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Math  /  Algebra

Question Let A=[0040−40−20−6]\text { Let } A=\left[\begin{array}{ccc} 0 & 0 & 4 \\ 0 & -4 & 0 \\ -2 & 0 & -6 \end{array}\right] (a) Determine the eigenvalues λ1\lambda_{1} and λ2\lambda_{2} of AA where λ1<λ2\lambda_{1}<\lambda_{2}. λ1=\lambda_{1}= □\square λ2=\lambda_{2}= □\square (b) Determine the algebraic multiplicity mm of each of the eigenvalues in part (a). m(λ1)=m\left(\lambda_{1}\right)= □\square m(λ2)=m\left(\lambda_{2}\right)= □\square

Studdy Solution

STEP 1

1. The matrix A A is a 3×3 3 \times 3 matrix.
2. Eigenvalues are found by solving the characteristic polynomial det⁡(A−λI)=0 \det(A - \lambda I) = 0 .
3. The algebraic multiplicity of an eigenvalue is the number of times it appears as a root of the characteristic polynomial.

STEP 2

1. Write the characteristic equation.
2. Calculate the determinant of A−λI A - \lambda I .
3. Solve the characteristic polynomial for eigenvalues.
4. Determine the algebraic multiplicity of each eigenvalue.

STEP 3

Write the characteristic equation for matrix A A :
The characteristic equation is given by:
det⁡(A−λI)=0 \det(A - \lambda I) = 0
where I I is the identity matrix of the same size as A A .

STEP 4

Calculate the determinant of A−λI A - \lambda I :
First, construct A−λI A - \lambda I :
A−λI=[0−λ040−4−λ0−20−6−λ]=[−λ040−4−λ0−20−6−λ] A - \lambda I = \begin{bmatrix} 0 - \lambda & 0 & 4 \\ 0 & -4 - \lambda & 0 \\ -2 & 0 & -6 - \lambda \end{bmatrix} = \begin{bmatrix} -\lambda & 0 & 4 \\ 0 & -4 - \lambda & 0 \\ -2 & 0 & -6 - \lambda \end{bmatrix}
Now, compute the determinant:
det⁡(A−λI)=∣−λ040−4−λ0−20−6−λ∣ \det(A - \lambda I) = \begin{vmatrix} -\lambda & 0 & 4 \\ 0 & -4 - \lambda & 0 \\ -2 & 0 & -6 - \lambda \end{vmatrix}
Using the cofactor expansion along the second row:
=(−4−λ)∣−λ4−2−6−λ∣ = (-4 - \lambda) \begin{vmatrix} -\lambda & 4 \\ -2 & -6 - \lambda \end{vmatrix}
Calculate the 2×2 2 \times 2 determinant:
=(−4−λ)((−λ)(−6−λ)−(4)(−2)) = (-4 - \lambda) ((-\lambda)(-6 - \lambda) - (4)(-2))
=(−4−λ)(λ2+6λ+8) = (-4 - \lambda) (\lambda^2 + 6\lambda + 8)

STEP 5

Simplify and solve the characteristic polynomial:
det⁡(A−λI)=(−4−λ)(λ2+6λ+8)=0 \det(A - \lambda I) = (-4 - \lambda) (\lambda^2 + 6\lambda + 8) = 0
This gives us two factors to solve:
1. −4−λ=0 -4 - \lambda = 0 which gives λ=−4 \lambda = -4 .
2. λ2+6λ+8=0 \lambda^2 + 6\lambda + 8 = 0 .
Solve the quadratic equation using the quadratic formula:
λ=−b±b2−4ac2a \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
where a=1,b=6,c=8 a = 1, b = 6, c = 8 .
λ=−6±62−4⋅1⋅82⋅1 \lambda = \frac{-6 \pm \sqrt{6^2 - 4 \cdot 1 \cdot 8}}{2 \cdot 1}
λ=−6±36−322 \lambda = \frac{-6 \pm \sqrt{36 - 32}}{2}
λ=−6±42 \lambda = \frac{-6 \pm \sqrt{4}}{2}
λ=−6±22 \lambda = \frac{-6 \pm 2}{2}
λ=−2orλ=−4 \lambda = -2 \quad \text{or} \quad \lambda = -4

STEP 6

Determine the algebraic multiplicity of each eigenvalue:
The eigenvalues are λ1=−4 \lambda_1 = -4 and λ2=−2 \lambda_2 = -2 .
From the characteristic polynomial:
1. λ1=−4 \lambda_1 = -4 appears twice (once from the linear factor and once from the quadratic factor).
2. λ2=−2 \lambda_2 = -2 appears once.
Thus, the algebraic multiplicities are:
m(λ1)=2 m(\lambda_1) = 2 m(λ2)=1 m(\lambda_2) = 1
The eigenvalues and their multiplicities are: λ1=−4,m(λ1)=2\lambda_1 = -4, \quad m(\lambda_1) = 2 λ2=−2,m(λ2)=1\lambda_2 = -2, \quad m(\lambda_2) = 1

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