Order of Eigenvectors in Matrix Generation: Does it Make a Difference?

[squaremeplz]

Homework Statement

When generating a matrix from eigenvectors, does it matter in which order
the columns are placed?

 

[Klockan3]

No, as long as the column of the eigenvector is the same as the column of the corresponding eigenvalue it should all be fine. This is due to how the diagonalization is made, the first matrix from the right transforms the vector into the eigenbasis of the matrix, the second matrix is the original matrix's diagonal form while the third transforms it back to the basis you started out from.

The reason you use eigenvectors and such is because it is often much easier to work with matrices in this form and it also tells you a lot about its properties.

 

[HallsofIvy]

If you use the eigenvectors as columns (and assuming that the eigenvectors form a basis for the space) the matrix will be a diagonal matrix with the eigenvalues on the diagonal. Changing the order of the eignvectors will change the order of those eigenvalues but you still get a matrix representing that linear transformation.

 

[squaremeplz]

Sorry I am trying to diagonalize the matrix h

I used MATLAB to check my results:

h =

2 1 0
1 2 1
0 0 2

>> [e,r] = eig(h)

e =

0.7071 -0.7071 -0.7071
0.7071 0.7071 0
0 0 0.7071r =

3 0 0
0 1 0
0 0 2The eigenvalues I get are 1;2;3 and and my eigenvector matrix is a bit differnt than e when I do it out by hand:

1 -1 -1
1 1 0
0 0 1

but since c[1;1;0] where c is any scalar; I am assuming it's the same thing due to the ratios. However, you are saying that I can use the matrix r instead? I know the arithmetic, this is just a bit confusing. Thanks again.
I think r might be my end result, but the way I am trying to get it is

e^(-1) * h * e = r

is this correct?

 

[Klockan3]

Matlab normalized the eigenvectors, other than that your solution is the same.

 

[squaremeplz]

I figured as much.. but why did it make the matrix in that order? Does it have to do with the diagonal result?

 

[HallsofIvy]

Sorry I am trying to diagonalize the matrix h

I used MATLAB to check my results:

h =

2 1 0
1 2 1
0 0 2

>> [e,r] = eig(h)

e =

0.7071 -0.7071 -0.7071
0.7071 0.7071 0
0 0 0.7071

This is, exactly,
\begin{bmatrix}\frac{\sqrt{2}}{2} & -\frac{\sqrt{2}}{2}& -\frac{\sqrt{2}}{2} \\ \frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} & 0 \\ 0 & 0 & \frac{\sqrt{2}}{2}\end{bmatrix}

r =

3 0 0
0 1 0
0 0 2


The eigenvalues I get are 1;2;3 and and my eigenvector matrix is a bit differnt than e when I do it out by hand:

1 -1 -1
1 1 0
0 0 1

and if you normalize the vectors first, this is
\begin{bmatrix} \frac{\sqrt{2}}{2} & -\frac{\sqrt{2}}{2} & -\frac{\sqrt{2}}{2} \\ \frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} & 0 \\ 0 & 0 & 1\end{bmatrix}

which is exactly as before.

but since c[1;1;0] where c is any scalar; I am assuming it's the same thing due to the ratios. However, you are saying that I can use the matrix r instead? I know the arithmetic, this is just a bit confusing. Thanks again.
I think r might be my end result, but the way I am trying to get it is

e^(-1) * h * e = r

is this correct?

Yes, did you try it? If e is
\begin{bmatrix}1 & -1 & -1 \\ 1 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}
what is e^{-1}? What is e^{-1}he?

 

[squaremeplz]

Yep, I did try it all out and i got the diagonal

it took some time to get the inverse correct by hand but looking back at it
it's pretty straight forward.

Thanks for your help :)

 

[squaremeplz]

I think the placement of the eigenvectors is also not crucial due to the nature of the eigenvectors.

e^-1 and e act to still give the diagonal I believe. but i will try that some other time

 

[HallsofIvy]

Each eigenvector is associated, of course, with a specific eigenvalue. Changing the order of the eigenvectors as rows will, as I said before, change the order of the eigenvalues on the diagonal. The matrix you give has columns <1, 1, 0>, <-1, 1, 0>, and <-1, 0, 1>, from left to right, which are eigenvectors corresponding to eigenvalues, 3, 1, and 2 in that order. Using that gives a diagonal matrix with3 in the top row, 1 in the second row, and 2 in the third row:
\begin{bmatrix}3 &amp; 0 &amp; 0 \\ 0 &amp; 1 &amp; 0 \\ 0 &amp; 0 &amp; 2\end{bmatrix}
If you swapped those columns around so that <-1, 1, 0>was the first column, <-1, 0, 1> the second column, and <1, 1, 0> the third column, then you would get the diagonal matrix
\begin{bmatrix}1 &amp; 0 &amp; 0 \\ 0 &amp; 2 &amp; 0 \\ 0 &amp; 0 &amp; 3\end{bmatrix}

 

[squaremeplz]

That is awesome.

Great explanation. I figure this will be the only useful linear algebra technique so this is very helpful.

Go your brain!