I Find the Eigenvalues and eigenvectors of 3x3 matrix

[Michael_0039]

Assume a table A(3x3) with the following:

A [ 1 2 1 ]^T = 6 [ 1 2 1 ]^T
A [ 1 -1 1 ]^T = 3 [ 1 -1 1 ]^T
A [ 2 -1 0]^T = 3 [ 1 -1 1]^T

Find the Eigenvalues and eigenvectors:

I have in mind to start with the Av=λv or det(A-λI)v=0....

Also, the first 2 equations seems to have the form Av=λv:
So maybe, u1= [ 1 2 1 ] with λ=6 and u2= [1 -1 1] with λ=3 .......but what about 3rd ?

 

[anuttarasammyak]

Why don't you get A explicitly at first ?

 

[Michael_0039]

Why don't you get A explicitly at first ?

Matrix A is not given, only those equations.

 

[anuttarasammyak]

Those equations are equivalent to a product relation of 3X3 matrices say
AB=C
So explicitly
A=CB^{-1}
...But I see getting explicit A is not a straight forward way

 

[fresh_42]

Assume a table A(3x3) with the following:

A [ 1 2 1 ]^T = 6 [ 1 2 1 ]^T
A [ 1 -1 1 ]^T = 3 [ 1 -1 1 ]^T
A [ 2 -1 0]^T = 3 [ 1 -1 1]^T

Find the Eigenvalues and eigenvectors:

I have in mind to start with the Av=λv or det(A-λI)v=0....

Also, the first 2 equations seems to have the form Av=λv:
So maybe, u1= [ 1 2 1 ] with λ=6 and u2= [1 -1 1] with λ=3 .......but what about 3rd ?

Correct.

The second and third means, that the eigenspace to the eigenvalue $3$ has dimension two. It is spanned by $(1,-1,1)^\tau$ and $(2,-1,0)^\tau.$

 

[Michael_0039]

So the answer will be:
λ₁=6 with u₁=[ 1 2 1 ]^T
λ₂=3 with u₂=[ 1 -1 1 ]^T and u₃=[ 2 -1 0 ]^T

(?)

 

[fresh_42]

So the answer will be:
λ₁=6 with u₁=[ 1 2 1 ]^T
λ₂=3 with u₂=[ 1 -1 1 ]^T and u₃=[ 2 -1 0 ]^T

(?)

Yes.

I mean, almost.
All linear combinations of $(1,2,1)$ (i.e. all multiples of) are eigenvectors to the eigenvalue $6,$ too.
All linear combinations of $(1,-1,1)$ and $(2,-1,0)$ are eigenvectors to the eigenvalue $3,$ too.

 

[fresh_42]

@Michael_0039 note my edit. You've been too fast.

If $v_1,\ldots,v_n$ are all eigenvectors to the same eigenvalue $\lambda$ then
$$
A\left(\sum_{k=1}^n \alpha_k v_k\right)=\sum_{k=1}^n\alpha_k A(v_k)=\sum_{k=1}^n\alpha_k \lambda v_k=\lambda \cdot \left(\sum_{k=1}^n\alpha_k v_k\right)
$$
all linear combinations are eigenvectors to the eigenvalue $\lambda,$ too.

 

[Michael_0039]

So for the 2^nd question: Is A reversible or/and diagonal ?
The answer if A^-1 exists is λ₁*λ₂≠0 and is non-diagonal because λ₂ has reached one time but has 2 eigenvectors (?)

 

[fresh_42]

So for the 2^2nd question: Is A reversible ...

What do you think? Hint: not reversible implies not injective. Now, what does not injective mean?

... or/and diagonal

Are those three vectors linearly independent? If so, they will be a basis. How does $A$ look like according to that basis?

 

[anuttarasammyak]

A [ 1 -1 1 ]^T = 3 [ 1 -1 1 ]^T
A [ 2 -1 0]^T = 3 [ 1 -1 1]^T

By subtraction I observe
A[-1,0,1]^T=[0,0,0]^T=0[-1,0,1]^T
which suggests $\lambda=0,3,6$ in all.

 

[pasmith]

Correct.

The second and third means, that the eigenspace to the eigenvalue $3$ has dimension two. It is spanned by $(1,-1,1)^\tau$ and $(2,-1,0)^\tau.$

The second and third equations tell us that (2, -1, 0)^T - (1, -1, 1)^T = (1,0,-1)^T \in \ker A.

 

[fresh_42]

The second and third equations tell us that (2, -1, 0)^T - (1, -1, 1)^T = (1,0,-1)^T \in \ker A.

Thank you. My mistake! I mistook it for $A(2,-1,0)=(2,-1,0).$

@Michael_0039 Forget my posts. I was thinking about a different situation.