What is the center of $GL_n(\mathbb{R})$?

[evinda]

Hello! (Wave)

I want to find the center of $GL_n(\mathbb{R})$.

$Z(GL_n(\mathbb{R}))=\{ c \in GL_n(\mathbb{R}): cg=gc \forall g \in GL_n(\mathbb{R}) \}$

I have thought the following.

Let $c \in GL_n(\mathbb{R})$. Then there is a $b \in GL_n(\mathbb{R})$ such that $cb=I$ where $I$ is the $n \times n$ identity matrix.
Then it also holds that $bc=I$ and so $cb=bc$.

But in this case the equality holds only for one $g$.

It would hold for all $g$ if we would consider as $c$ the identity matrix.

Is there also an other matrix that commutes with every other matrix? (Thinking)

 

[Deveno]

Yes, there is, in fact infinitely many.

For example, $kI$ commutes with any matrix, for any $k \in \Bbb R$.

I suggest doing the following-

1st, show that any matrix that commutes with all of $GL_n(\Bbb R)$ must be diagonal (that is, for any matrix that has a non-zero off-diagonal element, find a matrix it DOESN"T commute with-I would try one of the elementary matrices $E_{i,j}$ for a "wise" choice of $i$ and $j$).

2nd, show that if a diagonal matrix has two unequal diagonal elements, there is ALSO a matrix it doesn't commute with.

 

[evinda]

1st, show that any matrix that commutes with all of $GL_n(\Bbb R)$ must be diagonal (that is, for any matrix that has a non-zero off-diagonal element, find a matrix it DOESN"T commute with-I would try one of the elementary matrices $E_{i,j}$ for a "wise" choice of $i$ and $j$).

I haven't heard of elementary matrices yet, but I found the following in wikipedia:View attachment 5280

Can we use this one?

 

[Deveno]

The elementary matrix $E_{i,j}$ is the $n \times n$ matrix that has a $1$ as the $ij$-th entry and $0$'s everywhere else. For example in $GL_3(\Bbb R)$ the matrix $E_{1,3}$ is:

$E_{1,3} = \begin{bmatrix}0&0&1\\0&0&0\\0&0&0\end{bmatrix}$

These matrices behave differently (typically) when used on the left versus the right.

 

[evinda]

So you mean that we should pick a matrix that has a non-zero off-diagonal element and multiply by $E_{ij}$ and then multiply $E_{ij}$ by the above matrix?

Or do we choose $E_{ij}$ to be the matrix that has a non-zero off-diagonal element?

 

[Deveno]

Let's say a matrix $A$ has an off-diagonal element $a_{ij} \neq 0$.

Since $a_{ij}$ is off-diagonal we know $i \neq j$. This will be useful.

Now let's consider what (in general) right-multiplication by $E_{i,j}$ does.

Clearly, $E_{i,j}$ sends the $k$-th column of $A = (a_{ij})$ to a 0-column if $k \neq i$.

And $E_{i,j}$ sends the $i$-th column to the $j$-th column in the product $AE_{i,j}$.

so let's use $E_{j,i}$ where $j$ is the column that our off-diagonal element $a_{ij}$ occurs in, and $i$ is the row it occurs in.

If $(b_{ij}) = B = AE_{j,i}$, then we know that $b_{ii} = a_{ij} \neq 0$. Remember this.

Now let's look at what happens when we multiply $E_{j,i}A$.

What $E_{j,i}$ does, when multiplied on the left side of $A$ is zero out every row of $A$ except the $i$-th, which it puts in the $j$-th row.

In particular, the $i$-th row of $(c_{ij}) = C = E_{j,i}A$ is all zeros, since $i \neq j$, that is $c_{ii} = 0$, so $C \neq B$.

For example, suppose we have:

$A = \begin{bmatrix}\ast&\ast&\ast\\ \ast&\ast&\ast\\ a&\ast&\ast\end{bmatrix}$.

Here, our non-zero off-diagonal is $a_{31} = a$ (so $i = 3$ and $j = 1$).

If we multiply $A$ on the right by $E_{1,3}$, we get:

$\begin{bmatrix}\ast&\ast&\ast\\ \ast&\ast&\ast\\ a&\ast&\ast\end{bmatrix}\begin{bmatrix}0&0&1\\0&0&0\\0&0&0\end{bmatrix} = \begin{bmatrix}0&0&\ast\\0&0&\ast\\0&0&a\end{bmatrix}$

with $a$ in the $3,3$ position.

If we multiply $A$ on the left with $E_{1,3}$ we get:

$\begin{bmatrix}0&0&1\\0&0&0\\0&0&0\end{bmatrix}\begin{bmatrix}\ast&\ast&\ast\\ \ast&\ast&\ast\\ a&\ast&\ast\end{bmatrix} = \begin{bmatrix}a&\ast&\ast\\0&0&0\\0&0&0\end{bmatrix}$

and here you can see that the $3,3$ position is $0$, and it does not matter what the asterisks are, the two matrices cannot be the same.

 

[evinda]

Now let's consider what (in general) right-multiplication by $E_{i,j}$ does.

Clearly, $E_{i,j}$ sends the $k$-th column of $A = (a_{ij})$ to a 0-column if $k \neq i$.

And $E_{i,j}$ sends the $i$-th column to the $j$-th column in the product $AE_{i,j}$.

We have that $(AE_{ij})_{nm}=\sum_{k=1}^M A_{nk}(E_{ij})_{km}$, right?

Do we get from that the second statement that you said, "$E_{i,j}$ sends the $i$-th column to the $j$-th column in the product $AE_{i,j}$"?

so let's use $E_{j,i}$ where $j$ is the column that our off-diagonal element $a_{ij}$ occurs in, and $i$ is the row it occurs in.

If $(b_{ij}) = B = AE_{j,i}$, then we know that $b_{ii} = a_{ij} \neq 0$.

How did you get that?

What $E_{j,i}$ does, when multiplied on the left side of $A$ is zero out every row of $A$ except the $i$-th, which it puts in the $j$-th row.

What do you mean by "which it puts in the $j$-th row"?

In particular, the $i$-th row of $(c_{ij}) = C = E_{j,i}A$ is all zeros, since $i \neq j$, that is $c_{ii} = 0$, so $C \neq B$.

Why does it hold that $c_{ii} = 0$? (Thinking)

 

[Deveno]

Let's "fill in the blanks" in my last example, and I'll use a different elementary matrix.

Suppose $A = \begin{bmatrix}a&b&c\\d&e&f\\g&h&k\end{bmatrix}$.

Let us suppose further that the $2,3$ entry (the third entry in the second row, in this case, $f$) is non-zero.

Now $E_{3,2}$ (note how we reversed the indices) is the matrix:

$E_{3,2} = \begin{bmatrix}0&0&0\\0&0&0\\0&1&0\end{bmatrix}$

Watch what happens to $A$ when we multiply by $E_{3,2}$ on the right:

$AE_{3,2} = \begin{bmatrix}a&b&c\\d&e&f\\g&h&k\end{bmatrix}\begin{bmatrix}0&0&0\\0&0&0\\0&1&0\end{bmatrix}$

$=\begin{bmatrix}0&c&0\\0&f&0\\0&k&0\end{bmatrix}$

As I indicated, we killed all the columns except the 3rd, which wound up in the 2nd column.

In particular, our $f \neq 0$ moved from the $2,3$ position, to the $2,2$ position (on the main diagonal).

Now let's multiply $A$ on the left by $E_{3,2}$:

$E_{3,2}A = \begin{bmatrix}0&0&0\\0&0&0\\0&1&0\end{bmatrix}\begin{bmatrix}a&b&c\\d&e&f\\g&h&k\end{bmatrix}$

$=\begin{bmatrix}0&0&0\\0&0&0\\d&e&f\end{bmatrix}$.

Here, $E_{3,2}$ moved the second row (where the $f$ we're interested in) to the third row, the other two rows are $0$.

Do you see how that works? I did the 3x3 case so you can work it out yourself, but the same logic applies to any $n\times n$ matrix. $E_{i,j}$ takes the $i$-th column and puts it in the $j$-th column (and all the other columns become 0), when we use it on the right.

$E_{i,j}$ takes the $j$-th row, and moves it to the $i$-th row, making all other rows 0.

In my example here, we saw the $2,2$ entry of $AE_{3,2} = f \neq 0$.

However, the $2,2$ entry of $E_{3,2}A = 0$.

In the general case, if $a_{ij} \neq 0$, for our matrix $A = (a_{ij})$, if we hit it on the RIGHT by $E_{j,i}$ (again, note the reversal of indices), we will take the $j$-th column (which is where our non-zero entry $a_{ij}$ lives), and move it to the $i$-th column, and now the non-zero $a_{ij}$ is the $i$-the entry down in the $i$-th column, that is, it has moved to the diagonal, in the $i,i$-th position.

When we hit $A$ from the LEFT with $E_{j,i}$, it's again going to preserve the $i$-th row, but since $i \neq j$, it's not going to KEEP it in the $i$-th row, but move it to the $j$-th row.

And the $i$-th row(which contains the $i,i$-th diagonal position), along with any other row besides the $j$-th is going to be 0. So in our products, we have that the $i,i$-th entry of $AE_{j,i} = a_{ij} \neq 0$, but the $i,i$-th entry of $E_{j,i}A$ is 0.

In particular, $A$ and $E_{j,i}$ cannot commute.

This shows that if a matrix $A$ HAS an off-diagonal non-zero entry, we can find some matrix $A$ doesn't commute with, and thus $A$ cannot lie in the center of the general linear group.

So, we are left with diagonal matrices.