How Do You Calculate Eigenvalues for Specific Matrices?

[Davio]

1. 1 3 0
3 -2 -1
0 -1 1
What are the eingenvalues? (they are 1, 3 and -4)




2. 1-n 3 0
3 -2-n -1
0 -1 1-n




I basically get 4n-n^3 -3 = 0
I'm not sure what to do from there?

Working: (1-n). (n+2).(n-1) -1) - 3 (3 (1-n))


Thanks guys!

 

[HallsofIvy]

1. 1 3 0
3 -2 -1
0 -1 1
What are the eingenvalues? (they are 1, 3 and -4)




2. 1-n 3 0
3 -2-n -1
0 -1 1-n




I basically get 4n-n^3 -3 = 0

That is not at all the equation I get. Please show in detail how you arrived at that.

I'm not sure what to do from there?

Working: (1-n). (n+2).(n-1) -1) - 3 (3 (1-n))

Thanks guys!

 

[tiny-tim]

Hi Davio!

I basically get 4n-n^3 -3 = 0
I'm not sure what to do from there?

Working: (1-n). (n+2).(n-1) -1) - 3 (3 (1-n))

For a start, you can see that (1 - n) divides everything, so leave that on the outside …

(1 - n){(n+2)(n-1) - ? - ?} …

carry on from there!

 

[HallsofIvy]

Hi Davio!


For a start, you can see that (1 - n) divides everything, so leave that on the outside …

(1 - n){(n+2)(n-1) - ? - ?} …

carry on from there!

Yes, n= 1 satisfies that equation but clearly n= 3 and n= -4 do not. His real problem is he has the wrong equation.

 

[Davio]

Hi guys.
1-n 3 0
3 -2-n -1
0 -1 1-n

My working for determinant :
+ 1-n x (-2-n)(1-n) - (-1).(-1)
+ 1-n x (-2-n)(1-n) -1
+ 1-n x (n^2 +2n -n -2) -1
+ 1-n x (n^2 +2n -n -3)
+ 1-n x (n^2 +n -3)
(n^2 +n -3) - (n^3 -n^2 +3n)
(n^2 +n -3) - n^3 -n^2 +3n
- (n^3 +4n) -3
4n - n^3 -3
The original question is to prove the eigen values are 1, 3 and ? (negative 4 from online calculator)
Thanks in advance :S
I'm not sure how polynonial division would help me :S






-3 x (3 x 1-n) - 0
-3 x (3 - 3n)
-9 + 9n

 

[HallsofIvy]

Hi guys.
1-n 3 0
3 -2-n -1
0 -1 1-n

My working for determinant :
+ 1-n x (-2-n)(1-n) - (-1).(-1)

Please use parentheses! You mean (1- n)[(-2-n)((1-n)- (-1)(-1)]

+ 1-n x (-2-n)(1-n) -1
+ 1-n x (n^2 +2n -n -2) -1
+ 1-n x (n^2 +2n -n -3)
+ 1-n x (n^2 +n -3)
(n^2 +n -3) - (n^3 -n^2 +3n)

No, if you are keeping "-" outside that last parenthesis, it is -(n^3+ n^2- 3n)

(n^2 +n -3) - n^3 -n^2 +3n
- (n^3 +4n) -3
4n - n^3 -3

Even if you had not messed up the sign, this is just the top left corner, 1-n, times its cofactor
$$\left|\begin{array}{cc} -2-n & -1 \\ -1 & 1- n\end{array}\right|$$
If you are expanding on the first row, you should also have
$$-3\left|\begin{array}{cc}3 & -1 \\ 0 & 1-n\end{array}\right]$$
If instead you are expanding on the first column, you should have
$$-3\left|\begin{array}{cc}3 & 0 \\ -1 & 1-n\end{array}\right]$$
which is obviously the same thing.

The original question is to prove the eigen values are 1, 3 and ? (negative 4 from online calculator)
Thanks in advance :S
I'm not sure how polynonial division would help me :S

In general if you no "n= a" is a root of a cubic polynomial then you know the polynomial is equal to (x- a) times quadratic polynomial. Dividing the cubic polynomial by x-a gives the remaining quadratic polynomial which you can solve, for example, using the quadratic formula.

-3 x (3 x 1-n) - 0
-3 x (3 - 3n)
-9 + 9n

Oh, here's your missing term! You should have (n^2 +n -3)-(n^3+ n^2- 3n)- 9+ 9n=
-n^3+ 13n- 12= 0.

Yes, 1 satisfies that: -1^3+ 13(1)- 12= -1+ 13- 12= 0. Now divide -n^3+ 13n- 12 by n- 1 to get a quadratic.