282371

Find eigen values and associated eigen vectors of the matrix

A=[ 1 −3 3

−3 −5 3

6 −6 4 ] in the field R. Also find an invertible matrix p such that p−1 Ap is diagonal.


1
Expert's answer
2021-12-24T12:11:40-0500
AλI=[1λ3335λ3664λ]A-\lambda I=\begin{bmatrix} 1-\lambda & -3 & 3 \\ -3 & -5-\lambda & 3 \\ 6 & -6 & 4-\lambda \end{bmatrix}

The characteristic equation

det(AλI)=0\det(A-\lambda I)=0

(1λ)(5λ)(4λ)54+54(1-\lambda)(-5-\lambda)(4-\lambda)-54+54

18(5λ)+18(1λ)9(4λ)=0-18(-5-\lambda)+18(1-\lambda)-9(4-\lambda)=0

λ3+30λ+52=0-\lambda^3+30\lambda+52=0

λ2(λ+2)+2λ(λ+2)+26(λ+2)=0-\lambda^2(\lambda+2)+2\lambda(\lambda+2)+26(\lambda+2)=0

λ1=2\lambda_1=-2

λ22λ26=0\lambda^2-2\lambda-26=0

(λ1)2=27(\lambda-1)^2=27

λ2=133,λ3=1+33\lambda_2=1-3\sqrt{3}, \lambda_3=1+3\sqrt{3}

λ1=2,λ2=133,λ3=1+33\lambda_1=-2,\lambda_2=1-3\sqrt{3}, \lambda_3=1+3\sqrt{3}

These are eigenvalues.


λ=2\lambda=-2


[1λ3335λ3664λ]=[333333666]\begin{bmatrix} 1-\lambda & -3 & 3 \\ -3 & -5-\lambda & 3 \\ 6 & -6 & 4-\lambda \end{bmatrix}=\begin{bmatrix} 3 & -3 & 3 \\ -3 & -3 & 3 \\ 6 & -6 & 6 \end{bmatrix}

R1=R1/3R_1=R_1/3


[111333666]\begin{bmatrix} 1 & -1 & 1 \\ -3 & -3 & 3 \\ 6 & -6 & 6 \end{bmatrix}

R2=R2+3R1R_2=R_2+3R_1


[111066666]\begin{bmatrix} 1 & -1 & 1 \\ 0 & -6 & 6 \\ 6 & -6 & 6 \end{bmatrix}

R3=R36R1R_3=R_3-6R_1


[111066000]\begin{bmatrix} 1 & -1 & 1 \\ 0 & -6 & 6 \\ 0 & 0 & 0 \end{bmatrix}

R2=R2/(6)R_2=R_2/(-6)


[111011000]\begin{bmatrix} 1 & -1 & 1 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \end{bmatrix}

R1=R1+R2R_1=R_1+R_2


[100011000]\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \end{bmatrix}

If we take v3=t,v_3=t, then v1=0,v2=t,v3=t.v_1=0, v_2=t, v_3=t.

Thus the eigenvector is

v=[011]\bold v=\begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix}


λ=133\lambda=1-3\sqrt{3}

[1λ3335λ3664λ]\begin{bmatrix} 1-\lambda & -3 & 3 \\ -3 & -5-\lambda & 3 \\ 6 & -6 & 4-\lambda \end{bmatrix}

=[333336+333663+33]=\begin{bmatrix} 3\sqrt{3} & -3 & 3 \\ -3 & -6+3\sqrt{3} & 3 \\ 6 & -6 & 3+3\sqrt{3} \end{bmatrix}

R1=3R1/9R_1=\sqrt{3}R_1/9

[13/33/336+333663+33]\begin{bmatrix} 1 & -\sqrt{3}/3 & \sqrt{3}/3 \\ -3 & -6+3\sqrt{3} & 3 \\ 6 & -6 & 3+3\sqrt{3} \end{bmatrix}

R2=R2+3R1R_2=R_2+3R_1

[13/33/306+233+3663+33]\begin{bmatrix} 1 & -\sqrt{3}/3 & \sqrt{3}/3 \\ 0 & -6+2\sqrt{3} & 3+\sqrt{3} \\ 6 & -6 & 3+3\sqrt{3} \end{bmatrix}

R3=R36R1R_3=R_3-6R_1

[13/33/306+233+306+233+3]\begin{bmatrix} 1 & -\sqrt{3}/3 & \sqrt{3}/3 \\ 0 & -6+2\sqrt{3} & 3+\sqrt{3} \\ 0 & -6+2\sqrt{3} & 3+\sqrt{3} \end{bmatrix}

R2=R2/(6+23)R_2=R_2/(-6+2\sqrt{3})

[13/33/30113/206+233+3]\begin{bmatrix} 1 & -\sqrt{3}/3 & \sqrt{3}/3 \\ 0 & 1 & -1-\sqrt{3}/2 \\ 0 & -6+2\sqrt{3} & 3+\sqrt{3} \end{bmatrix}

R1=R1+R2/3R_1=R_1+R_2/\sqrt{3}

[101/20113/206+233+3]\begin{bmatrix} 1 & 0 & -1/2 \\ 0 & 1 & -1-\sqrt{3}/2 \\ 0 & -6+2\sqrt{3} & 3+\sqrt{3} \end{bmatrix}

R3=R3+(623)R2R_3=R_3+(6-2\sqrt{3})R_2


[101/20113/2000]\begin{bmatrix} 1 & 0 & -1/2 \\ 0 & 1 & -1-\sqrt{3}/2 \\ 0 & 0 & 0 \end{bmatrix}

If we take u3=t,u_3=t, then u1=t/2,u2=(2+3)t/2,u3=t.u_1=t/2, u_2=(2+\sqrt{3})t/2, u_3=t.

Thus the eigenvector is

u=[1/2(2+3)/21]\bold u=\begin{bmatrix} 1/2 \\ (2+\sqrt{3})/2 \\ 1 \end{bmatrix}



λ=1+33\lambda=1+3\sqrt{3}

[1λ3335λ3664λ]\begin{bmatrix} 1-\lambda & -3 & 3 \\ -3 & -5-\lambda & 3 \\ 6 & -6 & 4-\lambda \end{bmatrix}

=[33333633366333]=\begin{bmatrix} -3\sqrt{3} & -3 & 3 \\ -3 & -6-3\sqrt{3} & 3 \\ 6 & -6 & 3-3\sqrt{3} \end{bmatrix}

R1=3R1/9R_1=-\sqrt{3}R_1/9

[13/33/33633366333]\begin{bmatrix} 1 & \sqrt{3}/3 & -\sqrt{3}/3 \\ -3 & -6-3\sqrt{3} & 3 \\ 6 & -6 & 3-3\sqrt{3} \end{bmatrix}

R2=R2+3R1R_2=R_2+3R_1

[13/33/306233366333]\begin{bmatrix} 1 & \sqrt{3}/3 & -\sqrt{3}/3 \\ 0 & -6-2\sqrt{3} & 3-\sqrt{3} \\ 6 & -6 & 3-3\sqrt{3} \end{bmatrix}

R3=R36R1R_3=R_3-6R_1

[13/33/3062333062333]\begin{bmatrix} 1 & \sqrt{3}/3 & -\sqrt{3}/3 \\ 0 & -6-2\sqrt{3} & 3-\sqrt{3} \\ 0 & -6-2\sqrt{3} & 3-\sqrt{3} \end{bmatrix}

R2=R2/(623)R_2=R_2/(-6-2\sqrt{3})

[13/33/3011+3/2062333]\begin{bmatrix} 1 & \sqrt{3}/3 & -\sqrt{3}/3 \\ 0 & 1 & -1+\sqrt{3} /2\\ 0 & -6-2\sqrt{3} & 3-\sqrt{3} \end{bmatrix}

R1=R1R2/3R_1=R_1-R_2/\sqrt{3}

[101/2011+3/2062333]\begin{bmatrix} 1 & 0 & -1/2 \\ 0 & 1 & -1+\sqrt{3} /2\\ 0 & -6-2\sqrt{3} & 3-\sqrt{3} \end{bmatrix}

R3=R3+(6+23)R2R_3=R_3+(6+2\sqrt{3})R_2

[101/2011+3/2000]\begin{bmatrix} 1 & 0 & -1/2 \\ 0 & 1 & -1+\sqrt{3} /2\\ 0 & 0 & 0 \end{bmatrix}


If we take w3=t,w_3=t, then w1=t/2,w2=(23)t/2,w3=t.w_1=t/2, w_2=(2-\sqrt{3})t/2, w_3=t.

Thus the eigenvector is

w=[1/2(23)/21]\bold w=\begin{bmatrix} 1/2 \\ (2-\sqrt{3})/2 \\ 1 \end{bmatrix}


Form the matrix P,P, whose column ii is eigenvector no. i:i:


P=[01/21/211+3/213/2111]P=\begin{bmatrix} 0 & 1/2 & 1/2 \\ 1 & 1+\sqrt{3}/2 & 1-\sqrt{3} /2\\ 1 & 1 & 1 \end{bmatrix}

Form the matrix DD whose element at row i,i, column ii is eigenvalue no. i:i:


D=[20001330001+33]D=\begin{bmatrix} -2 & 0 & 0 \\ 0 & 1-3\sqrt{3} & 0\\ 0 & 0 & 1+3\sqrt{3} \end{bmatrix}

The matrices PP and DD  are such that the initial matrix


A=[133353664]=PDP1A=\begin{bmatrix} 1 & -3 & 3 \\ -3 & -5 & 3 \\ 6 & -6 & 4 \end{bmatrix}=PDP^{-1}


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