Eigenbasis 3 x 3 Matrix

[MikeDietrich]

Homework Statement

Find all real eigenvalues, the basis for each eigenspace, and an eigenbasis.

A = [ 1 0 0 $-5 0 2$ 0 0 1] Note: ## starts a new row

Homework Equations

det A = 0
E = N ($I-A) where $ = eigenvector

The Attempt at a Solution

So, after calculating detA = 0 I determined $_1=1, $_2=1, and $_3=0 where $ = eigenvector.

I then calculated E_1 = [1 1/5 2/5 $0 0 0$ 0 0 0] [ v_1 $v_2$ v_3] = [0 $0$ 0] therefore, E_1 = span [1 $-5$ 0]

E_0 = [1 0 0 $0 0 1$ 0 0 0][ v_1 $v_2$ v_3] = [0 $0$ 0] therefore, E_0 = span [0 $1$ 0]

At this point I assumed there is no eigenbasis since there are less unique eigenvalues then 3 (3 x 3 matrix). However, when I plug the matrix into an online eigenvalue calculator I get:
Eigenbasis: [0 $1$ 0], [1 $-5$ 0], [0 $2$ 1].

Where did the [0 $2$ 1] come from?

Thanks!

 

[Mark44]

Homework Statement

Find all real eigenvalues, the basis for each eigenspace, and an eigenbasis.

A = [ 1 0 0 $-5 0 2$ 0 0 1] Note: ## starts a new row

Homework Equations

det A = 0
E = N ($I-A) where $ = eigenvector

The Attempt at a Solution

So, after calculating detA = 0 I determined $_1=1, $_2=1, and $_3=0 where $ = eigenvector.

No, these are eigenvalues.

I then calculated E_1 = [1 1/5 2/5 $0 0 0$ 0 0 0] [ v_1 $v_2$ v_3] = [0 $0$ 0] therefore, E_1 = span [1 $-5$ 0]

E_0 = [1 0 0 $0 0 1$ 0 0 0][ v_1 $v_2$ v_3] = [0 $0$ 0] therefore, E_0 = span [0 $1$ 0]

At this point I assumed there is no eigenbasis since there are less unique eigenvalues then 3 (3 x 3 matrix). However, when I plug the matrix into an online eigenvalue calculator I get:
Eigenbasis: [0 $1$ 0], [1 $-5$ 0], [0 $2$ 1].

Where did the [0 $2$ 1] come from?

It must be one of the eigenvectors associated with the eigenvalue 1. The eigenvector <0, 1, 0> is associated with the eigenvalue 0.

The eigenspace for the eigenvalue 1 is two-dimensional, so there are two eigenvectors. The ones I get are <-1, 5, 0> and <2, 0, 5>. Other pairs are possible.

 

[MikeDietrich]

Thanks Mark... you are correct... I did mean eigenvalue (not eigenvector). I can now see how to get <2, 0, 5> but how did you know the eigenspace was two dimensional for the eigenvalue of 1? For example, the matrix A = [1 1 0 $0 1 1$ 0 0 1] has an eigenvalue equal to 1 and there is only one eigenvector (at least the only one I have is <1, 0, 0>).

 

[MikeDietrich]

Wait. I think I have it. It is 2D because that is the multiplicity of the eigenvalue, correct?

 

[Mark44]

Look at the matrix A - 1I, which is
[0 0 0]
[-5 -1 2]
[0 0 0]

(We're trying to solve (A - 1I)x = 0, for x.)

If you swap the 1st and 2nd rows and row reduce, you get
[1 1/5 -2/5]
[0 0 0]
[0 0 0]

The first row says
x₁ + (1/5)x₂ - (2/5)x₃ = 0
or equivalently,
x₁ = (-1/5)x₂ + (2/5)x₃
x₂ and x₃ are arbitrary, so I can write the system as

x₁ = (-1/5)x₂ + (2/5)x₃
x₂ = ...x₂
x₃ = ......x₃

Since there are two free variables, x₂ and x₃, the dimension of the eigenspace for $\lambda = 1$ is 2. The vectors <-1/5, 1, 0> and <2/5, 0, 1> form a basis for this space, and you might notice that I picked these coordinates off the equations just above.

Any multiples of these vectors also work, so the basis could also be the vectors <-1, 5, 0> and <2, 0, 5>.

You can check that these vectors work by showing that A*u₁ = 1*u₁ and A*u₂ = 1*u₂, where u₁ and u₂ are the vectors above (either pair).

 

[HallsofIvy]

There is a difference between "algebraic multiplicity" and "geometric multiplicity". The first, "algebraic multiplicity" is the multiplicity of the eigenvalue as a root of the characteristic equation. The second, "geometric multiplicity" is the dimension of the subspace of eigenvectors corresponding to a specific eigenvalue. They are NOT necessarily the same- if an eigenvalue has "algebraic multiplicity" n, then it may have "geometric multiplicty" any integer from 1 to n.
For example, the matrices
$$\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0\\0 & 0 & 1\end{bmatrix}$$
$$\begin{bmatrix}1 & 1 & 0 \\ 0 & 1 & 0\\0 & 0 & 1\end{bmatrix}$$
and
$$\begin{bmatrix}1 & 1 & 0 \\ 0 & 1 & 1\\0 & 0 & 1\end{bmatrix}$$
all have \(\displaystyle (\lambda- 1)^3= 0\) as characteristic equation so eigenvalue 1 with algebraic multiplicity 3. The first has all of R³ as "eigenspace", the second has a two dimensional eigenspace and the third a one dimensional eigenspace.

Personally, I prefer to find eigenvectors directly from the defintions: $Av= \lambda v$.

With
$$A= \begin{bmatrix} 1 & 0 & 0 \\ -5 & 0 & 2 \\ 0 & 0 & 1 \end{bmatrix}$$
we must have
$$Av= \begin{bmatrix} 1 & 0 & 0 \\ -5 & 0 & 2 \\ 0 & 0 & 1 \end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix}= \begin{bmatrix}x \\ -5x+ 2z \\ z\end{bmatrix}= \begin{bmatrix}x \\ y \\ z\end{bmatrix}$$

That is equivalent to the three equations x= x, -5x+ 2z= y, and z= z. The first and last tell us nothing, of course, so it reduces to -5x+ 2z= y. Then <x, y, z>= <x, -5x+ 2z, z>= <x, -5x, 0>+ <0, 2z, z>= x<1, -5, 0>+ z<0, 2, 1>.

Yes, the eigenspace corresponding to eigenvalue 1 is spanned by <1, -5, 0> and <0, 2, 1> and so has dimension 2. Either there is another eigenvalue or A cannot be diagonalized. Assuming there is no other eigenvalue, then it can be put into the "Jordan normal form"
$$\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 1 \\0 & 0 & 1\end{bmatrix}$$

To do that you need to use a matrix, P, whose first two columns are the eigenvectors found and whose third column is a "generalized eigenvector", a vector v, such that (A- I)v is an eigenvector.