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timeforchg · Oct 23, 2011, 04:16 AM

Really helpful. :)

By the way I got this question regarding matrix,

If A=[ a b c
c a b
b a c ]

X1 = [ 1 1 1]T (transpose)

Show that the vectors satisfy the linear equation A X1 = λ1.

Can we take A = λ for this case?

timeforchg · Oct 23, 2011, 04:17 AM

The sign I wanted to show λ is lamda

jcaron2 · Oct 23, 2011, 10:00 AM

I'm not exactly sure what you mean. If A is 3x3 and X1 is 3x1, then the right side of the equation should be a 3x1 vector. Lambda is typically used to represent a scalar (not that it has to be). Does lambda have some special meaning I'm not aware of, or are you perhaps missing an X1 on the right side of the equation?

If the latter, then the equation would read

$A \cdot X_1= \lambda \cdot X_1$ ,

which is another way of saying "Show that X1 is an eigenvector of A." This is pretty trivial to show, simply by doing out the matrix multiplication. The left side simply comes out to a 3x1 vector with elements (a+b+c), (c+a+b), and (b+a+c), respectively. Since addition is commutative, all three elements can be rewritten as (a+b+c), and thus you can say that the answer is $\lambda \cdot$ X1, where $\lambda$ =(a+b+c).

You could also forget that they gave you the answer for the eigenvector ahead of time and just go through the normal procedure to find the eigenvectors and eigenvalues of them matrix A. You'd see that [1 1 1] transpose is indeed an eigenvector, thus satisfying the question.

By the way, a couple of quick things: First, when asking a new question, it is preferred that you start a new thread, rather than posting it as a "reply" in an existing thread (even if it's one you started). This makes it easier for people searching for help in the future to find out if their question has been asked before, without having to read through long lists of replies to unrelated questions. It also makes it much more likely that you'll get an answer, since everyone can see new questions. Otherwise, most people will not notice that there's been a new reply to an old question from several weeks or months back. Only people who "subscribe" to that particular thread (which happens by default for people who answered it) will be notified that there's a new reply. Anyway, notice that the site admins were nice enough to move your previous question into its own new thread (albeit with an unrelated title "Matrices", since that's the title of the thread in which you originally posted it). They'll probably do it again with this latest question, but in the future please try to post all of these things as new questions.

The second thing I wanted to point out was something you may find helpful. There are two different "skins" for AskMeHelpDesk. The default skin is called "Go", but there's also an older "traditional" skin that many people find much more useful, especially when posting questions and answers regarding math. The big difference is that you have the ability to preview your post before submitting it, and you have the ability to edit it during the first 24 hours if you find you made a mistake. Both skins have their advantages and disadvantages, but you might want to check out the old skin for future posts. You switch to it by clicking on the "Settings" button at the top right of the page, then clicking "Take me back to the old style" in the lower left of the subsequent popup. You can preview your post by clicking on the "Go Advanced" button before submitting it.

AMHD keeps a cookie on your computer to remember which skin you want to use, so if you want to get back to the Go skin you have to either log out and then log back in, or simply type "go" after the URL: Ask Me Help Desk.

Finally, one other thing you may find helpful: Greek letters, fractions, integrals, and all the other math related functions you see here are entered using Latex encoding. You can Google it and find a lot of different "how to" pages or go to a page like this one where you can enter equations graphically and it will generate the proper latex code for you. At any rate, to use Latex code in your posts you have to start with the word "math" in square brackets and end with the word "/math" in square brackets. For example, if I type [m*th]\int_0^\infty x^2 dx=\frac{x^3}{3}+c[/m*th], but I change the * to an "a", I get this:

$\int_0^\infty x^2 dx=\frac{x^3}{3}+c$

Thought you might find that useful. :)

timeforchg · Oct 23, 2011, 06:29 PM

Yes A is a 3x3 matrix and X1 is 3x1 matrix.

So meaning to say the vector X1 (above) the corresponding values of $\lambda 1$ is
a+b+c?

What if the value of X1 is $\left [ 1, -\frac{1}{2}+\frac{\sqrt{3}}{2}j, -\frac{1}{2}-\frac{\sqrt{3}}{2}j \right ] T$

jcaron2 · Oct 24, 2011, 12:14 PM

Originally Posted by timeforchg

So meaning to say the vector X1 (above) the corresponding values of $\lambda 1$ is
a+b+c?

I guess so. I'm still unclear if there was supposed to be an additional X1 on the right side of the equation or not.

Originally Posted by timeforchg

What if the value of X1 is $\left [ 1, -\frac{1}{2}+\frac{\sqrt{3}}{2}j, -\frac{1}{2}-\frac{\sqrt{3}}{2}j \right ]^T$

Now I'm really confused. I guess I'm missing the meaning of $\lambda 1$ . This is definitely not my area of expertise, so maybe it has a common meaning that I'm not aware of. I see that all of the components of this new vector X1 have a magnitude of 1, so it seems somehow related to 3-dimensional rotation or affine transformations or something. Can you give me some context here? Is this a general linear algebra class or is it something more like 3D graphics or the like?

timeforchg · Oct 24, 2011, 06:25 PM

Let me try to explain the whole question.

1)Given a circular matrix
$ A=\begin{bmatrix} a& b& c\\ c& a& b\\ b& c& a \end{bmatrix} B=\begin{bmatrix} e& f& g\\ g& e& f\\ f& g& e \end{bmatrix} $

1) Show that C=AB is also a cicular matrix. Write the form of a circular matrix 4x4.

* I have done this part

2) Show the vector $x1= \left[1 1 1 \right ]T$ satisfy the linear equations $A x1 = \Lambda 1 x1$

*This part also has been done

3) For the vector x1 in part 2 find the corresponding value of $\Lambda1$
*I guess this part also have been solved.

Now here is my question, instead of $x1=\left[1 1 1 \right ]T$ now the vector x1 is replace by $x1 = \left [1, -\frac{1}{2}+\frac{\sqrt{}3}{2}j,-\frac{1}{2}-\frac{\sqrt{}3}{2}j \right ]T$

Lets assume the value for x1 is the same for the left and right side.
How do we do part 2 and 3 if we use $x1 = \left [1, -\frac{1}{2}+\frac{\sqrt{}3}{2}j,-\frac{1}{2}-\frac{\sqrt{}3}{2}j \right ]T$

jcaron2 · Oct 27, 2011, 07:48 PM

Aaaah. The last row of the matrix is [b c a], not [b a c] like in your original question. Now it makes MUCH more sense! So let's do out the multiplication using the new value of x1:

$ A=\begin{bmatrix} a& b& c\\ c& a& b\\ b& c& a \end{bmatrix} \begin{bmatrix} 1\\ -\frac{1}{2}+\frac{\sqrt 3}{2}j\\ -\frac{1}{2}-\frac{\sqrt 3}{2}j \end{bmatrix}= \begin{bmatrix} a +b$-\frac{1}{2}+\frac{\sqrt 3}{2}j $ +c $-\frac{1}{2}-\frac{\sqrt 3}{2}j $\\ c +a$-\frac{1}{2}+\frac{\sqrt 3}{2}j $ +b $-\frac{1}{2}-\frac{\sqrt 3}{2}j $\\ b +c$-\frac{1}{2}+\frac{\sqrt 3}{2}j $ +a $-\frac{1}{2}-\frac{\sqrt 3}{2}j $\\ \end{bmatrix} $

Now we can factor the elements of the vector X1 out of the corresponding elements of the right-hand side of the equation. If we get a common factor for all three elements, that means we can write the right-hand side as the product of that factor ( $\lambda_1$ ) and X1, which is what you're trying to show. Let's try:

$\begin{bmatrix} a +b$-\frac{1}{2}+\frac{\sqrt 3}{2}j $ +c $-\frac{1}{2}-\frac{\sqrt 3}{2}j $\\ c +a$-\frac{1}{2}+\frac{\sqrt 3}{2}j $ +b $-\frac{1}{2}-\frac{\sqrt 3}{2}j $\\ b +c$-\frac{1}{2}+\frac{\sqrt 3}{2}j $ +a $-\frac{1}{2}-\frac{\sqrt 3}{2}j $\\ \end{bmatrix} = \begin{bmatrix} $1$\frac{a +b$-\frac{1}{2}+\frac{\sqrt 3}{2}j $ +c $-\frac{1}{2}-\frac{\sqrt 3}{2}j $}{$1$}\\ $-\frac{1}{2}+\frac{\sqrt 3}{2}j $\frac{c +a$-\frac{1}{2}+\frac{\sqrt 3}{2}j $ +b $-\frac{1}{2}-\frac{\sqrt 3}{2}j $}{$-\frac{1}{2}+\frac{\sqrt 3}{2}j $}\\ $-\frac{1}{2}-\frac{\sqrt 3}{2}j $\frac{b +c$-\frac{1}{2}+\frac{\sqrt 3}{2}j $ +a $-\frac{1}{2}-\frac{\sqrt 3}{2}j $}{$-\frac{1}{2}-\frac{\sqrt 3}{2}j $}\\ \end{bmatrix} = \begin{bmatrix} $1$$a +b\(-\frac{1}{2}+\frac{\sqrt 3}{2}j $ +c $-\frac{1}{2}-\frac{\sqrt 3}{2}j $\)\\ $-\frac{1}{2}+\frac{\sqrt 3}{2}j $\frac{c$-\frac{1}{2}-\frac{\sqrt 3}{2}j $ +a$-\frac{1}{2}+\frac{\sqrt 3}{2}j $$-\frac{1}{2}-\frac{\sqrt 3}{2}j $ +b $-\frac{1}{2}-\frac{\sqrt 3}{2}j $$-\frac{1}{2}-\frac{\sqrt 3}{2}j $}{$-\frac{1}{2}+\frac{\sqrt 3}{2}j $$-\frac{1}{2}-\frac{\sqrt 3}{2}j $}\\ $-\frac{1}{2}-\frac{\sqrt 3}{2}j $\frac{b$-\frac{1}{2}+\frac{\sqrt 3}{2}j $ +c$-\frac{1}{2}+\frac{\sqrt 3}{2}j $$-\frac{1}{2}+\frac{\sqrt 3}{2}j $ +a $-\frac{1}{2}-\frac{\sqrt 3}{2}j $$-\frac{1}{2}+\frac{\sqrt 3}{2}j $}{$-\frac{1}{2}-\frac{\sqrt 3}{2}j $$-\frac{1}{2}+\frac{\sqrt 3}{2}j $}\\ \end{bmatrix} = $

$ \begin{bmatrix} $1$$a +b\(-\frac{1}{2}+\frac{\sqrt 3}{2}j $ +c $-\frac{1}{2}-\frac{\sqrt 3}{2}j $\)\\ $-\frac{1}{2}+\frac{\sqrt 3}{2}j $ $ c\(-\frac{1}{2}-\frac{\sqrt 3}{2}j $ +a +b $-\frac{1}{2}+\frac{\sqrt 3}{2}j $ \)\\ $-\frac{1}{2}-\frac{\sqrt 3}{2}j $$b\(-\frac{1}{2}+\frac{\sqrt 3}{2}j $ +c$-\frac{1}{2}-\frac{\sqrt 3}{2}j $ +a\)\\ \end{bmatrix} = $a +b\(-\frac{1}{2}+\frac{\sqrt 3}{2}j $ +c $-\frac{1}{2}-\frac{\sqrt 3}{2}j $\) \begin{bmatrix} 1\\ -\frac{1}{2}+\frac{\sqrt 3}{2}j\\ -\frac{1}{2}-\frac{\sqrt 3}{2}j\\ \end{bmatrix} = \lambda_1 \[X_1 \] $

So yes, the new value of X1 also works, giving you the above eigenvalue $\lambda_1$ .

timeforchg · Nov 1, 2011, 08:14 PM

Thanks again for your time.
Even though there are some missing symbol I manage to figure it out. :)