Question #248553

Diagonalize the matrix

10 −2 −5

−2 2 3

−5 3 5


1
Expert's answer
2021-10-11T02:58:32-0400
A=(1025223535)A=\begin{pmatrix} 10 & -2 & -5 \\ -2 & 2 & 3 \\ -5 & 3 & 5 \\ \end{pmatrix}

Find the eigen values


AλI=(10λ2522λ3535λ)A-\lambda I=\begin{pmatrix} 10-\lambda & -2 & -5 \\ -2 & 2-\lambda & 3 \\ -5 & 3 & 5-\lambda \\ \end{pmatrix}

det(AλI)=10λ2522λ3535λ\det(A-\lambda I)=\begin{vmatrix} 10-\lambda & -2 & -5 \\ -2 & 2-\lambda & 3 \\ -5 & 3 & 5-\lambda \\ \end{vmatrix}

=(10λ2λ335λ+22355λ=(10-\lambda\begin{vmatrix} 2-\lambda & 3 \\ 3 & 5-\lambda \end{vmatrix}+2\begin{vmatrix} -2 & 3 \\ -5 & 5-\lambda \end{vmatrix}

522λ53=(10λ)(107λ+λ29)-5\begin{vmatrix} -2 & 2-\lambda \\ -5 & 3 \end{vmatrix}=(10-\lambda)(10-7\lambda+\lambda^2-9)

+2(10+2λ+15)5(6+105λ)+2(-10+2\lambda+15)-5(-6+10-5\lambda)

=1070λ+10λ2λ+7λ2λ3+10+4λ=10-70\lambda+10\lambda^2-\lambda+7\lambda^2-\lambda^3+10+4\lambda

20+25λ=λ3+17λ242λ-20+25\lambda=-\lambda^3+17\lambda^2-42\lambda

=λ(λ3)(λ14)=-\lambda(\lambda-3)(\lambda-14)

det(AλI)=0=>λ(λ3)(λ14)=0\det(A-\lambda I)=0=>-\lambda(\lambda-3)(\lambda-14)=0



The roots are λ1=14,λ2=3,λ3=0.\lambda_1=14, \lambda_2=3, \lambda_3=0.

These are eigenvalues.

Find the eigenvectors

λ=14\lambda =14


AλI=(1014252214353514)A-\lambda I=\begin{pmatrix} 10-14 & -2 & -5 \\ -2 & 2-14 & 3 \\ -5 & 3 & 5-14 \\ \end{pmatrix}

=(4252123539)=\begin{pmatrix} -4 & -2 & -5 \\ -2 & -12 & 3 \\ -5 & 3 & -9 \\ \end{pmatrix}

R2=R2R1/2R_2=R_2-R_1/2


(42501111/2539)\begin{pmatrix} -4 & -2 & -5 \\ 0 & -11 & 11/2 \\ -5 & 3 & -9 \\ \end{pmatrix}

R3=R35R1/4R_3=R_3-5R_1/4


(42501111/2011/211/4)\begin{pmatrix} -4 & -2 & -5 \\ 0 & -11 & 11/2 \\ 0 & 11/2 & -11/4 \\ \end{pmatrix}

R3=R3+R2/2R_3=R_3+R_2/2


(42501111/2000)\begin{pmatrix} -4 & -2 & -5 \\ 0 & -11 & 11/2 \\ 0 & 0 & 0 \\ \end{pmatrix}

R2=R2/(11)R_2=R_2/(-11)


(425011/2000)\begin{pmatrix} -4 & -2 & -5 \\ 0 & 1 & -1/2 \\ 0 & 0 & 0 \\ \end{pmatrix}

R1=R1+2R2R_1=R_1+2R_2


(406011/2000)\begin{pmatrix} -4 & 0 & -6 \\ 0 & 1 & -1/2 \\ 0 & 0 & 0 \\ \end{pmatrix}

R1=R1/(4)R_1=R_1/(-4)


(103/2011/2000)\begin{pmatrix} 1 & 0 & 3/2 \\ 0 & 1 & -1/2 \\ 0 & 0 & 0 \\ \end{pmatrix}

If we take v3=t,v_3=t, then v1=32t,v2=12t.v_1=-\dfrac{3}{2}t, v_2=\dfrac{1}{2}t.

The eigenvector is v=(3/21/21)v=\begin{pmatrix} - 3/2 \\ 1/2 \\ 1 \end{pmatrix}


λ=3\lambda =3

AλI=(1032522335353)A-\lambda I=\begin{pmatrix} 10-3 & -2 & -5 \\ -2 & 2-3 & 3 \\ -5 & 3 & 5-3 \\ \end{pmatrix}

=(725213532)=\begin{pmatrix} 7 & -2 & -5 \\ -2 & -1 & 3 \\ -5 & 3 & 2 \\ \end{pmatrix}

R2=R2+2R1/7R_2=R_2+2R_1/7

(725011/711/7532)\begin{pmatrix} 7 & -2 & -5 \\ 0 & -11/7 & 11/7 \\ -5 & 3 & 2 \\ \end{pmatrix}

R3=R3+5R1/7R_3=R_3+5R_1/7

(725011/711/7011/711/7)\begin{pmatrix} 7 & -2 & -5 \\ 0 & -11/7 & 11/7 \\ 0 & 11/7 & -11/7 \\ \end{pmatrix}

R3=R3+R2R_3=R_3+R_2

(725011/711/7000)\begin{pmatrix} 7 & -2 & -5 \\ 0 & -11/7 & 11/7 \\ 0 & 0 & 0 \\ \end{pmatrix}

R2=7R2/11R_2=-7R_2/11

(725011000)\begin{pmatrix} 7 & -2 & -5 \\ 0 & 1 &-1 \\ 0 & 0 & 0 \\ \end{pmatrix}

R1=R1+2R2R_1=R_1+2R_2

(707011000)\begin{pmatrix} 7 & 0 & -7 \\ 0 & 1 &-1 \\ 0 & 0 & 0 \\ \end{pmatrix}

R1=R1/7R_1=R_1/7

(101011000)\begin{pmatrix} 1 & 0 & -1 \\ 0 & 1 &-1 \\ 0 & 0 & 0 \\ \end{pmatrix}

If we take u3=t,u_3=t, then u1=t,u2=t.u_1=t, u_2=t.

The eigenvector is u=(111)u=\begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}



λ=0\lambda =0

AλI=(1002522035350)A-\lambda I=\begin{pmatrix} 10-0 & -2 & -5 \\ -2 & 2-0 & 3 \\ -5 & 3 & 5-0 \\ \end{pmatrix}

=(1025223535)=\begin{pmatrix} 10 & -2 & -5 \\ -2 & 2 & 3 \\ -5 & 3 & 5 \\ \end{pmatrix}

R2=R2+R1/5R_2=R_2+R_1/5

(102508/52535)\begin{pmatrix} 10 & -2 & -5 \\ 0 & 8/5 &2 \\ -5 & 3 & 5 \\ \end{pmatrix}

R3=R3+R1/2R_3=R_3+R_1/2

(102508/52025/2)\begin{pmatrix} 10 & -2 & -5 \\ 0 & 8/5 &2 \\ 0 & 2 & 5/2 \\ \end{pmatrix}

R3=R35R2/4R_3=R_3-5R_2/4

(102508/52000)\begin{pmatrix} 10 & -2 & -5 \\ 0 & 8/5 &2 \\ 0 & 0 & 0 \\ \end{pmatrix}

R2=5R2/8R_2=5R_2/8

(1025015/4000)\begin{pmatrix} 10 & -2 & -5 \\ 0 & 1 & 5/4 \\ 0 & 0 & 0 \\ \end{pmatrix}

R1=R1+2R2R_1=R_1+2R_2

(1005/2015/4000)\begin{pmatrix} 10 & 0 & -5/2 \\ 0 & 1 & 5/4 \\ 0 & 0 & 0 \\ \end{pmatrix}

R1=R1/10R_1=R_1/10

(101/4015/4000)\begin{pmatrix} 1 & 0 & -1/4 \\ 0 & 1 & 5/4 \\ 0 & 0 & 0 \\ \end{pmatrix}

If we take w3=t,w_3=t, then w1=14t,w2=54t.w_1=\dfrac{1}{4}t, w_2=-\dfrac{5}{4}t.

The eigenvector is u=(1/45/41)u=\begin{pmatrix} 1/4\\ -5/4\\ 1 \end{pmatrix}

Form the matrix PP


P=(3/211/41/215/4111)P=\begin{pmatrix} -3/2 & 1 & 1/4 \\ 1/2 & 1 & -5/4 \\ 1 & 1 & 1 \end{pmatrix}

Form the diagonal matrix DD


D=(1400030000)D=\begin{pmatrix} 14 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 0 \end{pmatrix}

The matrices PP and DD are such that the initial matrix


A=(1025223535)=PDP1A=\begin{pmatrix} 10 & -2 & -5 \\ -2 & 2 & 3 \\ -5 & 3 & 5 \\ \end{pmatrix}=PDP^{-1}

P=(3/211/41/215/4111)P=\begin{pmatrix} -3/2 & 1 & 1/4 \\ 1/2 & 1 & -5/4 \\ 1 & 1 & 1 \end{pmatrix}

D=(1400030000)D=\begin{pmatrix} 14 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 0 \end{pmatrix}


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