Find basis of eigenspace

[Nemanjavuk69]

Hello

I am reading up on linear algebra and its applications 6 global edition by David C. Lay.

I am at chapter 5.3 where it talks about eigenvalues and eigenspace, and I am trying to solve excercise number 5 (Please look at the attached image file)




I also know that the answer is $$\lambda=5: \begin{bmatrix} 1 &\\ 1 &\\ 1 \end{bmatrix}$$
$$\lambda = 1: \begin{bmatrix} 1 &\\ 0 &\\ -1 \end{bmatrix}, \begin{bmatrix} 2 \\ -1 \\ 0 \end{bmatrix}$$
My question is now, where does the basis(answer) come from? Can anyone show me a computation of how the basis are being calculated? I seem to always get another basis when solving it myself.

 

[BigBeachBanana]

Hello

I am reading up on linear algebra and its applications 6 global edition by David C. Lay.

I am at chapter 5.3 where it talks about eigenvalues and eigenspace, and I am trying to solve excercise number 5 (Please look at the attached image file)

View attachment 34681


I also know that the answer is $$\lambda=5: \begin{bmatrix} 1 &\\ 1 &\\ 1 \end{bmatrix}$$
$$\lambda = 1: \begin{bmatrix} 1 &\\ 0 &\\ -1 \end{bmatrix}, \begin{bmatrix} 2 \\ -1 \\ 0 \end{bmatrix}$$
My question is now, where does the basis(answer) come from? Can anyone show me a computation of how the basis are being calculated? I seem to always get another basis when solving it myself.

To find the eigenspace $\bold{v_1}$ associated with $\lambda_1 = 5$ it must sastisfy the following:
$$\bold{A\cdot v_1} = \lambda_1 \cdot \bold{v_1}$$$$\bold{(A-\lambda_1)}\cdot \bold{v_1}=0$$$$\left(\begin{bmatrix} 2 & 2 & 1 \\ 1 & 3 & 1\\ 1& 2 & 2 \end{bmatrix} - \begin{bmatrix} 5 & 0 & 0 \\ 0 & 5 & 0\\ 0 & 0 & 5 \end{bmatrix}\right) \cdot \begin{bmatrix} v_1 \\ v_2 \\ v_3 \\ \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \\ \end{bmatrix}$$
Using REF or RREF to solve for $\bold{v_1} = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}$

 

[Dr.Peterson]

I also know that the answer is $$\lambda=5: \begin{bmatrix} 1 &\\ 1 &\\ 1 \end{bmatrix}$$
$$\lambda = 1: \begin{bmatrix} 1 &\\ 0 &\\ -1 \end{bmatrix}, \begin{bmatrix} 2 \\ -1 \\ 0 \end{bmatrix}$$
My question is now, where does the basis(answer) come from? Can anyone show me a computation of how the basis are being calculated? I seem to always get another basis when solving it myself.

Please show your own work; it is perfectly respectable to get a different answer for the eigenspace of $\lambda=1$, and your answer may well be valid. Or there could be an error we could correct.

As for how they got their answer for that eigenvalue, it isn't what I got (though mine is quite close), so I'd want to see how they have taught you to find the basis of an eigenspace. Do you have any examples from the book?

 

[Nemanjavuk69]

Please show your own work; it is perfectly respectable to get a different answer for the eigenspace of $\lambda=1$, and your answer may well be valid. Or there could be an error we could correct.

As for how they got their answer for that eigenvalue, it isn't what I got (though mine is quite close), so I'd want to see how they have taught you to find the basis of an eigenspace. Do you have any examples from the book?

I have attached a PDF snippet of the chapter where they talk about Diagonalization and the basis of eigenspace. As for my work, I will try again and calculate everything from the bottom and be extra careful when reading the Theorem. I will paste it here soon.

 

[Dr.Peterson]

I have attached a PDF snippet of the chapter where they talk about Diagonalization and the basis of eigenspace. As for my work, I will try again and calculate everything from the bottom and be extra careful when reading the Theorem. I will paste it here soon.

I see that in Example 6, which is the relevant one, they just say "Using the method in Section 5.1, we find a basis for each eigenspace." So you'll want to look there (and show that to us if you need more help).

 

[Nemanjavuk69]

Here is the chapter 5.1 if that helps

I see that in Example 6, which is the relevant one, they just say "Using the method in Section 5.1, we find a basis for each eigenspace." So you'll want to look there (and show that to us if you need more help).

 

[Nemanjavuk69]

Please show your own work; it is perfectly respectable to get a different answer for the eigenspace of $\lambda=1$, and your answer may well be valid. Or there could be an error we could correct.

As for how they got their answer for that eigenvalue, it isn't what I got (though mine is quite close), so I'd want to see how they have taught you to find the basis of an eigenspace. Do you have any examples from the book?

Dear @Dr.Peterson I have finally managed to solve this type of exercises correctly. After hours of looking at theorem and getting different answers for eigenspace, I can see, by using a calculator, that indeed I get the correct answer. Now lies a curious question, how come we can get the same answer with different answer for the eigenspace?

 

[Dr.Peterson]

Here is the chapter 5.1 if that helps

They are pointing you to example 4. Note, however, that at the end they make an arbitrary change in one of the eigenvectors in order to avoid fractions. In the problem you have been working on, it appears that they did the same thing (as their method is identical to mine), this time perhaps to avoid too many negatives.

I can see, by using a calculator, that indeed I get the correct answer. Now lies a curious question, how come we can get the same answer with different answer for the eigenspace?

A key idea is that their are many possible bases for any space; and in particular, multiplying each basis vector by a scalar will not change the space. So it is to be expected that people will come up with different specific answers. This is why I said, "it is perfectly respectable to get a different answer for the [basis of the] eigenspace of $\lambda=1$, and your answer may well be valid."

 

[Nemanjavuk69]

They are pointing you to example 4. Note, however, that at the end they make an arbitrary change in one of the eigenvectors in order to avoid fractions. In the problem you have been working on, it appears that they did the same thing (as their method is identical to mine), this time perhaps to avoid too many negatives.


A key idea is that their are many possible bases for any space; and in particular, multiplying each basis vector by a scalar will not change the space. So it is to be expected that people will come up with different specific answers. This is why I said, "it is perfectly respectable to get a different answer for the [basis of the] eigenspace of $\lambda=1$, and your answer may well be valid."

If it is not too much to ask, can I be allowed to ask this question. If you try to diagonalize a $2\times2$ matrix A and it appears that you only have one eigenvalue, let's say $(\lambda-4)\cdot(\lambda-4)$ would that tell you that the matrix A IS NOT diagonizable? I only ask to broaden my knowledge a little bit. Hope my question is clear enough.

 

[Dr.Peterson]

If it is not too much to ask, can I be allowed to ask this question. If you try to diagonalize a $2\times2$ matrix A and it appears that you only have one eigenvalue, let's say $(\lambda-4)\cdot(\lambda-4)$ would that tell you that the matrix A IS NOT diagonizable? I only ask to broaden my knowledge a little bit. Hope my question is clear enough.

Not necessarily. You can have a diagonal matrix whose eigenvalues are both 4, right?

 

[Nemanjavuk69]

Not necessarily. You can have a diagonal matrix whose eigenvalues are both 4, right?

Alright, that would be all of my questions. Thank you so much @Dr.Peterson for taking your time to teach me and elaborate on my questions. Have a wonderful day/ night forward