# Eigenvalues and Eigenvectors question

• I
Let's say i have the 3x3 matrix
Code:
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:
Code:
0  0  0
b  -a  0
1  2  1-a
after doing elementary operations i get:
Code:
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 ?

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andrewkirk
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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.

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
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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
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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. 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.

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andrewkirk
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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.

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

andrewkirk
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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
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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 therfore 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
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@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 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 aswell and there will be a solution. same problem as in post # 9

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 gotta put them in P such that
P'AP = Diagonal.

andrewkirk
$$bx=ay$$
$$y=\frac ba x$$