Eigenvectors help Watch

ManLike007
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Suppose 2 known 3x3 matrices A and B. The eigenvectors and eigenvalues of A are known. I need to determine the eigenvectors and eigenvalues of B through a relationship between A and B given that

A-B=kI where k is a real number. How do I do this? (Note: the eigenvectors are the same but the eigenvalues are different)

I've had a few attempts but I don't know where I'm going with this,

i.e. Multiply both sides with A^{-1}

A^{-1}(A-B)=A^{-1}k \Rightarrow -A^{-1}B=A^{-1}k and multiply both sides with eigenvector {\bf  v}, {\bf v} \neq 0,

-A^{-1}B{\bf v}=A^{-1}k{\bf v}

Any help is appreciated
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RDKGames
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(Original post by ManLike007)
Any help is appreciated
A \mathbf{v} = \lambda \mathbf{v}

(A-B) \mathbf{v} = \lambda \mathbf{v} - B \mathbf{v}

kI \mathbf{v} = \lambda \mathbf{v} - B \mathbf{v}

finish it off from there.
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ghostwalker
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(Original post by ManLike007)
i.e. Multiply both sides with A^{-1}

A^{-1}(A-B)=A^{-1}k \Rightarrow -A^{-1}B=A^{-1}k
On a side note:

A^{-1}(A-B)=A^{-1}k \not\Rightarrow -A^{-1}B=A^{-1}k

Rather, you would have had:

I-A^{-1}B=A^{-1}k
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ManLike007
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(Original post by RDKGames)
A \mathbf{v} = \lambda \mathbf{v}

(A-B) \mathbf{v} = \lambda \mathbf{v} - B \mathbf{v}

kI \mathbf{v} = \lambda \mathbf{v} - B \mathbf{v}

finish it off from there.
Sorry I had to go somewhere urgent when you posted this,

It still hasn't clicked yet, do you move everything to one side?

i.e. (kI+B-\lambda){\bf v}=0?

Honestly, I'm still a bit baffled, could you help me a bit more by any chance?

(Original post by ghostwalker)
On a side note:

A^{-1}(A-B)=A^{-1}k \not\Rightarrow -A^{-1}B=A^{-1}k

Rather, you would have had:

I-A^{-1}B=A^{-1}k
Ah ok fair enough, I forgot to do this
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RDKGames
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(Original post by ManLike007)
Sorry I had to go somewhere urgent when you posted this,

It still hasn't clicked yet, do you move everything to one side?

i.e. (kI+B-\lambda){\bf v}=0?

Honestly, I'm still a bit baffled, could you help me a bit more by any chance?
You want to get B\mathbf{v}=(...)\mathbf{v} then the eigenvalue of B with a corresponding eigenvector is obvious from the (...) bit
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ManLike007
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(Original post by RDKGames)
the eigenvalue of B with a corresponding eigenvector is obvious
Oh right I see what you mean, but what if you didn't know the eigenvector of B is the same as A? This is what I need to find out assuming I didn't know they both had the same eigenvectors.

Sorry if my original post confused you
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RDKGames
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(Original post by ManLike007)
Oh right I see what you mean, but what if you didn't know the eigenvector of B is the same as A? This is what I need to find out assuming I didn't know they both had the same eigenvectors.

Sorry if my original post confused you
Doesn't matter. We did not use that fact in our process, and only 'verified' it at the end.

The method begins under the assumption that \mathbf{v} is an eigenvector of A with a corresponding eigenvalue \lambda. You then introduce B\mathbf{v} which leads us to the end after using the relation fact of A-B=kI, and clearly shows that B\lambda = \Lambda \mathb{v} hence \mathbf{v} is an eigenvector of B as well, with a corresponding eigenvalue of \Lambda \in \mathbb{R}
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ManLike007
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(Original post by RDKGames)
Doesn't matter.
The method begins under the assumption that \mathbf{v} is an eigenvector of A with a corresponding eigenvalue \lambda. You then introduce B\mathbf{v} which leads us to the end after using the relation fact of A-B=kI, and clearly shows that B\lambda = \Lambda \mathb{v} hence \mathbf{v} is an eigenvector of B as well, with a corresponding eigenvalue of \Lambda \in \mathbb{R}
I get the first first of starting from A{\bf v}=\lambda{\bf v} etc, and using the fact that A-B=kI.

Then suppose B = \begin{pmatrix} 1 & 2 & 3 \\4 & 5 & 6 \\7 & 8 & 9 \end{pmatrix}.

B{\bf v} = (\lambda - kI){\bf v} \Rightarrow \begin{pmatrix} 1 & 2 & 3 \\4 & 5 & 6 \\7 & 8 & 9 \end{pmatrix}{\bf v} = \begin{pmatrix} \lambda -k & 0 & 0 \\0 & \lambda - k & 0 \\0 & 0 & \lambda - k \end{pmatrix}{\bf v}

Move everything to the RHS, divide both sides by {\bf v} then the determinant thingy all over again? I'm sorry this question has been on my mind since Sunday, I can't think this through
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RDKGames
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(Original post by ManLike007)
I get the first first of starting from A{\bf v}=\lambda{\bf v} etc, and using the fact that A-B=kI.

Then suppose B = \begin{pmatrix} 1 & 2 & 3 \\4 & 5 & 6 \\7 & 8 & 9 \end{pmatrix}.

B{\bf v} = (\lambda - kI){\bf v} \Rightarrow \begin{pmatrix} 1 & 2 & 3 \\4 & 5 & 6 \\7 & 8 & 9 \end{pmatrix}{\bf v} = \begin{pmatrix} \lambda -k & 0 & 0 \\0 & \lambda - k & 0 \\0 & 0 & \lambda - k \end{pmatrix}{\bf v}

Move everything to the RHS, divide both sides by {\bf v} then the determinant thingy all over again? I'm sorry this question has been on my mind since Sunday, I can't think this through
This is not correct. The 'eigenvalue' here is not a single value but a matrix!

In your working out, you need to use the fact that I\mathbf{v}=\mathbf{v}, then you can deduce the actual eigenvalue.

Clearly, if you know your k and \lambda for any two matrices A and B with the given property in your OP, then you can just read off the eigenvalue for B here.

P.S. "divide by the \mathbf{v}" makes no sense.
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ManLike007
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(Original post by RDKGames)
Clearly, if you know your k and \lambda for any two matrices A and B with the given property in your OP, then you can just read off the eigenvalue for B here.
Yh I get that.

(Original post by RDKGames)
In your working out, you need to use the fact that I\mathbf{v}=\mathbf{v}, then you can deduce the actual eigenvalue.
What do you mean? Can you give an example?

(Sorry for taking up your time)
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RDKGames
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(Original post by ManLike007)
What do you mean? Can you give an example?
From

kI \mathbf{v} = \lambda \mathbf{v} - B \mathbf{v}

you say

k \mathbf{v}= \lambda \mathbf{v} - B \mathbf{v}

and THEN get the B\mathbf{v}=(...)\mathbf{v} form where (...) is the eigenvalue, NOT a matrix.

It is extremely simple to verify that I \mathbf{v} = \mathbf{v} if you don't know why it's true - just find what I\mathbf{v} is when \mathbf{v}=(v_1,v_2,v_3)
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ManLike007
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(Original post by RDKGames)
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You went from kI to k, how though? from a matrix to a single value?

Besides that, you'll have B{\bf v}=(\lambda-k){\bf v}

B is a matrix whereas k and \lambda are single values. \lambda is the unknown, k and B are known BUT k and \lambda are integers and B is a matrix..? I can't compare them
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RDKGames
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(Original post by ManLike007)
You went from kI to k, how though? from a matrix to a single value?

Besides that, you'll have B{\bf v}=(\lambda-k){\bf v}

B is a matrix whereas k and \lambda are single values. \lambda is the unknown, k and B are knowns BUT k is an integer and B is a matrix..? I can't compare them
Firstly, kI\mathbf{v}=k(I\mathbf{v})=k( \mathbf{v})=k\mathbf{v}

Secondly, that is exactly what we get, and what we WANT to get. You asked in your OP about finding what the eigenvalues for B are, well THERE'S the eigenvalue for the corresponding eigenvector \mathbf{v} of B. It's k-\lambda

Lastly, I don't know what you are trying to compare or WHY. In fact, \lambda IS known, because it's an eigenvalue of A and you said we know those along with their eigenvectors!
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ManLike007
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(Original post by RDKGames)
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I'm very very sorry if I worded it wrong, I meant to say how do I find out the eigenvectors and eigenvalues of B using the relationship that A-B=kI.

The eigenvectors and eigenvalues and matrix A are all known, Matrix B is known but not its eigenvectors or eigenvalues.

The end result should be that A and B eigenvectors are the same but we don't know just yet. (Again I'm very sorry if I confused you)
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RDKGames
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(Original post by ManLike007)
I'm very very sorry if I worded it wrong, I meant to say how do I find out the eigenvectors and eigenvalues of B using the relationship that A-B=kI.

The eigenvectors and eigenvalues and matrix A are all known, Matrix B is known but not its eigenvectors or eigenvalues.

The end result should be that A and B eigenvectors are the same but we don't know just yet. (Again I'm very sorry if I confused you)
I believe you are extremely confused yourself and aren't expressing yourself properly. Look, this is what we get from doing the whole process:

1. We know that A has some eigenvalue \lambda and a corresponding eigenvector \mathbf{v}

2. We are looking for the eigenvalues and eigenvectors of B, knowing that the following relationship between A and B is satisfied: A-B=kI for some k \in \mathbb{R}

3. We begin with the definition of what it means for A to have an eigenvector/value, so we start with A\mathbf{v} = \lambda \mathbf{v}

4. We follow the process through and end up with the result of B\mathbf{v}=(\lambda - k)\mathbf{v} which is in fact the definition form of what it means for B to have eigenvectors/values. This form proves that the eigenvectors of A are also the eigenvectors of B! So, now we know the eigenvectors of B.

5. Along with the previous deduction, we also know that the eigenvalue for a corresponding eigenvector \mathbf{v} of B is deduced from the formula \lambda - k.



Get it??

I can give you an example of how we utilise these results if it would help you understand.
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ManLike007
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(Original post by rdkgames)
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yes yes yes I get it now!!

Thank you so much and sorry for taking up your time!!
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ManLike007
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(Original post by RDKGames)
I can give you an example of how we utilise these results if it would help you understand.
Sorry I just saw this now, if you have time then sure I'm interested but if you feel like I've consumed too much of your time then it's totally fine.
(It's just that I've been stuck on this since the post I made about AB=kI on Sunday if you remember and been stuck on this problem A-B=kI since )
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RDKGames
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(Original post by ManLike007)
Sorry I just saw this now, if you have time than sure I'm interested but if you feel like I've consumed too much of your time then it's totally fine.
(It's just that I've been stuck on this since the post I made about AB=kI on Sunday if you remember and been stuck on this problem A-B=kI since )
Yeah I remember you asking that, but I couldn't reply as it was Christmas Eve after all

Anyhow, consider:

\displaystyle A=\begin{pmatrix} 1 & 2 \\ 4 & 3 \end{pmatrix} which as eigenvalues \lambda_1=5, \lambda_2=-1 and corresponding eigenvectors \mathbf{v}_1=(1,2) and \mathbf{v}_2=(-1,1)

Let \displaystyle B=\begin{pmatrix} -2 & 2 \\ 4 & 0 \end{pmatrix}

Here, A-B=3I=kI hence k=3

Let's say the eigenvalues of B are \mu_1 and \mu_2, with corresponding eigenvectors being \mathbf{w}_1 and \mathbf{w}_2

Next we will apply our result.

For the first set of eigenvalues/vectors, we have:
\mathbf{w}_1=\mathbf{v}_1=(1,2)
\mu_1=\lambda_1 - k=2

For the second set of eigenvalues/vectors, we have:
\mathbf{w}_2=\mathbf{v}_2
\mu_2=\lambda_2-k=-4


Hence, B has eigenvalues \mu_1=2 and \mu_2=-4 with corresponding eigenvectors \mathbf{w}_1=(1,2) and \mathbf{w}_2=(-1,1)
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ManLike007
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(Original post by RDKGames)
Yeah I remember you asking that, but I couldn't reply as it was Christmas Eve after all
I completely respect that, no worries!

(Original post by RDKGames)
Anyhow, consider: .....
I appreciate your help and everything, I understand this a lot better now. Thanks
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