Can someone explain this proof?
http://math.stackexchange.com/questions/554957/center-of-the-orthogonal-group-and-special-orthogonal-group

- thomas5267

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- schrodinger

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- Kainui

Well I guess you answered your own question, haha

- thomas5267

Why does \(A=EAE^{-1}\) imply all the diagonal entries are the same?

- thomas5267

In fact, what does it mean to right multiply a elementary matrix with two rows switched?

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## More answers

- thomas5267

\(E^T=E^{-1}\) for all row switching elementary matrix if that helps.

- Kainui

Well it looks like you're really you're just taking the identity matrix and flipping two rows or columns. So when you transpose a matrix where you've flipped two columns it's the same as if you flipped two rows now.
The difference between left or right multiplying is I believe right multiplying will flip the columns and left multiplying will flip rows.
I haven't really played with this sort of stuff in a while but intuitively I'm familiar with the concept of what you're trying to prove since you can geometrically just think of these as being rotation matrices. But that's not quite a proof I understand.

- thomas5267

\[
E=\begin{bmatrix}
0 & 0 & 1 & 0 \\
0 & 1 & 0 & 0 \\
1 & 0 & 0 & 0 \\
0 & 0 & 0 & 1
\end{bmatrix}\\
A=\begin{bmatrix}
a_{1,1} & a_{1,2} & a_{1,3} & a_{1,4} \\
a_{2,1} & a_{2,2} & a_{2,3} & a_{2,4} \\
a_{3,1} & a_{3,2} & a_{3,3} & a_{3,4} \\
a_{4,1} & a_{4,2} & a_{4,3} & a_{4,4}
\end{bmatrix}\\
A=EAE^{-1}=EAE=\begin{bmatrix}
a_{3,3} & a_{3,2} & a_{3,1} & a_{3,4} \\
a_{2,3} & a_{2,2} & a_{2,1} & a_{2,4} \\
a_{1,3} & a_{1,2} & a_{1,1} & a_{1,4} \\
a_{4,3} & a_{4,2} & a_{4,1} & a_{4,4} \\
\end{bmatrix}
\]
The diagonals are permuted.

- thomas5267

EAE is equivalent to exchanging row 1 and row 3 then exchange column 1 and column 3 in this case.

- Kainui

Right or you could say that the other way around, that it's exchanging column 1 and 3 and then exchanging row 1 and 3 since matrix multiplication is associative.

- ganeshie8

By considering \(AD=DA\), it is established first that \(A\) is a diagonal matrix :
\[A=\begin{bmatrix} a&0\\0&d\end{bmatrix}\]
Next conjugate this with any one row switching matrix to further establish that the diagonal entries must be same

- thomas5267

Okay. Let \(D_k\) be the matrix with -1 on \(\left[D\right]_{kk}\) and otherwise same as the identity matrix. Note that \(D_k=D_k^{-1}=D_k^T\). Conjugation of \(A\) by \(D_k\) multiplies the column k of \(A\) by -1 and row k by -1. It leaves the diagonal unchanged.
For example.
\[
D_2=\begin{bmatrix}
1 & 0 & 0 & 0 \\
0 & -1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1
\end{bmatrix}\\
A=D_2AD_2^{-1}=D_2AD_2=\begin{bmatrix}a_{1,1} & -a_{1,2} & a_{1,3} & a_{1,4} \\
-a_{2,1} & a_{2,2} & -a_{2,3} & -a_{2,4} \\
a_{3,1} & -a_{3,2} & a_{3,3} & a_{3,4} \\
a_{4,1} & -a_{4,2} & a_{4,3} & a_{4,4}
\end{bmatrix}
\]
By comparing the entries, we can see that \(a_{2,1}=-a_{2,1}=0\) and so on. By considering all the possible \(D_k\), we can see that all entries outside the diagonal are zero.

- thomas5267

I think I understand why he only considers the row switching elementary matrices and matrices with -1 on diagonal now. Any invertible matrix are generated by a chain of elementary matrices and these two kind of elementary matrices are the only two satisfying det(E)=1 or -1 and \(E^T=E^{-1}\).

- thomas5267

Those are the requirements for matrices in the orthogonal group.

- ganeshie8

Yes, since the inverse of a row switching matrix is itself, we may try this
If \(EA\) switches the rows \(i,j\), then it changes four elements in \(A\) :
\(A_{ij} = A_{jj}\tag{1}\)
\(A_{ji} = A_{ii}\tag{2}\)
\(A_{ii} = A_{jj}=0\)
Also,\(AE\) switches the columns \(i,j\), so it changes four elements in \(A\) :
\(A_{ij} = A_{ii}\tag{1'}\)
\(A_{ji} = A_{jj}\tag{2'}\)
\(A_{ii} = A_{jj}=0\)
from \((1)\) and \((1')\) it follows that \(A_{jj} = A_{ii}\) because we want \(AE=EA\) for the center

- thomas5267

If such is the case, wouldn't we end up with a zero matrix? That seems wrong as the center of O(n) certainly contains the identity matrix.

- ganeshie8

How would you end up with 0 ?

- ganeshie8

Notice that \(i\) and \(j\) are specific rows/columns that are affected by \(E\)

- thomas5267

But if \(A_{ii}=A_{jj}=0\) we will certainly have a matrix with 0 on the diagonals right?

- ganeshie8

i, j are specific two rows that are affected by doing EA.

- ganeshie8

consider a 2x2 example, we must have \(AE-EA = 0\) for \(A\) to be an element in center :
http://www.wolframalpha.com/input/?i=%7B%7Ba%2C0%7D%2C%7B0%2Cd%7D%7D*%7B%7B0%2C1%7D%2C%7B1%2C0%7D%7D-%7B%7B0%2C1%7D%2C%7B1%2C0%7D%7D*%7B%7Ba%2C0%7D%2C%7B0%2Cd%7D%7D
as expected, we must have \(a=d\)

- thomas5267

I should interpret \(A_{ii}=A_{jj}=0\) as \((EA)_{ii}=(EA)_{jj}=0\) right?

- ganeshie8

Nope, as I said, \(i,j\) are the specific two rows in \(A\) that are affected by doing \(EA\)
\(EA\) swaps two particular rows in \(A\), for example, \(i=2, j=3\)

- thomas5267

What exactly do you mean by \(A_{ii}=A_{jj}=0\)? Clearly you don't mean by \(A_{22}=A_{33}=0\) as that would mean the diagonals are zero?

- ganeshie8

I mean exactly that, by doing \(EA\), the diagonal elements in rows \(2,3\) become \(0\)

- ganeshie8

ofcourse \(i\ne j\)

- ganeshie8

http://www.wolframalpha.com/input/?i=%7B%7B1%2C0%2C0%7D%2C%7B0%2C0%2C1%7D%2C%7B0%2C1%2C0%7D%7D*%7B%7Ba%2C0%2C0%7D%2C%7B0%2Cb%2C0%7D%2C%7B0%2C0%2Cc%7D%7D

- thomas5267

So by referring to \(A_{ii}=A_{jj}=0\) you are referring to the elements of EA instead of A itself? That is absolutely confusing. So the equality are more like assignment operators in computer programming?

- ganeshie8

Yes sorry about that, i can't think of a better notation at the moment because I also don't know much about groups

- thomas5267

Okay I get it now. As \(A_{ii}=A_{jj}\) for all i and j and 0 otherwise, it must be a multiple of identity matrix. Moreover, as A is required to be in the orthogonal group, for which all element have determinant 1 or -1, so \(A_{ii}=A_{jj}=\pm 1\).

- thomas5267

So the only elements in center of O(n) is \(\pm I\).

- ganeshie8

looks good to me!

- thomas5267

I was just trying to find the biggest group in a subset of orthogonal matrices. The subset consists of orthogonal matrices that are symmetric. As the product of two symmetric matrices A and B is symmetric if they commute, I have to find the center of the orthogonal group. Turns out that the group is almost trivial.
\[
AB=(AB)^T=B^TA^T=BA \quad \text{as A and B are both symmetric.}
\]

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