Eigenvalues and Eigen-vectors for Botched Homework | Characteristic Equation

[SadSack]

Homework Statement

(a) Find the characteristic equation and
(b) the eigenvalues and corresponding eigen-vectors

[-5 0 0
3 7 0
4 -2 3]

Homework Equations

det($$\lambda$$*I - A)

The Attempt at a Solution

Finding the characteristic equation wasn't that challenging, I took lambda, $$\lambda$$ , and subtracted the given matrix values from it for every diagonal entry. The determiant of a 3x3 matrix is just the product of these diagonals...I think so my characteristic equation is ($$\lambda$$ + 5)($$\lambda$$-7)($$\lambda$$ - 3)
Next the eigenvalues are the values for $$\lambda$$ that when plugged in would make the characteristic equation equal zero...so the eigenvalues are -5, 7, 3
This is where I'm confused.
To find the eigen-vectors I'm supposed to multiply each eigenvalue by the identity matrix and subtract the original given matrix: so for the eigenvalue of -5...

[-5 0 0
0 -5 0
0 0 -5]
MINUS
[-5 0 0
3 7 0
4 -2 3]
which equals
[0 0 0
-3 -12 0
-4 2 -8]
I took the reduced row echelon form of this matrix and got...
[1 0 1.777
0 1 -.4444
0 0 0]
Is my eigen vector in the matrix above? I am so confused...I also tried to solve the system generated by multiplying a 3x1 vector of variable components x, y, z times
[0 0 0
-3 -12 0
-4 2 -8]
I appear to have failed miserably at this...because the vector I got did not work. Can someone help walk me through this?

 

[HallsofIvy]

Homework Statement

(a) Find the characteristic equation and
(b) the eigenvalues and corresponding eigen-vectors

[-5 0 0
3 7 0
4 -2 3]

Homework Equations

det($$\lambda$$*I - A)

The Attempt at a Solution

Finding the characteristic equation wasn't that challenging, I took lambda, $$\lambda$$ , and subtracted the given matrix values from it for every diagonal entry. The determiant of a 3x3 matrix is just the product of these diagonals...I think so my characteristic equation is ($$\lambda$$ + 5)($$\lambda$$-7)($$\lambda$$ - 3)

Next the eigenvalues are the values for $$\lambda$$ that when plugged in would make the characteristic equation equal zero...so the eigenvalues are -5, 7, 3

Yes, in an "upper triangular" or "lower triangular" matrix like this, the eigenvalues are just the numbers on the main diagonal.

This is where I'm confused.
To find the eigen-vectors I'm supposed to multiply each eigenvalue by the identity matrix and subtract the original given matrix: so for the eigenvalue of -5...

[-5 0 0
0 -5 0
0 0 -5]
MINUS
[-5 0 0
3 7 0
4 -2 3]
which equals
[0 0 0
-3 -12 0
-4 2 -8]
I took the reduced row echelon form of this matrix and got...
[1 0 1.777
0 1 -.4444
0 0 0]
Which is equivalent to the equations x+ 16/9 z= 0 and y- 4/9 z= 0 so x= -16/9 z and y= 4/9 z. Any eigenvector corresponding to eigenvalue -5 is of the form <-16/9 z, 4/9 z, z>. Since you only need one, you can take z= 9 to get rid of the fractions: <-16, 4, 9>.

Is my eigen vector in the matrix above? I am so confused...I also tried to solve the system generated by multiplying a 3x1 vector of variable components x, y, z times
[0 0 0
-3 -12 0
-4 2 -8]

So your equations would be 0= 0, -3x- 12y= 0, and -4x+ 2y- 8z= 0. From -3x- 12y= 0, x= -4y. Putting that into -4x+ 2y- 8z= 0 gives 16y+ 2y- 8z= 0 so 18y= 8z and y= (4/9)z and x= -4y= (-16/9)z. That gives the same answer as before, <-16, 4, 9>.

[quoteI appear to have failed miserably at this...because the vector I got did not work. Can someone help walk me through this?

 

[Architeuthis Dux]

Great job on finding the characteristic equation and the eigenvalues! Your characteristic equation is correct and the eigenvalues are indeed -5, 7, and 3.

To find the corresponding eigenvectors, you are on the right track. However, the steps you took to find the eigenvector for an eigenvalue of -5 are incorrect. Let me guide you through the correct steps:

1. Start by setting up the equation (\lambda * I - A) * x = 0, where A is the given matrix and x is the eigenvector we are trying to find.

2. Plug in the eigenvalue -5 for \lambda and the given matrix A to get (-5 * I - A) * x = 0. This gives us the following equation:

[0 0 0
-3 -12 0
-4 2 -8] * x = 0

3. Solve this system of equations to find the components of x. You can do this by using row operations or by setting up a system of equations and solving it using substitution or elimination.

4. Once you have found the components of x, you can write the eigenvector as [x1 x2 x3] (where x1, x2, x3 are the components you found in step 3).

5. Repeat this process for the other eigenvalues (7 and 3) to find their corresponding eigenvectors.

I hope this helps clarify the steps for finding eigenvectors. If you are still having trouble, don't hesitate to ask for further clarification or help from your teacher or classmates. Remember, practice makes perfect! Keep working at it and you will get the hang of it. Good luck!