Question #153072
3. Reduce the quadratic form x 1 ^ 2 +5 x 2 ^ 2 + x 3 ^ 2 +2x 1 x 2 +2x 2 x 3 +6x 3 x 1 to canonical form through an orthogonal transformation and also find its rank, index, signature and nature of the quadratic form.
1
Expert's answer
2020-12-29T17:11:58-0500

The matrix of the quadratic form is


A=[113151311]A=\begin{bmatrix} 1 & 1 & 3 \\ 1 & 5 & 1 \\ 3 & 1 & 1 \\ \end{bmatrix}

Find the eigenvalues and eigenvectors of A


AλI=[1λ1315λ1311λ]A-\lambda I=\begin{bmatrix} 1-\lambda & 1 & 3 \\ 1 & 5-\lambda & 1 \\ 3 & 1 & 1-\lambda \\ \end{bmatrix}

1λ1315λ1311λ=(1λ)5λ111λ\begin{vmatrix} 1-\lambda & 1 & 3 \\ 1 & 5-\lambda & 1 \\ 3 & 1 & 1-\lambda \\ \end{vmatrix}=(1-\lambda)\begin{vmatrix} 5-\lambda & 1 \\ 1 & 1-\lambda \end{vmatrix}

(1)1131λ+(3)15λ31-(1)\begin{vmatrix} 1 & 1 \\ 3 & 1-\lambda \end{vmatrix}+(3)\begin{vmatrix} 1 & 5-\lambda \\ 3 & 1 \end{vmatrix}

=(1λ)2(5λ)1+λ1+λ+3+345+9λ=(1-\lambda)^2(5-\lambda)-1+\lambda-1+\lambda+3+3-45+9\lambda

=λ3+7λ236=-\lambda^3+7\lambda^2-36

This is a characteristic polynomial.

Solve the equation


λ3+7λ236=0-\lambda^3+7\lambda^2-36=0

λ2(λ6)+(λ6)(λ+6)=0-\lambda^2(\lambda-6)+(\lambda-6)(\lambda+6)=0

(λ6)(λ3)(λ+2)=0-(\lambda-6)(\lambda-3)(\lambda+2)=0

The roots are: λ1=6,λ2=3,λ3=2.\lambda_1=6,\lambda_2=3,\lambda_3=-2.

These are the eigenvalues.

Find the eigenvectors.

λ1=6\lambda_1=6


[1λ1315λ1311λ]=[513111315]\begin{bmatrix} 1-\lambda & 1 & 3 \\ 1 & 5-\lambda & 1 \\ 3 & 1 & 1-\lambda \\ \end{bmatrix}=\begin{bmatrix} -5 & 1 & 3 \\ 1 & -1 & 1 \\ 3 & 1 & -5 \\ \end{bmatrix}

R2=R2+(1/5)R1R_2=R_2+(1/5)R_1


[51304/58/5315]\begin{bmatrix} -5 & 1 & 3 \\ 0 & -4/5 & 8/5 \\ 3 & 1 & -5 \\ \end{bmatrix}

R3=R3+(3/5)R1R_3=R_3+(3/5)R_1


[51304/58/508/516/5]\begin{bmatrix} -5 & 1 & 3 \\ 0 & -4/5 & 8/5 \\ 0 & 8/5 & -16/5 \\ \end{bmatrix}

R3=R3+(2)R2R_3=R_3+(2)R_2


[51304/58/5000]\begin{bmatrix} -5 & 1 & 3 \\ 0 & -4/5 & 8/5 \\ 0 & 0 & 0 \\ \end{bmatrix}

R2=(5/4)R2R_2=(-5/4)R_2


[513012000]\begin{bmatrix} -5 & 1 & 3 \\ 0 & 1 & -2 \\ 0 & 0 & 0 \\ \end{bmatrix}

R1=R1R2R_1=R_1-R_2


[505012000]\begin{bmatrix} -5 & 0 & 5 \\ 0 & 1 & -2 \\ 0 & 0 & 0 \\ \end{bmatrix}

R1=(1/5)R1R_1=(-1/5)R_1


[101012000]\begin{bmatrix} 1 & 0 & -1 \\ 0 & 1 & -2 \\ 0 & 0 & 0 \\ \end{bmatrix}

Solve the matrix equation


[101012000][v1v2v3]=[000]\begin{bmatrix} 1 & 0 & -1 \\ 0 & 1 & -2 \\ 0 & 0 & 0 \\ \end{bmatrix}\begin{bmatrix} v _1 \\ v_2 \\ v_3 \end{bmatrix}=\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}

If we take v3=t,v_3=t, then v1=t,v2=2t,v3=t.v_1=t,v_2=2t,v_3=t.


v=[t2tt]=[121]t\text{v}=\begin{bmatrix} t \\ 2t \\ t \end{bmatrix}=\begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix}t

λ2=3\lambda_2=3

[1λ1315λ1311λ]=[213121312]\begin{bmatrix} 1-\lambda & 1 & 3 \\ 1 & 5-\lambda & 1 \\ 3 & 1 & 1-\lambda \\ \end{bmatrix}=\begin{bmatrix} -2 & 1 & 3 \\ 1 & 2 & 1 \\ 3 & 1 & -2 \\ \end{bmatrix}

R2=R2+(1/2)R1R_2=R_2+(1/2)R_1

[21305/25/2312]\begin{bmatrix} -2 & 1 & 3 \\ 0 & 5/2 & 5/2 \\ 3 & 1 & -2 \\ \end{bmatrix}

R3=R3+(3/2)R1R_3=R_3+(3/2)R_1

[21305/25/205/25/2]\begin{bmatrix} -2 & 1 & 3 \\ 0 & 5/2 & 5/2 \\ 0 & 5/2 & 5/2 \\ \end{bmatrix}

R3=R3R2R_3=R_3-R_2

[21305/25/2000]\begin{bmatrix} -2 & 1 & 3 \\ 0 & 5/2 & 5/2 \\ 0 & 0 & 0 \\ \end{bmatrix}

R2=(2/5)R2R_2=(2/5)R_2

[213011000]\begin{bmatrix} -2 & 1 & 3 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \\ \end{bmatrix}

R1=R1R2R_1=R_1-R_2

[202011000]\begin{bmatrix} -2 & 0 & 2 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \\ \end{bmatrix}

R1=(1/2)R1R_1=(-1/2)R_1

[101011000]\begin{bmatrix} 1 & 0 & -1 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \\ \end{bmatrix}

Solve the matrix equation


[101011000][v1v2v3]=[000]\begin{bmatrix} 1 & 0 & -1 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \\ \end{bmatrix}\begin{bmatrix} v _1 \\ v_2 \\ v_3 \end{bmatrix}=\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}

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

v=[ttt]=[111]t\text{v}=\begin{bmatrix} t \\ -t \\ t \end{bmatrix}=\begin{bmatrix} 1 \\ -1 \\ 1 \end{bmatrix}t



λ3=2\lambda_3=-2

[1λ1315λ1311λ]=[313171313]\begin{bmatrix} 1-\lambda & 1 & 3 \\ 1 & 5-\lambda & 1 \\ 3 & 1 & 1-\lambda \\ \end{bmatrix}=\begin{bmatrix} 3 & 1 & 3 \\ 1 & 7 & 1 \\ 3 & 1 & 3 \\ \end{bmatrix}

R2=R2(1/3)R1R_2=R_2-(1/3)R_1

[313020/30313]\begin{bmatrix} 3 & 1 & 3 \\ 0 & 20/3 & 0 \\ 3 & 1 & 3 \\ \end{bmatrix}

R3=R3R1R_3=R_3-R_1

[313020/30000]\begin{bmatrix} 3 & 1 & 3 \\ 0 & 20/3 & 0 \\ 0 & 0 & 0 \\ \end{bmatrix}


R2=(3/20)R2R_2=(3/20)R_2

[313010000]\begin{bmatrix} 3 & 1 & 3 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \\ \end{bmatrix}

R1=R1R2R_1=R_1-R_2

[303010000]\begin{bmatrix} 3 & 0 & 3 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \\ \end{bmatrix}

R1=(1/3)R1R_1=(1/3)R_1

[101010000]\begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \\ \end{bmatrix}


Solve the matrix equation


[101010000][v1v2v3]=[000]\begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \\ \end{bmatrix}\begin{bmatrix} v _1 \\ v_2 \\ v_3 \end{bmatrix}=\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}

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

v=[t0t]=[101]t\text{v}=\begin{bmatrix} -t \\ 0 \\ t \end{bmatrix}=\begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix}t


The Normalized matrix


N=[1/61/31/22/61/301/61/31/2]N=\begin{bmatrix} 1/\sqrt{6} & 1/\sqrt{3} & -1/\sqrt{2} \\ 2/\sqrt{6} & -1/\sqrt{3} & 0 \\ 1/\sqrt{6} & 1/\sqrt{3} & 1/\sqrt{2} \\ \end{bmatrix}

NT=[1/62/61/61/31/31/31/201/2]N^T=\begin{bmatrix} 1/\sqrt{6} & 2/\sqrt{6} & 1/\sqrt{6} \\ 1/\sqrt{3} & -1/\sqrt{3} & 1/\sqrt{3} \\ -1/\sqrt{2} & 0 & 1/\sqrt{2} \\ \end{bmatrix}

[1/62/61/61/31/31/31/201/2]A[1/61/31/22/61/301/61/31/2]=\begin{bmatrix} 1/\sqrt{6} & 2/\sqrt{6} & 1/\sqrt{6} \\ 1/\sqrt{3} & -1/\sqrt{3} & 1/\sqrt{3} \\ -1/\sqrt{2} & 0 & 1/\sqrt{2} \\ \end{bmatrix}A\begin{bmatrix} 1/\sqrt{6} & 1/\sqrt{3} & -1/\sqrt{2} \\ 2/\sqrt{6} & -1/\sqrt{3} & 0 \\ 1/\sqrt{6} & 1/\sqrt{3} & 1/\sqrt{2} \\ \end{bmatrix}=

=[600030002]=\begin{bmatrix} 6 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & -2 \\ \end{bmatrix}

[y1y2y3][600030002][y1y2y3]=\begin{bmatrix} y_1 & y_2 & y_3 \end{bmatrix}\begin{bmatrix} 6 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & -2 \\ \end{bmatrix}\begin{bmatrix} y_1 \\ y_2 \\ y_3 \end{bmatrix}=

=6y12+3y222y32=6y_1^2+3y_2^2-2y_3^2

 which is the required canonical form.


Rank=3Rank=3 (the number of non-zero eigenvalues).

Index=2Index=2 (the number of positive eigenvalues).

Signature=21=1Signature=2-1=1 (the difference between the number of positive eigenvalues and the number of negative eigenvalues ).

 

Nature: the eigenvalues are 6>0,3>0,2<0.6>0,3>0,-2<0. Therefore the given quadratic form is indefinite in nature.



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