Finding Eigenvalues for 3x3 Matrix

[geft]

$$\left( \begin{array}{ccc} 3 - \lambda & 1 & -1 \\ -4 & 2 - \lambda & 2 \\ -2 & 2 & 2 - \lambda \end{array} \right)$$

$$(3 - \lambda) \left| \begin{array}{cc} 2 - \lambda & 2 \\ 2 & 2 - \lambda \end{array} \right| + 4 \left| \begin{array}{cc} 1 & -1 \\ 2 & 2 - \lambda \end{array} \right| - 2 \left| \begin{array}{cc} 1 & -1 \\ 2 - \lambda & 2 \end{array} \right|$$

$$det(A - \lambda I) = (3 - \lambda)[(2 - \lambda)(2 - \lambda) - 4] + 4[(2 - \lambda) + 2] - 2[4 - \lambda] = 0$$

$$-3x^3 + 7x^2 -14x + 8 = 0$$

?

 

[Pengwuino]

You can factor out a $$\lambda - 4$$ to determine one eigenvalue.

 

[geft]

Can you elaborate? I don't know how to work out cubic polynomials.

 

[lanedance]

pengwuino means that 4 is a solution to the charcteristic equation, so you can re-write it as

$$-3x^3 + 7x^2 -14x + 8 = (\lambda -4)(ax^2 +bx + c) = 0$$

multiply out & equate co-efficients

 

[geft]

pengwuino means that 4 is a solution to the charcteristic equation, so you can re-write it as

$$-3x^3 + 7x^2 -14x + 8 = (\lambda -4)(ax^2 +bx + c) = 0$$

multiply out & equate co-efficients

Yes, but I want to know how he got that 4.

 

[lanedance]

test that putting 4 into your equation gives 0

 

[geft]

Never mind. Somehow I solved it by not opening too many brackets.

$$(3-x)((2-x)(2-x)-4)+4((2-x)+2)-2(4-x)$$
$$(3-x)(4-4x+x^2-4)+4(4-x)-2(4-x)$$
$$(3-x)(-4x+x^2)+(4-2)(4-x)$$
$$(3-x)(-x)(4-x)+2(4-x)$$
$$(3-x)(2-x)(4-x)$$

 

[lanedance]

Yes, but I want to know how he got that 4.

in general I don't think its that easy, though you could probably attempt polynomial long division, one trick is say you have the following polynomial and p is a factor

$$ax^3 + bx^2 + cx + d = (x+p)(qx^2 + rx + s)$$

multplying out gives
$$ax^3 + bx^2 + cx + d = (x+p)(qx^2 + rx + s) = qx^3 + (pq +r)x^2 (pr+s)x + ps$$

note by equating coefficients
$$p = \frac{d}{s}$$

so if there are integer factors and everything stays nice, p must divide d - giving options here of 8,4,2,1,-1,-2,-4,-8. You could test these reasonably quickly to see if any work

in your simplification step though you could have done the following, though not exactly obvious
$$det(A - \lambda I) = (3 - \lambda)[(2 - \lambda)(2 - \lambda) - 4] + 4[(2 - \lambda) + 2] - 2[4 - \lambda] = 0$$

$$= (3 - \lambda)[(\lambda^2-4 \lambda ] - 4[(\lambda-4)] + 2[\lambda-4] = 0$$

$$= (3 - \lambda)[(\lambda-4 )\lambda ] - 4[(\lambda-4)] + 2[\lambda-4] = 0$$

 

[lanedance]

Never mind. Somehow I solved it by not opening too many brackets.

$$(3-x)((2-x)(2-x)-4)+4((2-x)+2)-2(4-x)$$
$$(3-x)(4-4x+x^2-4)+4(4-x)-2(4-x)$$
$$(3-x)(-4x+x^2)+(4-2)(4-x)$$
$$(3-x)(-x)(4-x)+2(4-x)$$

the next step you take doesn't look right

$$(3-x)(2-x)(4-x)$$

it should be
$$(3-x)(-x)(4-x)+2(4-x)$$
$$=(x^2-3x)(4-x)+2(4-x)$$
$$=(x^2-3x+2)(4-x)$$

note if your initial multiplication was correct, then all the factors should solve the equation...

though I'm not so convinced about you original polynomial now - i don't see how you get a -3x^3 term

 

[lanedance]

assuming your determinant is correct, then expanding this as a check
$$det(A - \lambda I) = (3 - \lambda)[(2 - \lambda)(2 - \lambda) - 4] + 4[(2 - \lambda) + 2] - 2[4 - \lambda] = 0$$

$$= (3 - \lambda)(\lambda^2 -4 \lambda) + 4(4 - \lambda) - 2(4 - \lambda)$$

$$=(-\lambda^3+7\lambda^2 -12 \lambda) + 2(4 - \lambda)$$

$$=-\lambda^3+7 \lambda^2 -14 \lambda +8 = 0$$

checking 4 is a solution
$$=-(4^3)+7.(4^2) -10.(4) +8 = 4.(-4^2 +7.4-10+2) = 4(-16+28-8) = 4.0 = 0$$

 

[geft]

Thanks! I'm horridly careless at these meticulous calculations.

 

[Mark44]

Yes, but I want to know how he got that 4.

He probably used the rational root theorem. If p/q is a root of the polynomial a_nxⁿ + a_n-1x^n-1 + ... + a₁x + a₀ = 0, then p must be a divisor of a₀ and q must be a divisor of a_n.

For the polynomial equation in question, $-3x^3 + 7x^2 -14x + 8 = 0$, any rational root p/q must be such that p divides 8, and q divides -3. Possible choices for p are 1, -1, 2, -2, 4, -4, 8, and -8. Possible choices for q are 1, -1, 3, -3. Possible values for p/q are any of the 32 possibilities.