Proving End($K^2$)=K for A-Module $K^2$

[Fermat1]

Let A be the K-algebra of 2 x 2 upper triangular matrices. Thinking of $K^2$ as an A module with the action of matrix multiplication, prove that End($K^2$)=K.

You may use the fact that the centraliser of A consists of all scalar multiples of identity matrix

 

[Deveno]

Could you please specify if you mean:

$\text{End}_K(K^2)$
$\text{End}_{\Bbb Z}(K^2)$ or:
$\text{End}_A(K^2)$

as all of these are different structures...

I suspect you mean the latter, that is you want $A$-linear maps.

 

[Fermat1]

Could you please specify if you mean:

$\text{End}_K(K^2)$
$\text{End}_{\Bbb Z}(K^2)$ or:
$\text{End}_A(K^2)$

as all of these are different structures...

I suspect you mean the latter, that is you want $A$-linear maps.

You suspect correctly.

 

[Deveno]

Here is my idea (I haven't worked through all the details):

1) Show that any $A$-linear map is a $K$-linear map by using the fact that if:

$(x,y) \in K^2, r\in K, L \in \text{End}_A(K^2)$

$L(r(x,y)) = L(\alpha.(x,y)) = \alpha.L(x,y) = r(L(x,y))$

where $\alpha = \begin{bmatrix}r&0\\0&r \end{bmatrix} \in A$.

This means we can represent an $A$-linear map as multiplication by a 2x2 matrix over $K$.

2) Show such a matrix must lie in $A$.

(hint: it suffices to show the map $(x,y) \mapsto (0,kx)$ (for $k \in K \setminus \{0\}$) is not $A$-linear).

3) Use what you are given in the hint concerning $Z(A)$:

If $L(x,y) = \beta.(x,y)$ for some $\beta \in A$ then $A$-linearity means that:

$\alpha.L(x,y) = L(\alpha.(x,y))$ for ANY $\alpha \in A$

$\alpha.(\beta.(x,y)) = \beta.(\alpha.(x,y))$

$(\alpha\beta).(x,y) = (\beta\alpha).(x,y)$

so that $\alpha\beta = \beta\alpha$, that is: $\beta \in Z(A)$.

Note that you wind up with something *isomorphic* to $K$, not equal to it.

 

[Fermat1]

Here is my idea (I haven't worked through all the details):

1) Show that any $A$-linear map is a $K$-linear map by using the fact that if:

$(x,y) \in K^2, r\in K, L \in \text{End}_A(K^2)$

$L(r(x,y)) = L(\alpha.(x,y)) = \alpha.L(x,y) = r(L(x,y))$

where $\alpha = \begin{bmatrix}r&0\\0&r \end{bmatrix} \in A$.

This means we can represent an $A$-linear map as multiplication by a 2x2 matrix over $K$.

2) Show such a matrix must lie in $A$.

(hint: it suffices to show the map $(x,y) \mapsto (0,kx)$ (for $k \in K \setminus \{0\}$) is not $A$-linear).

3) Use what you are given in the hint concerning $Z(A)$:

If $L(x,y) = \beta.(x,y)$ for some $\beta \in A$ then $A$-linearity means that:

$\alpha.L(x,y) = L(\alpha.(x,y))$ for ANY $\alpha \in A$

$\alpha.(\beta.(x,y)) = \beta.(\alpha.(x,y))$

$(\alpha\beta).(x,y) = (\beta\alpha).(x,y)$

so that $\alpha\beta = \beta\alpha$, that is: $\beta \in Z(A)$.

Note that you wind up with something *isomorphic* to $K$, not equal to it.

Hi Deveno, I have been following out a thought of my own (for once) and tried to show that the map $f:K->End_{A}(M)$ given by $f(t)=w_{t}$ where $w_{t}(x,y)=t(x,y)$ is an isomorphism
Its easy to show that $w_{t}$ is in $End_{A}(M)$. So we need to show $f$ is an isomorphism. Well, it's easy to show $f$ is an algebra homomorphism. The only bit I'm struggling with is showing $f$ is surjective.

 

[Deveno]

Establishing surjectivity is essentially the "whole problem", because: we don't know much specifically about $\text{End}_A(K^2)$.

That is, it is hard to say what a "typical" $A$-linear endomorphism of $K^2$ looks like.

The fact that an isomorphic copy of $K$ lies within $Z(A)$ is trivial, because $A$ is a $K$-algebra.

 

[Fermat1]

Establishing surjectivity is essentially the "whole problem", because: we don't know much specifically about $\text{End}_A(K^2)$.

That is, it is hard to say what a "typical" $A$-linear endomorphism of $K^2$ looks like.

The fact that an isomorphic copy of $K$ lies within $Z(A)$ is trivial, because $A$ is a $K$-algebra.

what has the centraliser got to do with surjectivity?

 

[Deveno]

Ok, to prove your mapping is surjective, you basically have to show that EVERY endomorphism in $\text{End}_A(K^2)$ is of the form $w_t$ (using your notation).

To avoid confusion with the $K$-scalar product, I use $\alpha.(x,y)$ for the $A$-scalar product.

Note that $w_t$ is just the same as multiplying by the $A$ element:

$tI = \begin{bmatrix}t&0\\0&t \end{bmatrix}$.

That is: $w_t(x,y) = t(x,y) = (tI).(x,y)$.

To apply what we know about the centralizer (which is actually kind of a weird thing to call it, since the centralizer of $A$ in $A$ is just the center of $A: Z_A(A) = Z(A)$), we need to know that any $A$-linear map is of the form: $(x,y) \to \alpha.(x,y)$ for some $\alpha \in A$.

An alternative approach: we could use the fact that the centralizer of $A$ in the full $K$-algebra of 2x2 matrices is also $KI$. But to exploit this fact, we need to know FIRST that $A$-linear maps are $K$-linear maps.

But this is easily shown:

For $L \in \text{End}_A(K^2)$ we have:

$tL(x,y) = (tI).L(x,y) = L((tI).(x,y))$ (by $A$-linearity)

$ = L(tx,ty) = L(t(x,y))$.

Now working with $K$-linear maps is much more straight-forward, we already know there is an isomorphism:

$\text{Mat}_{2 \times 2}(K) \cong \text{End}_K(K^2)$

And $A$-linearity imposes upon us the condition that the matrix that represents an $A$-linear map must commute with any matrix representing a $K$-linear map.

If you remain unconvinced, prove to yourself by direct computation that the $K$-linear maps:

$E_{12}:\begin{bmatrix}x\\y \end{bmatrix} \mapsto \begin{bmatrix}0&1\\0&0 \end{bmatrix} \begin{bmatrix}x\\y \end{bmatrix}$

$E_{21}:\begin{bmatrix}x\\y \end{bmatrix} \mapsto \begin{bmatrix}0&0\\1&0 \end{bmatrix} \begin{bmatrix}x\\y \end{bmatrix}$

are not $A$-linear, and that the map:

$k_1E_{11} + k_2E_{22}:\begin{bmatrix}x\\y \end{bmatrix} \mapsto \begin{bmatrix}k_1&0\\0&k_2 \end{bmatrix} \begin{bmatrix}x\\y \end{bmatrix}$

is $A$-linear if and only if $k_1 = k_2$

(you don't have to check the additive property, since these are $K$-linear, you just have to check the $A$-scalar multiplication).