Question #162957

To determine the eigenvalues and eigenvectors of the following matrix :

2 -2 0

P = -2 1 -2

0 -2 0


1
Expert's answer
2021-02-24T07:29:38-0500

P=(2−20−21−20−20)P=\left(\begin{array}{rrr}2&-2&0\\-2&1&-2\\0&-2&0\end{array}\right)


Eigenvalues λ1\lambda_1, λ2\lambda_2, λ3\lambda_3 are the roots of the characteristic polynomial of PP .

Obtain the characteristic polynomial taking the determinant


∣2−λ−20−21−λ−20−2−λ∣=\left|\begin{array}{rrr}2-\lambda&-2&0\\-2&1-\lambda&-2\\0&-2&-\lambda\end{array}\right|=


(2−λ)(1−λ)(−λ)+(−2)⋅(−2)⋅0+(2-\lambda)(1-\lambda)(-\lambda)+(-2)\cdot(-2)\cdot0+ 0⋅(−2)⋅(−2)−0\cdot(-2)\cdot(-2)-

0⋅(1−λ)⋅0−(−2)⋅(−2)⋅(−λ)−0\cdot(1-\lambda)\cdot0-(-2)\cdot(-2)\cdot(-\lambda)- (2−λ)⋅(−2)⋅(−2)=(2-\lambda)\cdot(-2)\cdot(-2)=


(2−λ−2λ+λ2)(−λ)+0+0−0+4λ−(2−λ)⋅4=(2-\lambda-2\lambda+\lambda^2)(-\lambda)+0+0-0+4\lambda-(2-\lambda)\cdot4=


(2−3λ+λ2)(−λ)+4λ−8+4λ=(2-3\lambda+\lambda^2)(-\lambda)+4\lambda-8+4\lambda=


−2λ+3λ2−λ3+8λ−8=-2\lambda+3\lambda^2-\lambda^3+8\lambda-8=


−λ3+3λ2+6λ−8-\lambda^3+3\lambda^2+6\lambda-8


Find the roots factoring the polynomial


−(λ3−3λ2−6λ+8)=-(\lambda^3-3\lambda^2-6\lambda+8)=


−(λ3+8−3λ2−6λ)=-(\lambda^3+8-3\lambda^2-6\lambda)=


−((λ3+8)−(3λ2+6λ))=-((\lambda^3+8)-(3\lambda^2+6\lambda))=


−((λ3+23)−3λ(λ+2))=-((\lambda^3+2^3)-3\lambda(\lambda+2))=


−((λ+2)(λ2−2λ+22)−3λ(λ+2))=-((\lambda+2)(\lambda^2-2\lambda+2^2)-3\lambda(\lambda+2))=


−((λ+2)(λ2−2λ+4)−3λ(λ+2))=-((\lambda+2)(\lambda^2-2\lambda+4)-3\lambda(\lambda+2))=


−(λ+2)(λ2−2λ+4−3λ)=-(\lambda+2)(\lambda^2-2\lambda+4-3\lambda)=


−(λ+2)(λ2−5λ+4)=-(\lambda+2)(\lambda^2-5\lambda+4)=


−(λ+2)(λ2−4λ−λ+4)=-(\lambda+2)(\lambda^2-4\lambda-\lambda+4)=


−(λ+2)((λ2−4λ)−(λ−4))=-(\lambda+2)((\lambda^2-4\lambda)-(\lambda-4))=


−(λ+2)(λ(λ−4)−(λ−4))=-(\lambda+2)(\lambda(\lambda-4)-(\lambda-4))=


−(λ+2)(λ−1)(λ−4)-(\lambda+2)(\lambda-1)(\lambda-4)


So the roots of polynomial are


λ1=−2\lambda_1=-2,

λ2=1\lambda_2=1,

λ3=4\lambda_3=4.


Find eigenvectors solving the homogeneous system substituting λ\lambda by −2-2, 11, 44


{(2−λ)x1−2x2=0−2x1+(1−λ)x2−2x3=0−2x2−λx3=0\left\{\begin{array}{rrrr}(2-\lambda)x_1&-2x_2&&=0\\-2x_1&+(1-\lambda)x_2&-2x_3&=0\\&-2x_2&-\lambda x_3&=0\end{array}\right.


λ=λ1=−2\lambda=\lambda_1=-2


{4x1−2x2=0−2x1+3x2−2x3=0−2x2+2x3=0\left\{\begin{array}{rrrr}4x_1&-2x_2&&=0\\-2x_1&+3x_2&-2x_3&=0\\&-2x_2&+2x_3&=0\end{array}\right.

Use Gaussian elimination modifying the augmented matrix

(4−20−23−20−22∣000)∼\left(\begin{array}{rrr}4&-2&0\\-2&3&-2\\0&-2&2\end{array}\right|\left.\begin{array}{r}0\\0\\0\end{array}\right)\sim L2+12L1→L2∼L_2+\frac12L_1\to L_2\sim


(4−20−2+12⋅43+12⋅(−2)−2+12⋅00−22∣000)=\left(\begin{array}{lll}4&-2&0\\-2+\frac12\cdot4&3+\frac12\cdot(-2)&-2+\frac12\cdot0\\0&-2&2\end{array}\right|\left.\begin{array}{r}0\\0\\0\end{array}\right)=


(4−2002−20−22∣000)∼\left(\begin{array}{rrr}4&-2&0\\0&2&-2\\0&-2&2\end{array}\right|\left.\begin{array}{r}0\\0\\0\end{array}\right)\sim L3+L2→L3∼L_3+L_2\to L_3\sim


(4−2002−20+0−2+22−2∣000)=\left(\begin{array}{rrr}4&-2&0\\0&2&-2\\0+0&-2+2&2-2\end{array}\right|\left.\begin{array}{r}0\\0\\0\end{array}\right)=


(4−2002−2000∣000)\left(\begin{array}{rrr}4&-2&0\\0&2&-2\\0&0&0\end{array}\right|\left.\begin{array}{r}0\\0\\0\end{array}\right).

System represented as the matryx is

{4x1−2x2=0+2x2−2x3=0\left\{\begin{array}{rrrr}4x_1&-2x_2&&=0\\&+2x_2&-2x_3&=0\end{array}\right.

Let x3x_3 be free while x1x_1 and x2x_2 be dependent on x3x_3 . Then

{4x1=2x22x2=2x3⇒\left\{\begin{array}{rrrr}4x_1&&&=2x_2\\&2x_2&&=2x_3\end{array}\right.\Rightarrow {4x1=2x3x2=x3⇒\left\{\begin{array}{rrrr}4x_1&&&=2x_3\\&x_2&&=x_3\end{array}\right.\Rightarrow {x1=12x3x2=x3\left\{\begin{array}{rrrr}x_1&&&=\frac12x_3\\&x_2&&=x_3\end{array}\right.

Thus, the eigenvector corresponding to the eigenvalue λ1=−2\lambda_1=-2 is

v1=(x1x2x3)=\bold{v}_1=\left(\begin{array}{r}x_1\\x_2\\x_3\end{array}\right)= (12x3x3x3)\left(\begin{array}{r}\frac12x_3\\x_3\\x_3\end{array}\right).

Here x3x_3 is allowed to take any value. In particular, if x3=1x_3=1

v1=(1211)\bold{v}_1=\left(\begin{array}{r}\frac12\\1\\1\end{array}\right).


λ=λ2=1\lambda=\lambda_2=1


{x1−2x2=0−2x1−2x3=0−2x2−x3=0\left\{\begin{array}{rrrr}x_1&-2x_2&&=0\\-2x_1&&-2x_3&=0\\&-2x_2&-x_3&=0\end{array}\right.

Use Gaussian elimination modifying the augmented matrix

(1−20−20−20−2−1∣000)∼\left(\begin{array}{rrr}1&-2&0\\-2&0&-2\\0&-2&-1\end{array}\right|\left.\begin{array}{r}0\\0\\0\end{array}\right)\sim L2+2L1→L2∼L_2+2L_1\to L_2\sim


(1−20−2+2⋅10+2⋅(−2)−2+2⋅00−2−1∣000)=\left(\begin{array}{lll}1&-2&0\\-2+2\cdot1&0+2\cdot(-2)&-2+ 2\cdot0\\0&-2&-1\end{array}\right|\left.\begin{array}{r}0\\0\\0\end{array}\right)=


(1−200−4−20−2−1∣000)∼\left(\begin{array}{rrr}1&-2&0\\0&-4&-2\\0&-2&-1\end{array}\right|\left.\begin{array}{r}0\\0\\0\end{array}\right)\sim L3−12L2→L3∼L_3-\frac12L_2\to L_3\sim


(1−200−4−20−12⋅0−2−12⋅(−4)−1−12⋅(−2)∣000)=\left(\begin{array}{lll}1&-2&0\\0&-4&-2\\0-\frac12\cdot0&-2-\frac12\cdot(-4)&-1-\frac12\cdot(-2)\end{array}\right|\left.\begin{array}{r}0\\0\\0\end{array}\right)=


(1−200−4−2000∣000)\left(\begin{array}{rrr}1&-2&0\\0&-4&-2\\0&0&0\end{array}\right|\left.\begin{array}{r}0\\0\\0\end{array}\right)

System represented as the matryx is

{x1−2x2=0−4x2−2x3=0\left\{\begin{array}{rrrr}x_1&-2x_2&&=0\\&-4x_2&-2x_3&=0\end{array}\right.

Let x3x_3 be free while x1x_1 and x2x_2 be dependent on x3x_3 . Then

{x1=2x2−4x2=2x3⇒\left\{\begin{array}{rrrr}x_1&&&=2x_2\\&-4x_2&&=2x_3\end{array}\right.\Rightarrow {x1=2(−12x3)x2=−12x3⇒\left\{\begin{array}{rrrl}x_1&&&=2\left(-\frac12x_3\right)\\&x_2&&=-\frac12x_3\end{array}\right.\Rightarrow {x1=−x3x2=−12x3\left\{\begin{array}{rrrl}x_1&&&=-x_3\\&x_2&&=-\frac12x_3\end{array}\right.

Thus, the eigenvector corresponding to the eigenvalue λ2=1\lambda_2=1 is

v2=(x1x2x3)=\bold{v}_2=\left(\begin{array}{r}x_1\\x_2\\x_3\end{array}\right)= (−x3−12x3x3)\left(\begin{array}{r}-x_3\\-\frac12x_3\\x_3\end{array}\right).

Here x3x_3 is allowed to take any value. In particular, if x3=1x_3=1

v2=(−1−121)\bold{v}_2=\left(\begin{array}{r}-1\\-\frac12\\1\end{array}\right).


λ=λ3=4\lambda=\lambda_3=4


{−2x1−2x2=0−2x1−3x2−2x3=0−2x2−4x3=0\left\{\begin{array}{rrrr}-2x_1&-2x_2&&=0\\-2x_1&-3x_2&-2x_3&=0\\&-2x_2&-4x_3&=0\end{array}\right.

Use Gaussian elimination modifying the augmented matrix

(−2−20−2−3−20−2−4∣000)∼\left(\begin{array}{rrr}-2&-2&0\\-2&-3&-2\\0&-2&-4\end{array}\right|\left.\begin{array}{r}0\\0\\0\end{array}\right)\sim L2−L1→L2∼L_2-L_1\to L_2\sim


(−2−20−2−(−2)−3−(−2)−2−00−2−4∣000)=\left(\begin{array}{lll}-2&-2&0\\-2-(-2)&-3-(-2)&-2-0\\0&-2&-4\end{array}\right|\left.\begin{array}{r}0\\0\\0\end{array}\right)=


(−2−200−1−20−2−4∣000)∼\left(\begin{array}{rrr}-2&-2&0\\0&-1&-2\\0&-2&-4\end{array}\right|\left.\begin{array}{r}0\\0\\0\end{array}\right)\sim L3−2L2→L3∼L_3-2L_2\to L_3\sim


(−2−200−1−20−2⋅0−2−2⋅(−1)−4−2⋅(−2)∣000)=\left(\begin{array}{lll}-2&-2&0\\0&-1&-2\\0-2\cdot0&-2-2\cdot(-1)&-4-2\cdot(-2)\end{array}\right|\left.\begin{array}{r}0\\0\\0\end{array}\right)=


(−2−200−1−2000∣000)\left(\begin{array}{rrr}-2&-2&0\\0&-1&-2\\0&0&0\end{array}\right|\left.\begin{array}{r}0\\0\\0\end{array}\right) .

System represented as the matryx is

{−2x1−2x2=0−x2−2x3=0\left\{\begin{array}{rrrr}-2x_1&-2x_2&&=0\\&-x_2&-2x_3&=0\end{array}\right.

Let x3x_3 be free while x1x_1 and x2x_2 be dependent on x3x_3 . Then

{−2x1=2x2−x2=2x3⇒\left\{\begin{array}{rrrr}-2x_1&&&=2x_2\\&-x_2&&=2x_3\end{array}\right.\Rightarrow {x1=−x2x2=−2x3⇒\left\{\begin{array}{rrrl}x_1&&&=-x_2\\&x_2&&=-2x_3\end{array}\right.\Rightarrow {x1=2x3x2=−2x3\left\{\begin{array}{rrrl}x_1&&&=2x_3\\&x_2&&=-2x_3\end{array}\right.

Thus, the eigenvector corresponding to the eigenvalue λ2=1\lambda_2=1 is

v3=(x1x2x3)=\bold{v}_3=\left(\begin{array}{r}x_1\\x_2\\x_3\end{array}\right)= (2x3−2x3x3)\left(\begin{array}{r}2x_3\\-2x_3\\x_3\end{array}\right).

Here x3x_3 is allowed to take any value. In particular, if x3=1x_3=1

v3=(2−21)\bold{v}_3=\left(\begin{array}{r}2\\-2\\1\end{array}\right).



Answer:

Eigenvalues

λ1=−2\lambda_1=-2, λ2=1\lambda_2=1, λ3=4\lambda_3=4.

Eigenvectors

v1=(1211)t1\bold{v}_1=\left(\begin{array}{r}\frac12\\1\\1\end{array}\right)t_1, v2=(−1−121)t2\bold{v}_2=\left(\begin{array}{r}-1\\-\frac12\\1\end{array}\right)t_2, v3=(2−21)t3\bold{v}_3=\left(\begin{array}{r}2\\-2\\1\end{array}\right)t_3,

where factors t1t_1 , t2t_2 , t3t_3 may take any value.


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