Eigenvalues and Eigenvectors question

[Dank2]

Let's say i have the 3x3 matrix

a  0  0
b  0  0
1  2  1

it's eigenvalues are e1 =a, e2 = 0, e3 = 1.
now if a = / = 0, 1
i have 3 distinct eigenvalues and the matrix is surely can be Diagonalizable.
so if i try to solve for the eigenvector for the eigenvalue e1 =a:

0  0  0
b  -a  0
1  2  1-a

after doing elementary operations i get:

1        2      1-a
0   -a-2b    -b(1-a)
0      0        0

now if -a-2b =0, then -b(1-a) =/=0, and vice versa. how do i find the eigenvector for a=/=0,1, if i don't know which one is equal to zero A22 or A23 ?

 

[andrewkirk]

Your second code panel gives you two equations in the three unknowns (x,y,z) that are the components of the sought eigenvector. Set any of the three components to 1 to remove one unknown and then solve the system of two equations in two unknowns to find what the corresponding values of the other two components must be. Eigenvectors with eigenvalue a will be any multiple of that vector.

 

[Dank2]

Your second code panel gives you two equations in the three unknowns (x,y,z) that are the components of the sought eigenvector. Set any of the three components to 1 to remove one unknown and then solve the system of two equations in two unknowns to find what the corresponding values of the other two components must be. Eigenvectors with eigenvalue a will be any multiple of that vector.

You mean should i put values to a and b and then solve numerically? because if i just multiply by (x,1,z) i get that a = 0:
1 2 1-a
0 -a-2b -b(1-a)
0 0 0
i got the eigenvector v= ((3-a)/(a-1) , 1 , t) t in R.
but it doesn't worth the matrix * v= a * v

 

[andrewkirk]

You mean should i put values to a and b and then solve numerically? because if i just multiply by (x,1,z) i get that a = 0:
1 2 1-a
0 -a-2b -b(1-a)
0 0 0

You seem to be using the matrix in your third code panel. My suggestion was to use the matrix in the second code panel, which I am confident is correct. I have not checked the matrix in the third panel and do not know whether it is correct.

Also, it sometimes happens that setting a particular component to 1 in the candidate eigenvector does not lead to a solution. In that case, one just chooses a different component to set to 1. There will be at least one component that can generate a solution by setting it to 1. But in this case I'm fairly confident there will be no problem with using the second component, provided the correct matrix is used.

 

[Dank2]

If i just solve i got
0 = a*x
bx-ay = y*a
x+2y+z-az = z*a

from first eq i get, x = 0, since a=/=0, 1
then y = 0, since -ay = ay and a=/=0, 1
any more suggestions how can i continue ?

 

[andrewkirk]

I'm afraid I don't understand what calculations you are doing. Based on the matrix in the second code panel, the multiplication to find the eigenvector is:

$$\begin{pmatrix}
0 & 0 & 0 \\
b & -a & 0 \\
1 & 2 & (1-a)
\end{pmatrix}
\begin{pmatrix}x\\1\\z\end{pmatrix}=
\begin{pmatrix}0\\0\\0\end{pmatrix}
$$
That does not generate the same equations as you have written.

Ah, I just noticed. I think the problem might be that you are setting the result of the multiplication equal to $\begin{pmatrix}x\\1\\z\end{pmatrix}$, whereas actually it should be $\begin{pmatrix}0\\0\\0\end{pmatrix}$, as the former vector has already been deducted from both sides to obtain the matrix in the second panel. Also, remember to set y equal to 1 so that there are only two variables (x and z) to be found from the two equations.

 

[Dank2]

I think you thought i solve for a = 0 . I need to find eigenvalue a that is different from 0 or 1. I've uploaded a pic
do i must put numerical values in a and b in order to get the eigenvector that belongs to the eigenvalue a? i wanted to find the eigenvector in terms of a and b.

 

[andrewkirk]

That equation you have drawn is incorrect. Correct versions are the one I posted in my previous post or, alternatively (and equivalently):

$$
\begin{pmatrix}
a & 0 & 0 \\
b & 0 & 0 \\
1 & 2 & 1
\end{pmatrix}
\begin{pmatrix}x\\1\\z\end{pmatrix}=
a\cdot\begin{pmatrix}x\\1\\z\end{pmatrix}
$$

You solve this for $x$ and $z$, with the solutions being written as formulas that use $a$ and $b$. You do not replace $a$ or $b$ by numerical values.

 

[Dank2]

i get in my second equation bx = a, but i can't divide by b since b can be equal to 0. only a is not equal to 1 or 0. b in R .

 

[andrewkirk]

If you set b=0 and solve the equations you will deduce that a=0. Since that contradicts the premise that a is nonzero, you can conclude (by the contrapositive) that b must also be nonzero. Hence you can divide by b.

 

[Dank2]

but for any if a = 2, b = 0, the matrix is still diagonalizable, if i set b =/= 0 i exclude all the possiblities where b= 0 and a in R =/= 0,1.
so it's not the all of the solution.

 

[andrewkirk]

My mistake. If $b=0$ our conclusion is not that $a=0$ but rather that the eigenvector corresponding to eigenvalue $a$ cannot have 1 as the second component. So, as foreshadowed in post #4, let's instead set the third component equal to 1 and solve
$$
\begin{pmatrix}
a & 0 & 0 \\
b & 0& 0 \\
1 & 2 & 1
\end{pmatrix}
\begin{pmatrix}x\\y\\1\end{pmatrix}=
a \begin{pmatrix}x\\y\\1\end{pmatrix}
$$

 

[Dank2]

My mistake. If $b=0$ our conclusion is not that $a=0$ but rather that the eigenvector corresponding to eigenvalue $a$ cannot have 1 as the second component. So, as foreshadowed in post #4, let's instead set the third component equal to 1 and solve
$$
\begin{pmatrix}
a & 0 & 0 \\
b & 0& 0 \\
1 & 2 & 1
\end{pmatrix}
\begin{pmatrix}x\\y\\1\end{pmatrix}=
a \begin{pmatrix}x\\y\\1\end{pmatrix}
$$

It did work now, can you please explain me why i could set z to 1? because the eigenvector is subspace and therefore i can multiply the eigenvector by 1/x or 1/y or 1/z and it would still be eigenvector ? is that the reason? since A*v = a*v <==> A*1/x*v = 1/x*a*v ?

 

[andrewkirk]

@Dank2 Yes you can multiply an eigenvector by any nonzero scalar and it remains an eigenvector. So in searching for the eigenvector corresponding to a known eigenvalue, we can reduce the number of unknowns in our system of equations by setting a component of the candidate eigenvector to 1 as long as that component is not zero in the solution.

 

[Dank2]

@Dank2 Yes you can multiply an eigenvector by any nonzero scalar and it remains an eigenvector. So in searching for the eigenvector corresponding to a known eigenvalue, we can reduce the number of unknowns in our system of equations by setting a component of the candidate eigenvector to 1 as long as that component is not zero in the solution.

One last thing that i have noticed.
In getting the solution :
x = (a^2-a)/(a+2b)
y = (ba-b)/(a+2b)
I have noticed that i devided by b in order to solve for x. but b is in R, it can be 0 as well and there will be a solution. same problem as in post # 9

 

[Dank2]

I think since i have a and b and in the matrix , i cannot find one eigenvector for both of them, since ill get different ones for different values, and i need to find a matrix that diagonalize A, and so i think i must put numeric values in a and b for that . , since i need only one eigenvector for the eigenvalue a =/= 0, 1. and two other eigenvectors for other eigenvalues i have. i got to put them in P such that
P'AP = Diagonal.

 

[andrewkirk]

I have noticed that i devided by b in order to solve for x.

You shouldn't need to do that. From the second component equation you get
$$bx=ay$$
But we can divide by $a$ instead of $b$, since we are told that $a\neq 0$. That gives us
$$y=\frac ba x$$