Find Eigenvalues and normalised EigenVectors of a 3x3 matirx

**GreenyMcDuff** · October 5th 2011, 07:38 AM

Hey guys,

I'm new to the forum so please be gentle

Done a BSc in Maths (many moons ago), looking to brush up my skills so as to start a masters next year.

I have the matrix A:

6 -2 2
-2 5 0
2 0 7

This is how far I have got:

Calculated the determinant as (took me a while to remember what a determinant was!): -y^3 + 18y^2 - 99y +162 = 0

where y is a scalar

Firstly, does this look correct because I have struggled to factorise it.
Secondly, is there and easy way to write formulas in a post?

I have a couple more questions regarding the Eigenvectors and showing that A is invertible but I figured I would leave those until I have managed to work out the eigenvalues.

I appreciate any help you can give me.

Thanks all

Chris

**HallsofIvy** · October 5th 2011, 08:15 AM

To put formulae on here, use "LaTeX". see http://www.mathhelpforum.com/math-help/f47/

Now, to solve the equation, the first thing you do is cross your fingers and hope there is an rational number solution! By the "rational root theorem", any rational root of the polynomial equation $a_nx^n+ a_{n-1}x^{n-1}+ \cdot\cdot\cdot+ a_1x+ a_0= 0$ is of the form $\frac{a}{b}$ where a is an integer that evenly divides the "constant term" $a_0$ and b is an integer that evenly divides the "leading coefficient" $a_n$ . Here, [tex]a_n[tex] is 1 so any rational root must be an integer that evenly divides 162. Fortunately, 162 does not have too many factors: $162=2(81)= 2(3^4)$ .
So possible roots are $\pm 1, \pm 2, \pm 3, \pm 6, \pm 9, \pm 18, \pm 27, \pm 54, \pm 81, \pm 162$ . It is tedious, but not impossible, to try each of those.

(Good news! There is a simple integer root. Search for it. Once you have one factor, divide by it to get a quadratic equation you may be able to factor, or use the quadratic formula or complete the square to solve.)

(If there were no rational root, there is a "Cardano's formula", Cubic function - Wikipedia, the free encyclopedia
but it is very complex and tedious to use!)

**Opalg** · October 5th 2011, 11:46 AM

Originally Posted by GreenyMcDuff

I have the matrix A:

$\begin{bmatrix} 6 & -2 & 2\\ -2 & 5 & 0\\ 2 & 0 & 7\end{bmatrix}$

This is how far I have got:

Calculated the determinant as (took me a while to remember what a determinant was!): $-y^3 + 18y^2 - 99y +162 = 0$

where y is a scalar

Firstly, does this look correct because I have struggled to factorise it.

Where possible, try to avoid multiplying factors out. That way, you may never have to solve the cubic at all. I assume that you started by evaluating the determinant $\begin{vmatrix} 6-y & -2 & 2\\ -2 & 5-y & 0\\ 2 & 0 & 7-y\end{vmatrix}$ . If you expand that along the top row, you get

$(6-y)(5-y)(7-y) - 4(7-y) - 4(5-y).$

Now combine the second and third terms in that expression, to get

${\color{red}(6-y)}(5-y)(7-y) - 8{\color{red}(6-y)}.$

A common factor has magically appeared! So you can take that out, and then you only have a quadratic to solve.

There is no guarantee that a method like that will work. But eigenvalue problems are often rigged so that the answers come out as integers. You can often manipulate a determinant by using row and column operations (remember what those are?!) to simplify the determinant and get a common factor in a row or column. Then you can take that factor out and it gives you one of the eigenvalues.

Originally Posted by GreenyMcDuff

Secondly, is there and easy way to write formulas in a post?

In addition to following HallsofIvy's useful link, a good way to learn TeX is to hold the cursor over any of the formulas on this site, and the TeX input will appear in a popup.

**GreenyMcDuff** · October 5th 2011, 10:47 PM

Thanks for the quick replies guys, appreciate it - it is coming back to me.... slowly.

OK so I managed to find the eigenvalues (3, -6, 9) and their corresponding eigenvectors (dug out my old maths notes).

I also managed to remember what normalising meant: Essentially dividing the eigenvector by its length.

Now I have hit another snag that my notes don't seem to help me with (not sure if I should start a new thread for this but it seems relevant as it leads on from the previous matrix)

Hence, or otherwise, find the limit $\frac{1}{c^n}A^n$ for $n \rightarrow \infinity$ for all $c>0$ (if it exists)

Where:
A is the Matrix (specified earlier)
n is an integer
and I'm not sure what c is (question doesn't specify)

Not sure where to start with this one really :S a nudge in the right direction would be appreciated.

Thanks

Chris

**Opalg** · October 6th 2011, 12:30 AM

First, check what you have done so far. (I think that the eigenvalues should be 3, 6, 9, not 3, –6, 9.)

Next, look through those old notes to see that the normalised eigenvectors form the columns of an orthogonal matrix P with the property the $A = PDP^{-1},$ where D is the diagonal matrix whose diagonal entries are the eigenvalues. Then $A^n = PD^nP^{-1}.$

So to find $A^n$ it is sufficient to find $D^n$ . But if $D = \begin{bmatrix}3&0&0\\0&6&0\\0&0&9\end{bmatrix}$ then $D^n = \begin{bmatrix}3^n&0&0\\0&6^n&0\\0&0&9^n \end{bmatrix}$ . So the question is asking you to look at the limit as n goes to infinity of $\tfrac1{c^n}D^n = \begin{bmatrix}\tfrac{3^n}{c^n}&0&0\\0&\tfrac{6^n} {c^n}&0\\0&0&\tfrac{9^n}{c^n}\end{bmatrix},$ for all positive values of c.

If c<9 then the limit will not exist, because at least one of the fractions in that matrix will go to infinity. If c>9 then the limit is the zero matrix. Finally, the only interesting case occurs when c=9, in which case $\lim_{n\to\infty}D^n = \begin{bmatrix}0&0&0\\0&0&0\\0&0&1\end{bmatrix},$ and $\lim_{n\to\infty}A^n = P\begin{bmatrix}0&0&0\\0&0&0\\0&0&1\end{bmatrix}P^ {-1}.$

**GreenyMcDuff** · October 6th 2011, 11:05 PM

Originally Posted by Opalg

First, check what you have done so far. (I think that the eigenvalues should be 3, 6, 9, not 3, –6, 9.)

Next, look through those old notes to see that the normalised eigenvectors form the columns of an orthogonal matrix P with the property the $A = PDP^{-1},$ where D is the diagonal matrix whose diagonal entries are the eigenvalues. Then $A^n = PD^nP^{-1}.$

So to find $A^n$ it is sufficient to find $D^n$ . But if $D = \begin{bmatrix}3&0&0\\0&6&0\\0&0&9\end{bmatrix}$ then $D^n = \begin{bmatrix}3^n&0&0\\0&6^n&0\\0&0&9^n \end{bmatrix}$ . So the question is asking you to look at the limit as n goes to infinity of $\tfrac1{c^n}D^n = \begin{bmatrix}\tfrac{3^n}{c^n}&0&0\\0&\tfrac{6^n} {c^n}&0\\0&0&\tfrac{9^n}{c^n}\end{bmatrix},$ for all positive values of c.

If c<9 then the limit will not exist, because at least one of the fractions in that matrix will go to infinity. If c>9 then the limit is the zero matrix. Finally, the only interesting case occurs when c=9, in which case $\lim_{n\to\infty}D^n = \begin{bmatrix}0&0&0\\0&0&0\\0&0&1\end{bmatrix},$ and $\lim_{n\to\infty}A^n = P\begin{bmatrix}0&0&0\\0&0&0\\0&0&1\end{bmatrix}P^ {-1}.$

Thanks Opalg this is a very good explanation.

You were right about the eigenvalues of course.

OK, so I have looked through my notes and am now a little confused.

So my notes tell me that for a real nxn matrix $P^T=P^{-1}$ and so it is trivial to see that from $A=PDP^{-1}$ A is similar to D.

However, I have also found a theorem that states:

for non singular P that is constructed from the corresponding eigenvectors of A the following holds:

$P^{-1}AP=\begin{bmatrix}y_{1}&0&0\\0&y_{2}&0\\0&0&y_{3 }\end{bmatrix}$

Looking at both of these now they seem to say the same thing, is that correct?

Thanks

**Opalg** · October 7th 2011, 03:52 AM

Originally Posted by GreenyMcDuff

So my notes tell me that for a real nxn matrix $P^T=P^{-1}$

No, that is only true for an orthogonal matrix P. In fact, the definition of an orthogonal matrix is one that satisfies the condition $P^{\textsc t}=P^{-1}.$ If you form a matrix P by taking its columns to be eigenvectors of a (diagonalisable) matrix, then P will be invertible. If you normalise the eigenvectors then that will make P orthogonal.

Originally Posted by GreenyMcDuff

and so it is trivial to see that from $A=PDP^{-1}$ A is similar to D.

However, I have also found a theorem that states:

for non singular P that is constructed from the corresponding eigenvectors of A the following holds:

$P^{-1}AP=\begin{bmatrix}y_{1}&0&0\\0&y_{2}&0\\0&0&y_{3 }\end{bmatrix}$

Looking at both of these now they seem to say the same thing, is that correct

That is correct. The conditions $A=PDP^{-1}$ and $D=P^{-1}AP$ are equivalent. You just have to remember which way round the $P$ and $P^{-1}$ go in the two formulas. My method is to remember the single formula $AP = PD.$ If you multiply both sides of that on the left by $P^{-1}$ then you get one formula. If do the multiplications on the right then you get the other formula.

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