Linear Algebra Matrix Eigenvalues

[lina29]

Homework Statement

The matrix A has 3 distinct eigenvalues t₁< t₂< t₃. Let v_i be the unique eigenvector associated to t_i with a 1 as its first nonzero component. Let

D= [t₁ 0 0
0 t₂ 0
0 0 t₃]

and P= [v₁|v₂|v₃] so that the ith column of P is the eigenvector v_i associated to t_i

A=
7 2 -8
0 1 0
4 2 -5

a) Find D
b) Find P
c) Find P^-1

Homework Equations

The Attempt at a Solution

My thought to find D was to find the characteristic equation of A which I found to be (t-3)(t+1)=0 so the eigenvalues would be t=-3, t=1 I then plugged in these values into the matrix D so D became
3 0 0
0 -1 0
0 0 0

but it was counting it wrong. What did I mess up on?

 

[I like Serena]

Hi lina29

I'm afraid your characteristic equation is a little off.
How did you get it?

And your eigenvalues are not the solution of the equation you have.
They are off by a minus sign.

You're supposed to get 3 distinct eigenvalues.
(They should be 3, -1, 1.)

Oh, and according to your problem statement your eigenvalues should be sorted low to high on your diagonal.

 

[lina29]

To find the determinant I did aei+bfg+cdh-ceg-bdi-afh. So I got (t-7)(t-1)(t+5)+0+0-(8)(t-1)(-4)-0-0. I simplified that to (t-7)(t-1)(t+5)-(-32)(t-1). My mistake was that I accidently crossed out the t-1 on both sides. From there how would I get to P?

 

[I like Serena]

To find the determinant I did aei+bfg+cdh-ceg-bdi-afh. So I got (t-7)(t-1)(t+5)+0+0-(8)(t-1)(-4)-0-0. I simplified that to (t-7)(t-1)(t+5)-(-32)(t-1). My mistake was that I accidently crossed out the t-1 on both sides. From there how would I get to P?

Good! Find the eigenvectors v1, v2, v3.
P is the matrix with these eigenvectors as columns.

 

[lina29]

So the eigenvectors would be -1, 1, and 3. Then the matrix P would be
-1
1
3
is that it?

 

[lina29]

or I guess that wouldn't be right since I need to find P inverse so P would have to be a square matrix

 

[I like Serena]

So the eigenvectors would be -1, 1, and 3. Then the matrix P would be
-1
1
3
is that it?

No. There's a difference between eigenvalues and eigenvectors.
t₁=-1 is your first eigenvalue.
v₁ is the vector belonging to this eigenvalue.

To find v₁ you need to solve A v₁ = t₁ v₁.

Note that any scalar multiple of v₁ also satisfies the equation, but its direction is unique.Edit: I just saw that your problem statement says that v₁ should have "1 as its first nonzero component".

 

[lina29]

ohh I got it. Thank you so much!

 

[I like Serena]

Good!