Eigenvector, Eigenvalue and diagonalising help

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March 18th 2009, 06:33 AM
Redeemer_Pie

Eigenvector, Eigenvalue and diagonalising help

Hi guys, i've got some issues at the moment regarding eigenvector and eigenvalue Your help will be greatly appreciated. Thanks =)

First the matrix c is given as:

3 1 0
2 2 0
1 6 5 (Note: i have shown that the eigenvalues are 1,4 and 5).

It then asked me to write out the polynomial of C, namely write out: determinant(C - lamda multiply the identity matrix) as a polynomial of lamda. Any ideas whats this about?

Lastly, there are 4 column vectors:

a:

3
2
-4

b:

-6
-4
8

c:
0
-5
0

d:
0
0
-5

The sets {ac}, {ad}, {bc},d}, {bd}, and {cd} are all linearly independent.

The question asked me to find a basis of the vector space R^3 which lies in {a b c d}

any ideas guys?

thanks. Your help is much appreciated.
March 18th 2009, 09:07 AM
Showcase_22

It wants you to first find the characteristic polynomial. This is found by doing $|C-xI_3|=0$ .

ie. $\begin{vmatrix} {3}&{1}&{0}\\ {2}&{2}&{0}\\ {1}&{6}&{5} \end{vmatrix}-x\begin{vmatrix} {1}&{0}&{0}\\ {0}&{1}&{0}\\ {0}&{0}&{1} \end{vmatrix}=\begin{vmatrix} {3-x}&{1}&{0}\\ {2}&{2-x}&{0}\\ {1}&{6}&{5-x} \end{vmatrix}$

Normally this would be the way to do it however, you've already found the eigenvalues. The eigenvalues are the roots of the characteristic polynomial. Therefore the characteristic polynomial is $p_C(\lambda)=(\lambda-1)(\lambda-4)(\lambda-5)=0$ .

In the case of the four column vectors it's important to notice that since you're dealing with $\mathbb{R}^3$ and you have 4 vectors which span. As a result one of the vectors must be a linear combination of the others.

Sifting the vectors gives a,c and d. b is removed because it is $-2\begin{pmatrix} {3}\\ {2}\\ {-4} \end{pmatrix}$ or -2a (ie. a linear combination of a vector you already have as part of your basis).
March 19th 2009, 03:27 AM
Redeemer_Pie

Diagonalisation help!

Hi guys,

i though i got it correct. Turns out i've done it wrong! If you could show me what to do that'll be great! :) Thanks guys.

Firstly let matrix A be:

3 1 0
2 2 0
1 6 5

I have found the eigenvalues of 1,4 and 5 are the eigenvalues of A and the eigenvectors:

x:
4
-8
11

y:
1
1
-7

and w:
0
0
1

are corresponding to the eigenvalues 1,4,5 respectively.

It then asked me to find a matrix lets say, T such that T^(-1)AT is diagonal. The T^(-1) is the inverse of the matrix T. Please note that the equation T^(-1)AT stands for multiplication ie. T^(-1) multiply A multiply T.

Does anyone know how to do this? Im really stuck on this :( Again, your contribution is much appreciated!

Regards =)
March 19th 2009, 03:36 AM
Moo

Hello,

Is there something in your notes that says that T is the matrix formed by the eigenvectors ?

Let D be the diagonal matrix associated to A.

$D=\begin{pmatrix} 1 &0&0 \\ 0&4&0 \\ 0&0&5 \end{pmatrix}$

Then there is some matrix P such that $A=P^{-1}DP$ , where P is formed by the eigenvectors (in the same order as the eigenvalues) in columns :
$P=\begin{pmatrix} 4&1&0 \\ -8&1&0 \\ 11&-7&1 \end{pmatrix}$

Thus $D=PAP^{-1}$ , which is a diagonal matrix.

Now I bet you can find $P^{-1}$ :p

(actually, $T=P^{-1}$ in your problem)
March 19th 2009, 03:46 AM
Redeemer_Pie

Thank you sooooo much!!!! :)
March 19th 2009, 08:17 AM
Redeemer_Pie

Quote:

Originally Posted by Showcase_22

Sifting the vectors gives a,c and d. b is removed because it is $-2\begin{pmatrix} {3}\\ {2}\\ {-4} \end{pmatrix}$ or -2a (ie. a linear combination of a vector you already have as part of your basis).

So what would be the basis of the vector space of R^3 which lies in a,c and d given that d is discarded? Is there an equation i have to put in? I looked through the meaning in my textbooks but its waaay advanced and dont understand the jargon :(
March 19th 2009, 08:54 AM
Showcase_22

Quote:

Originally Posted by Redeemer_Pie

So what would be the basis of the vector space of R^3 which lies in a,c and d given that d is discarded?

Do you mean "b is discarded"?

The vector space has a basis of a,c and d.
ie. say the basis of $\mathbb{R}$ is denoted as $B$ . Then $B=\{a,c,d \}$ where a,c and d were defined in your first post.

I don't think you have to put in an equation. You've found three vectors that form a basis of $\mathbb{R}^3$ , you can't really put an equation to that.

Quote:

I looked through the meaning in my textbooks but its waaay advanced and dont understand the jargon :(

What meaning did you look up?
March 19th 2009, 01:42 PM
Redeemer_Pie

Quote:

Originally Posted by Showcase_22

Do you mean "b is discarded"?

The vector space has a basis of a,c and d.
ie. say the basis of $\mathbb{R}$ is denoted as $B$ . Then $B=\{a,c,d \}$ where a,c and d were defined in your first post.

I don't think you have to put in an equation. You've found three vectors that form a basis of $\mathbb{R}^3$ , you can't really put an equation to that.

What meaning did you look up?

I looked up bases and dimensions in my textbook but never really understood it.

At first i thought we have to put it into some sort of equation like in my textbook like: c1v1 + c2v2 +...cnvn = (0,0,0) where v are the vectors and c a scalar or something like that. I dont know if you encountered that equation before.

Again. Thanks for your help :)
March 20th 2009, 01:28 AM
Showcase_22

Suppose you have a set of vectors which span a vector space.

For example, in $\mathbb{R}^3$ $(1,0,0),(0,1,0)$ and $(0,0,1)$ span the vector space. This means that using any combination of these you can get any vector in the vector space.
The equation you wrote down is a means of proving that these vectors are linearly independent.

$\alpha_1(1,0,0)+\alpha_2(0,1,0)+\alpha_3(0,0,1)=(0 ,0,0)$ .
Just reading across the rows show $\alpha_1=\alpha_2=\alpha_3=0$ which shows that $(1,0,0),(0,1,0)$ and $(0,0,1)$ are linearly independent. If a set of vectors span a vector space and are linearly independent, then they form a basis.

The number of vectors in a basis is the dimension of the vector space. Since we have three vectors in this case, $\mathbb{R}^3$ has a dimension of 3.

The same ideas can be applied to other vector spaces in any dimension. I just used the standard basis in $\mathbb{R}^3$ as an example.
March 20th 2009, 08:59 AM
Redeemer_Pie

ok :) Great :)

Thanks again for the help! Much appreciated =D