MHB Linear Algebras or k-algebras - Cohn p. 53-54 - SIMPLE CLARIFICATION

[Math Amateur]

I am reading "Introduction to Ring Theory" by P. M. Cohn (Springer Undergraduate Mathematics Series)

In Chapter 2: Linear Algebras and Artinian Rings we read the following on pages 53-54:

View attachment 3132
View attachment 3133In the above text we read:

" … … The multiplication in A is completely determined by the products of the basis elements. Thus we have the equations

$$u_i u_j = \sum_k \gamma_{ijk} u_k$$ … … … (2.1)

where the elements $$\gamma_{ijk}$$ are called the multiplications constants of the algebra … … "

Can someone please explain how equation (2.1) follows?

Peter

 

[Euge]

Hi Peter,

The the products $u_i u_j$ are elements of $A$, so they have unique expressions as linear combinations of the basis elements $u_k$. This leads to (2.1).

 

[Math Amateur]

Hi Peter,

The the products $u_i u_j$ are elements of $A$, so they have unique expressions as linear combinations of the basis elements $u_k$. This leads to (2.1).

Yes, understand that Euge, but still bit puzzled … how shall I explain ...

well … I keep thinking … … if the algebra has n dimensions … then ...

$$u_i = 0.u_1 + 0.u_2 + \ … \ … \ + 1.u_i + \ … \ … \ + 0.u_n$$

and

$$u_j = 0.u_1 + 0.u_2 + \ … \ … \ + 1.u_j + \ … \ … \ + 0.u_n$$

and so it seems (I think)

$$u_i.u_j = 0$$

? can you clarify ?

Peter

EDIT Maybe my 'multiplication' is wrong?

 

[Euge]

Yes, understand that Euge, but still bit puzzled … how shall I explain ...

well … I keep thinking … … if the algebra has n dimensions … then ...

$$u_i = 0.u_1 + 0.u_2 + \ … \ … \ + 1.u_i + \ … \ … \ + 0.u_n$$

and

$$u_j = 0.u_1 + 0.u_2 + \ … \ … \ + 1.u_j + \ … \ … \ + 0.u_n$$

and so it seems (I think)

$$u_i.u_j = 0$$

? can you clarify ?

Peter
EDIT Maybe my 'multiplication' is wrong?

Peter, how do know that $u_i u_j = 0$ for all $i$ and $j$? The identities you have do not imply this since multiplication in $A$ has not been specified. In fact, what you're implying is that multiplication in $A$ is trivial. I'll give two examples with nontrivial multiplication.

$\textbf{Example 1}$. The polynomial ring $\Bbb R[x]$ is a linear algebra over $\Bbb R$ with basis $\{1, x, x^2,\ldots\}$ and multiplication defined by the usual ring multiplication of polynomials. For each $k \ge 0$, let $u_k = x^k$. Then $u_i u_j = u_{i+j}$ for all $i$ and $j$. So the multiplication constants are given by $\gamma_{ijk} = \delta_k^{i+j}$, where $\delta$ denotes the Kronecker delta.

$\textbf{Example 2}$. Consider $\Bbb R^3$ with multiplication given by the cross product. It is a non-associative linear algebra over $\Bbb R$ with standard basis $\{e_1, e_2, e_3\}$. Since $e_1e_2 = e_3$, $e_2e_3 = e_1$, $e_3e_1 = e_2$, and $e_ie_j = -e_je_i$ for all $i$ and $j$, the multiplication constants are given by

$\gamma_{111} = 0$, $\gamma_{112} = 0$, $\gamma_{113} = 0$,

$\gamma_{121} = 0$, $\gamma_{122} = 0$, $\gamma_{123} = 1$,

$\gamma_{131} = 0$, $\gamma_{132} = -1$, $\gamma_{133} = 0$,

$\gamma_{211} = 0$, $\gamma_{212} = 0$, $\gamma_{213} = -1$

$\gamma_{221} = 0$, $\gamma_{222} = 0$, $\gamma_{223} = 0$,

$\gamma_{231} = 1$, $\gamma_{232} = 0$, $\gamma_{233} = 0$,

$\gamma_{311} = 0$, $\gamma_{312} = 1$, $\gamma_{313} = 0$,

$\gamma_{321} = -1$, $\gamma_{322} = 0$, $\gamma_{323} = 0$,

$\gamma_{331} = 0$, $\gamma_{332} = 0$, $\gamma_{333} = 0$.

Observe that $\gamma_{ijk}$ equals 1 when $(i,j,k)$ is a cyclic permutation $(1 2 3)$ and -1 when $(i,j,k)$ is a cyclic permutation of $(132)$. Also, $\gamma_{ijk} = 0$ whenever $i, j$, and $k$ are not all distinct. So in fact, $\gamma_{ijk} = \epsilon_{ijk}$, the Levi-Civita symbol.

 

[Deveno]

It's somewhat of a trivial example, but it may be helpful to consider the $\Bbb Q$-algebra $\text{Mat}_n(\Bbb Q)$, which has a basis consisting of the elementary matrices $E_{ij} = (a_{km})$ where:

$a_{km} = 1, k = i,m = j$
$a_{km} = 0,$ otherwise.

So let's look at what happens when we multiply $E_{ij}E_{i'j'}$:

$E_{ij}E_{i'j'} = E_{ij'}, j = i'$
$E_{ij}E_{i'j'} = 0$, otherwise.

To get a handle on what this really means, let's specify $n = 2$, so our 2x2 matrices can be seen as being "$\Bbb Q^4$ with a multiplication". So our basis is, explicitly:

$u_1 = \begin{bmatrix}1&0\\0&0\end{bmatrix};\ u_2 = \begin{bmatrix}0&1\\0&0\end{bmatrix};\ u_3 = \begin{bmatrix}0&0\\1&0\end{bmatrix};\ u_4 = \begin{bmatrix}0&0\\0&1\end{bmatrix}$

We have:

$u_1u_1 = u_1$, so $\gamma_{111} = 1, \gamma_{11k} = 0, k = 2,3,4$.
$u_1u_2 = u_2$, so $\gamma_{121} = 0, \gamma_{122} = 1, \gamma_{12k} = 0, k = 3,4$
$u_1u_3 = 0$ (all the $\gamma$'s are 0)

and so on (there's 13 more products to compute, and thus 52 more multiplication constants).

Another important $k$-algebra is $k[x]$ (polynomials over $k$) which has as one possible basis:

$u_i = x^i$.

Since $u_iu_j = (x^i)(x^j) = x^{i+j} = u_{i+j}$, we see that:

$\gamma_{ijk} = 1$ when $k = i+j$
$\gamma_{ijk} = 0$, otherwise.

Another often-used example: we have $\Bbb R^2$ as an $\Bbb R$-algebra given the basis:

$\{e_1,e_2\} = \{(1,0),(0,1)\}$ with multiplication constants:

$\gamma_{111} = 1$
$\gamma_{112} = 0$
$\gamma_{121} = 0$
$\gamma_{122} = 1$
$\gamma_{211} = 0$
$\gamma_{212} = 1$
$\gamma_{221} = -1$
$\gamma_{222} = 0$.

Since $\gamma_{ijk} = \gamma_{jik}$, this forms a commutative $\Bbb R$-algebra, more commonly known as $\Bbb C$ (this is a DIVISION ALGEBRA since $U(\Bbb C) = \Bbb C - \{0\}$, that is: all non-zero elements are invertible).

Interestingly enough, we actually have this as a sub-algebra of $\text{Mat}_2(\Bbb R)$, with basis:

$\{v_1,v_2\} = \{E_{11} + E_{22},E_{21}-E_{12}\}$.

This is because the basis elements when multiplied give a linear combination of these two basis elements:

$v_1v_1 = v_1$
$v_2v_1 = v_1v_2 = v_2$
$v_2v_2 = -v_1$, as can be readily verified.

***************

Given a basis for $\text{Mat}_n(k)$, we can use this to define an isomorphism with $\text{Hom}_{k}(k^n,k^n)$, so the study of the (particular) linear algebra $\text{Mat}_n(k)$ is typically the focus of a course called "linear algebra".

It is important not to confuse the $k$-action (scalar multiplication) of a $k$-algebra with the ring multiplication, but typically, if our algebra $A$ is unital, we can often consider it as an extension ring of $k$ via the map:

$\alpha \mapsto \alpha\cdot 1_A$

For example, with $A = \text{Mat}_n(k)$ we have the embedding:

$\alpha \mapsto \alpha I_n$

and with $A = k[x]$ we have the natural embedding of $k$ as constant polynomials

(note that $k \cong k[x]/(x)$ which essentially amounts to "evaluating $p(x)$ at 0", for any polynomial $p$).

 

[Math Amateur]

It's somewhat of a trivial example, but it may be helpful to consider the $\Bbb Q$-algebra $\text{Mat}_n(\Bbb Q)$, which has a basis consisting of the elementary matrices $E_{ij} = (a_{km})$ where:

$a_{km} = 1, k = i,m = j$
$a_{km} = 0,$ otherwise.

So let's look at what happens when we multiply $E_{ij}E_{i'j'}$:

$E_{ij}E_{i'j'} = E_{ij'}, j = i'$
$E_{ij}E_{i'j'} = 0$, otherwise.

To get a handle on what this really means, let's specify $n = 2$, so our 2x2 matrices can be seen as being "$\Bbb Q^4$ with a multiplication". So our basis is, explicitly:

$u_1 = \begin{bmatrix}1&0\\0&0\end{bmatrix};\ u_2 = \begin{bmatrix}0&1\\0&0\end{bmatrix};\ u_3 = \begin{bmatrix}0&0\\1&0\end{bmatrix};\ u_4 = \begin{bmatrix}0&0\\0&1\end{bmatrix}$

We have:

$u_1u_1 = u_1$, so $\gamma_{111} = 1, \gamma_{11k} = 0, k = 2,3,4$.
$u_1u_2 = u_2$, so $\gamma_{121} = 0, \gamma_{122} = 1, \gamma_{12k} = 0, k = 3,4$
$u_1u_3 = 0$ (all the $\gamma$'s are 0)

and so on (there's 13 more products to compute, and thus 52 more multiplication constants).

Another important $k$-algebra is $k[x]$ (polynomials over $k$) which has as one possible basis:

$u_i = x^i$.

Since $u_iu_j = (x^i)(x^j) = x^{i+j} = u_{i+j}$, we see that:

$\gamma_{ijk} = 1$ when $k = i+j$
$\gamma_{ijk} = 0$, otherwise.

Another often-used example: we have $\Bbb R^2$ as an $\Bbb R$-algebra given the basis:

$\{e_1,e_2\} = \{(1,0),(0,1)\}$ with multiplication constants:

$\gamma_{111} = 1$
$\gamma_{112} = 0$
$\gamma_{121} = 0$
$\gamma_{122} = 1$
$\gamma_{211} = 0$
$\gamma_{212} = 1$
$\gamma_{221} = -1$
$\gamma_{222} = 0$.

Since $\gamma_{ijk} = \gamma_{jik}$, this forms a commutative $\Bbb R$-algebra, more commonly known as $\Bbb C$ (this is a DIVISION ALGEBRA since $U(\Bbb C) = \Bbb C - \{0\}$, that is: all non-zero elements are invertible).

Interestingly enough, we actually have this as a sub-algebra of $\text{Mat}_2(\Bbb R)$, with basis:

$\{v_1,v_2\} = \{E_{11} + E_{22},E_{21}-E_{12}\}$.

This is because the basis elements when multiplied give a linear combination of these two basis elements:

$v_1v_1 = v_1$
$v_2v_1 = v_1v_2 = v_2$
$v_2v_2 = -v_1$, as can be readily verified.

***************

Given a basis for $\text{Mat}_n(k)$, we can use this to define an isomorphism with $\text{Hom}_{k}(k^n,k^n)$, so the study of the (particular) linear algebra $\text{Mat}_n(k)$ is typically the focus of a course called "linear algebra".

It is important not to confuse the $k$-action (scalar multiplication) of a $k$-algebra with the ring multiplication, but typically, if our algebra $A$ is unital, we can often consider it as an extension ring of $k$ via the map:

$\alpha \mapsto \alpha\cdot 1_A$

For example, with $A = \text{Mat}_n(k)$ we have the embedding:

$\alpha \mapsto \alpha I_n$

and with $A = k[x]$ we have the natural embedding of $k$ as constant polynomials

(note that $k \cong k[x]/(x)$ which essentially amounts to "evaluating $p(x)$ at 0", for any polynomial $p$).

Thanks so much for the example Deveno … the examples you post are extremely helpful and informative ...

Working through the details of your post very soon ...

Peter

 

[Math Amateur]

It's somewhat of a trivial example, but it may be helpful to consider the $\Bbb Q$-algebra $\text{Mat}_n(\Bbb Q)$, which has a basis consisting of the elementary matrices $E_{ij} = (a_{km})$ where:

$a_{km} = 1, k = i,m = j$
$a_{km} = 0,$ otherwise.

So let's look at what happens when we multiply $E_{ij}E_{i'j'}$:

$E_{ij}E_{i'j'} = E_{ij'}, j = i'$
$E_{ij}E_{i'j'} = 0$, otherwise.

To get a handle on what this really means, let's specify $n = 2$, so our 2x2 matrices can be seen as being "$\Bbb Q^4$ with a multiplication". So our basis is, explicitly:

$u_1 = \begin{bmatrix}1&0\\0&0\end{bmatrix};\ u_2 = \begin{bmatrix}0&1\\0&0\end{bmatrix};\ u_3 = \begin{bmatrix}0&0\\1&0\end{bmatrix};\ u_4 = \begin{bmatrix}0&0\\0&1\end{bmatrix}$

We have:

$u_1u_1 = u_1$, so $\gamma_{111} = 1, \gamma_{11k} = 0, k = 2,3,4$.
$u_1u_2 = u_2$, so $\gamma_{121} = 0, \gamma_{122} = 1, \gamma_{12k} = 0, k = 3,4$
$u_1u_3 = 0$ (all the $\gamma$'s are 0)

and so on (there's 13 more products to compute, and thus 52 more multiplication constants).

Another important $k$-algebra is $k[x]$ (polynomials over $k$) which has as one possible basis:

$u_i = x^i$.

Since $u_iu_j = (x^i)(x^j) = x^{i+j} = u_{i+j}$, we see that:

$\gamma_{ijk} = 1$ when $k = i+j$
$\gamma_{ijk} = 0$, otherwise.

Another often-used example: we have $\Bbb R^2$ as an $\Bbb R$-algebra given the basis:

$\{e_1,e_2\} = \{(1,0),(0,1)\}$ with multiplication constants:

$\gamma_{111} = 1$
$\gamma_{112} = 0$
$\gamma_{121} = 0$
$\gamma_{122} = 1$
$\gamma_{211} = 0$
$\gamma_{212} = 1$
$\gamma_{221} = -1$
$\gamma_{222} = 0$.

Since $\gamma_{ijk} = \gamma_{jik}$, this forms a commutative $\Bbb R$-algebra, more commonly known as $\Bbb C$ (this is a DIVISION ALGEBRA since $U(\Bbb C) = \Bbb C - \{0\}$, that is: all non-zero elements are invertible).

Interestingly enough, we actually have this as a sub-algebra of $\text{Mat}_2(\Bbb R)$, with basis:

$\{v_1,v_2\} = \{E_{11} + E_{22},E_{21}-E_{12}\}$.

This is because the basis elements when multiplied give a linear combination of these two basis elements:

$v_1v_1 = v_1$
$v_2v_1 = v_1v_2 = v_2$
$v_2v_2 = -v_1$, as can be readily verified.

***************

Given a basis for $\text{Mat}_n(k)$, we can use this to define an isomorphism with $\text{Hom}_{k}(k^n,k^n)$, so the study of the (particular) linear algebra $\text{Mat}_n(k)$ is typically the focus of a course called "linear algebra".

It is important not to confuse the $k$-action (scalar multiplication) of a $k$-algebra with the ring multiplication, but typically, if our algebra $A$ is unital, we can often consider it as an extension ring of $k$ via the map:

$\alpha \mapsto \alpha\cdot 1_A$

For example, with $A = \text{Mat}_n(k)$ we have the embedding:

$\alpha \mapsto \alpha I_n$

and with $A = k[x]$ we have the natural embedding of $k$ as constant polynomials

(note that $k \cong k[x]/(x)$ which essentially amounts to "evaluating $p(x)$ at 0", for any polynomial $p$).

Hi Deveno … just working through your post … and need help ...You write:

" … … Another often-used example: we have $\Bbb R^2$ as an $\Bbb R$-algebra given the basis:

$\{e_1,e_2\} = \{(1,0),(0,1)\}$ with multiplication constants:

$\gamma_{111} = 1$
$\gamma_{112} = 0$
$\gamma_{121} = 0$
$\gamma_{122} = 1$
$\gamma_{211} = 0$
$\gamma_{212} = 1$
$\gamma_{221} = -1$
$\gamma_{222} = 0$. … … "

I am assuming (rather tentatively) that multiplication is of the form

$$(x_1 y_1) \cdot (x_2 y_2) = (x_1x_2, y_1y_2) $$ (component wise)

and that

$$e_i \cdot e_j = \sum_{k = 1}^2 \gamma_{ijk} e_k = \gamma_{111} e_1 +\gamma_{112} e_2$$Is that right?
BUT … if it is correct then$$e_1 \cdot e_1 = (1,0) \cdot (1,0) = (1,0) = (1,0) + (0,0) = 1 \cdot e_1 + 0 \cdot e_2 $$

so

$$\gamma_{111} =1 \text{ and } \gamma_{112} = 0$$Similarly

$$e_1 \cdot e_2 = (1,0) \cdot (0,1) = (0,0) = (0,0) + (0,0) = 0 \cdot e_1 + 0 \cdot e_2 $$

so

$$\gamma_{121} = 0 \text{ and } \gamma_{122} = 0$$
BUT … in your analysis we have$$\gamma_{121} =0$$ and $$\gamma_{122} = 1$$Can you please explain what is wrong in my analysis above?Peter

 

[Deveno]

Hi Deveno … just working through your post … and need help ...You write:

" … … Another often-used example: we have $\Bbb R^2$ as an $\Bbb R$-algebra given the basis:

$\{e_1,e_2\} = \{(1,0),(0,1)\}$ with multiplication constants:

$\gamma_{111} = 1$
$\gamma_{112} = 0$
$\gamma_{121} = 0$
$\gamma_{122} = 1$
$\gamma_{211} = 0$
$\gamma_{212} = 1$
$\gamma_{221} = -1$
$\gamma_{222} = 0$. … … "

I am assuming (rather tentatively) that multiplication is of the form

Why?

$$(x_1 y_1) \cdot (x_2 y_2) = (x_1x_2, y_1y_2) $$ (component wise)

and that

$$e_i \cdot e_j = \sum_{k = 1}^2 \gamma_{ijk} e_k = \gamma_{111} e_1 +\gamma_{112} e_2$$Is that right?
BUT … if it is correct then$$e_1 \cdot e_1 = (1,0) \cdot (1,0) = (1,0) = (1,0) + (0,0) = 1 \cdot e_1 + 0 \cdot e_2 $$

so

$$\gamma_{111} =1 \text{ and } \gamma_{112} = 0$$Similarly

$$e_1 \cdot e_2 = (1,0) \cdot (0,1) = (0,0) = (0,0) + (0,0) = 0 \cdot e_1 + 0 \cdot e_2 $$

This is incorrect. You are assuming that:

$(a,b)\cdot(c,d) = (ac,bd)$.

We have:

$(a,b)\cdot(c,d) = (ae_1 + be_2)(ce_1 + de_2) = ac(e_1e_1) + ad(e_1e_2) + bc(e_2e_1) + bd(e_2e_2)$

Now:

$e_1e_1 = \gamma_{111}e_1 + \gamma_{112}e_2$

so to evaluate this, we need to know the multiplicative constants BEFOREHAND. Let's find them by looking at the parent algebra these come from:

$e_1e_1 = (E_{11} + E_{22})(E_{11} + E_{22}) = E_{11}E_{11} + E_{11}E_{22} + E_{22}E_{11} + E_{22}E_{22}$

$= \begin{bmatrix}1&0\\0&0\end{bmatrix}\begin{bmatrix}1&0\\0&0\end{bmatrix} + \begin{bmatrix}1&0\\0&0\end{bmatrix}\begin{bmatrix}0&0\\0&1\end{bmatrix} + \begin{bmatrix}0&0\\0&1\end{bmatrix}\begin{bmatrix}1&0\\0&0\end{bmatrix} + \begin{bmatrix}0&0\\0&1\end{bmatrix}\begin{bmatrix}0&0\\0&1\end{bmatrix}$

$= \begin{bmatrix}1&0\\0&0\end{bmatrix} + \begin{bmatrix}0&0\\0&0\end{bmatrix} + \begin{bmatrix}0&0\\0&0\end{bmatrix} + \begin{bmatrix}0&0\\0&1\end{bmatrix}$

$= \begin{bmatrix}1&0\\0&0\end{bmatrix} + \begin{bmatrix}0&0\\0&1\end{bmatrix}$

$= E_{11} + E_{22} = e_1 = 1e_1 + 0e_2$, so $\gamma_{111} = 1,\gamma_{112} = 0$.

(Basically, in the product $E_{ij}E_{km}$ if $j = k$ we get $E_{im}$, otherwise we get the 0-matrix).

We have to do the same thing for $e_1e_2$:

$e_1e_2 = \gamma_{121}e_1 + \gamma_{122}e_2$

and working through the matrix multiplication we have:

$e_1e_2 = (E_{11} + E_{22})(E_{21} - E_{12}) = E_{11}E_{21} - E_{11}E_{12} +E_{22}E_{21} - E_{22}E_{12}$

$= 0 - E_{12} + E_{21} - 0 = E_{21} - E_{12} = e_2 = 0e_1 + 1e_2$

so that $\gamma_{121} = 0$ and $\gamma_{122} = 1$.

The other multiplicative constants can be verified the same way. It's worthwhile to do this for yourself, once.

$$\gamma_{121} = 0 \text{ and } \gamma_{122} = 0$$
BUT … in your analysis we have$$\gamma_{121} =0$$ and $$\gamma_{122} = 1$$Can you please explain what is wrong in my analysis above?Peter

This product is NOT "the usual component-wise product" of $\Bbb R \times \Bbb R$.

So what we wind up with is:

$(a,b)(c,d) = (ae_1 + be_2)(ce_1 + de_2) = ac(e_1e_1) + ad(e_1e_2) + bc(e_2e_1) + bd(e_2e_2)$

$= ac(\gamma_{111}e_1 + \gamma_{112}e_2) + ad(\gamma_{121}e_1 + \gamma_{122}e_2) + bc(\gamma_{211}e_1 + \gamma_{212}e_2) + bd(\gamma_{221}e_1 + \gamma_{222}e_2)$

$= (ac\gamma_{111} + ad\gamma_{121} + bc\gamma_{211} + bd\gamma_{221})e_1 + (ac\gamma_{112} + ad\gamma_{122} + bc\gamma_{212} + bd\gamma_{222})e_2$

$= (ac + 0 + 0 - bd)e_1 + (0 + ad + bc + 0)e_2 = (ac - bd,ad + bc)$.

Note that if $b = d = 0$, everything goes away except the $ac$ term, that is, it IS true that:

$(a,0)(c,0) = (ac,0)$. This gives us an embedding of $\Bbb R$ as: $a \mapsto (a,0)$, which is a ring-homomorphism (indeed a monomorphism).

Recall that in our "parent algebra" this is:

$ae_1 + 0e_2 = a(E_{11} + E_{22}) = \begin{bmatrix}a&0\\0&0\end{bmatrix} + \begin{bmatrix}0&0\\0&a\end{bmatrix} = aI$ so that this is the "same embedding" we had before (such matrices are often called "scalar matrices").

What is more interesting, is if $a = c = 0$, and $b = d = 1$, we have:

$(0,1)(0,1) = (-1,0)$ <--note how the 0 "jumps coordinates". Evidently in this algebra, $(0,1)$ is a square root of $(-1,0)$, which we are identifying with $-1$.

I urge you to think about what kind of mapping of $\Bbb R^2$ we get by sending:

$(x,y)\mapsto (0,1)(x,y) = (-y,x)$. Draw some pictures. It may help to think of this map as the composition of two reflections:

$(x,y) \mapsto (y,x)$ (reflecting about the diagonal line $x = y$) <---this one first.

$(a,b) \mapsto (-a,b)$ (reflecting about the $x$ axis) <---this one last (they don't commute).

Use this to convince yourself "dilation-rotations" are a good name for complex numbers (considered as a subalgebra of the algebra of real 2x2 matrices).

 

[Math Amateur]

Why?
This is incorrect. You are assuming that:

$(a,b)\cdot(c,d) = (ac,bd)$.

We have:

$(a,b)\cdot(c,d) = (ae_1 + be_2)(ce_1 + de_2) = ac(e_1e_1) + ad(e_1e_2) + bc(e_2e_1) + bd(e_2e_2)$

Now:

$e_1e_1 = \gamma_{111}e_1 + \gamma_{112}e_2$

so to evaluate this, we need to know the multiplicative constants BEFOREHAND. Let's find them by looking at the parent algebra these come from:

$e_1e_1 = (E_{11} + E_{22})(E_{11} + E_{22}) = E_{11}E_{11} + E_{11}E_{22} + E_{22}E_{11} + E_{22}E_{22}$

$= \begin{bmatrix}1&0\\0&0\end{bmatrix}\begin{bmatrix}1&0\\0&0\end{bmatrix} + \begin{bmatrix}1&0\\0&0\end{bmatrix}\begin{bmatrix}0&0\\0&1\end{bmatrix} + \begin{bmatrix}0&0\\0&1\end{bmatrix}\begin{bmatrix}1&0\\0&0\end{bmatrix} + \begin{bmatrix}0&0\\0&1\end{bmatrix}\begin{bmatrix}0&0\\0&1\end{bmatrix}$

$= \begin{bmatrix}1&0\\0&0\end{bmatrix} + \begin{bmatrix}0&0\\0&0\end{bmatrix} + \begin{bmatrix}0&0\\0&0\end{bmatrix} + \begin{bmatrix}0&0\\0&1\end{bmatrix}$

$= \begin{bmatrix}1&0\\0&0\end{bmatrix} + \begin{bmatrix}0&0\\0&1\end{bmatrix}$

$= E_{11} + E_{22} = e_1 = 1e_1 + 0e_2$, so $\gamma_{111} = 1,\gamma_{112} = 0$.

(Basically, in the product $E_{ij}E_{km}$ if $j = k$ we get $E_{im}$, otherwise we get the 0-matrix).

We have to do the same thing for $e_1e_2$:

$e_1e_2 = \gamma_{121}e_1 + \gamma_{122}e_2$

and working through the matrix multiplication we have:

$e_1e_2 = (E_{11} + E_{22})(E_{21} - E_{12}) = E_{11}E_{21} - E_{11}E_{12} +E_{22}E_{21} - E_{22}E_{12}$

$= 0 - E_{12} + E_{21} - 0 = E_{21} - E_{12} = e_2 = 0e_1 + 1e_2$

so that $\gamma_{121} = 0$ and $\gamma_{122} = 1$.

The other multiplicative constants can be verified the same way. It's worthwhile to do this for yourself, once.
This product is NOT "the usual component-wise product" of $\Bbb R \times \Bbb R$.

So what we wind up with is:

$(a,b)(c,d) = (ae_1 + be_2)(ce_1 + de_2) = ac(e_1e_1) + ad(e_1e_2) + bc(e_2e_1) + bd(e_2e_2)$

$= ac(\gamma_{111}e_1 + \gamma_{112}e_2) + ad(\gamma_{121}e_1 + \gamma_{122}e_2) + bc(\gamma_{211}e_1 + \gamma_{212}e_2) + bd(\gamma_{221}e_1 + \gamma_{222}e_2)$

$= (ac\gamma_{111} + ad\gamma_{121} + bc\gamma_{211} + bd\gamma_{221})e_1 + (ac\gamma_{112} + ad\gamma_{122} + bc\gamma_{212} + bd\gamma_{222})e_2$

$= (ac + 0 + 0 - bd)e_1 + (0 + ad + bc + 0)e_2 = (ac - bd,ad + bc)$.

Note that if $b = d = 0$, everything goes away except the $ac$ term, that is, it IS true that:

$(a,0)(c,0) = (ac,0)$. This gives us an embedding of $\Bbb R$ as: $a \mapsto (a,0)$, which is a ring-homomorphism (indeed a monomorphism).

Recall that in our "parent algebra" this is:

$ae_1 + 0e_2 = a(E_{11} + E_{22}) = \begin{bmatrix}a&0\\0&0\end{bmatrix} + \begin{bmatrix}0&0\\0&a\end{bmatrix} = aI$ so that this is the "same embedding" we had before (such matrices are often called "scalar matrices").

What is more interesting, is if $a = c = 0$, and $b = d = 1$, we have:

$(0,1)(0,1) = (-1,0)$ <--note how the 0 "jumps coordinates". Evidently in this algebra, $(0,1)$ is a square root of $(-1,0)$, which we are identifying with $-1$.

I urge you to think about what kind of mapping of $\Bbb R^2$ we get by sending:

$(x,y)\mapsto (0,1)(x,y) = (-y,x)$. Draw some pictures. It may help to think of this map as the composition of two reflections:

$(x,y) \mapsto (y,x)$ (reflecting about the diagonal line $x = y$) <---this one first.

$(a,b) \mapsto (-a,b)$ (reflecting about the $x$ axis) <---this one last (they don't commute).

Use this to convince yourself "dilation-rotations" are a good name for complex numbers (considered as a subalgebra of the algebra of real 2x2 matrices).

Thanks Deveno … at first glance this post looks EXTREMELY helpful to my gaining an understanding of k-algebras … …

Obviously I have to be very careful over assuming things regarding the multiplication …

Working through your post now ...

Thanks again,

Peter

 

[Math Amateur]

Why?
This is incorrect. You are assuming that:

$(a,b)\cdot(c,d) = (ac,bd)$.

We have:

$(a,b)\cdot(c,d) = (ae_1 + be_2)(ce_1 + de_2) = ac(e_1e_1) + ad(e_1e_2) + bc(e_2e_1) + bd(e_2e_2)$

Now:

$e_1e_1 = \gamma_{111}e_1 + \gamma_{112}e_2$

so to evaluate this, we need to know the multiplicative constants BEFOREHAND. Let's find them by looking at the parent algebra these come from:

$e_1e_1 = (E_{11} + E_{22})(E_{11} + E_{22}) = E_{11}E_{11} + E_{11}E_{22} + E_{22}E_{11} + E_{22}E_{22}$

$= \begin{bmatrix}1&0\\0&0\end{bmatrix}\begin{bmatrix}1&0\\0&0\end{bmatrix} + \begin{bmatrix}1&0\\0&0\end{bmatrix}\begin{bmatrix}0&0\\0&1\end{bmatrix} + \begin{bmatrix}0&0\\0&1\end{bmatrix}\begin{bmatrix}1&0\\0&0\end{bmatrix} + \begin{bmatrix}0&0\\0&1\end{bmatrix}\begin{bmatrix}0&0\\0&1\end{bmatrix}$

$= \begin{bmatrix}1&0\\0&0\end{bmatrix} + \begin{bmatrix}0&0\\0&0\end{bmatrix} + \begin{bmatrix}0&0\\0&0\end{bmatrix} + \begin{bmatrix}0&0\\0&1\end{bmatrix}$

$= \begin{bmatrix}1&0\\0&0\end{bmatrix} + \begin{bmatrix}0&0\\0&1\end{bmatrix}$

$= E_{11} + E_{22} = e_1 = 1e_1 + 0e_2$, so $\gamma_{111} = 1,\gamma_{112} = 0$.

(Basically, in the product $E_{ij}E_{km}$ if $j = k$ we get $E_{im}$, otherwise we get the 0-matrix).

We have to do the same thing for $e_1e_2$:

$e_1e_2 = \gamma_{121}e_1 + \gamma_{122}e_2$

and working through the matrix multiplication we have:

$e_1e_2 = (E_{11} + E_{22})(E_{21} - E_{12}) = E_{11}E_{21} - E_{11}E_{12} +E_{22}E_{21} - E_{22}E_{12}$

$= 0 - E_{12} + E_{21} - 0 = E_{21} - E_{12} = e_2 = 0e_1 + 1e_2$

so that $\gamma_{121} = 0$ and $\gamma_{122} = 1$.

The other multiplicative constants can be verified the same way. It's worthwhile to do this for yourself, once.
This product is NOT "the usual component-wise product" of $\Bbb R \times \Bbb R$.

So what we wind up with is:

$(a,b)(c,d) = (ae_1 + be_2)(ce_1 + de_2) = ac(e_1e_1) + ad(e_1e_2) + bc(e_2e_1) + bd(e_2e_2)$

$= ac(\gamma_{111}e_1 + \gamma_{112}e_2) + ad(\gamma_{121}e_1 + \gamma_{122}e_2) + bc(\gamma_{211}e_1 + \gamma_{212}e_2) + bd(\gamma_{221}e_1 + \gamma_{222}e_2)$

$= (ac\gamma_{111} + ad\gamma_{121} + bc\gamma_{211} + bd\gamma_{221})e_1 + (ac\gamma_{112} + ad\gamma_{122} + bc\gamma_{212} + bd\gamma_{222})e_2$

$= (ac + 0 + 0 - bd)e_1 + (0 + ad + bc + 0)e_2 = (ac - bd,ad + bc)$.

Note that if $b = d = 0$, everything goes away except the $ac$ term, that is, it IS true that:

$(a,0)(c,0) = (ac,0)$. This gives us an embedding of $\Bbb R$ as: $a \mapsto (a,0)$, which is a ring-homomorphism (indeed a monomorphism).

Recall that in our "parent algebra" this is:

$ae_1 + 0e_2 = a(E_{11} + E_{22}) = \begin{bmatrix}a&0\\0&0\end{bmatrix} + \begin{bmatrix}0&0\\0&a\end{bmatrix} = aI$ so that this is the "same embedding" we had before (such matrices are often called "scalar matrices").

What is more interesting, is if $a = c = 0$, and $b = d = 1$, we have:

$(0,1)(0,1) = (-1,0)$ <--note how the 0 "jumps coordinates". Evidently in this algebra, $(0,1)$ is a square root of $(-1,0)$, which we are identifying with $-1$.

I urge you to think about what kind of mapping of $\Bbb R^2$ we get by sending:

$(x,y)\mapsto (0,1)(x,y) = (-y,x)$. Draw some pictures. It may help to think of this map as the composition of two reflections:

$(x,y) \mapsto (y,x)$ (reflecting about the diagonal line $x = y$) <---this one first.

$(a,b) \mapsto (-a,b)$ (reflecting about the $x$ axis) <---this one last (they don't commute).

Use this to convince yourself "dilation-rotations" are a good name for complex numbers (considered as a subalgebra of the algebra of real 2x2 matrices).

Hi Deveno … thanks for your extensive help ...

Sorry to be slow, but I need your further assistance ...

In your example of $$\mathbb{R}^2$$ as an $$\mathbb{R}$$-algebra, you write:

" … … Given the basis $$\{ e_1, e_2 \} = \{ (1,0), (0,1) \} $$ … … "

So we have $$e_1 = (1,0)$$ and $$e_2 = (0,1)$$ … …

Then you write:

" … … $$e_1e_1 = ( E_{11} + E_{22} ) ( E_{11} + E_{22} )$$ … … "

so, that is $$e_1 = ( E_{11} + E_{22} )$$

BUT … …

$$ E_{11} + E_{22} = \begin{bmatrix} 1&0 \\ 0&0 \end{bmatrix} + \begin{bmatrix}0&0\\0&1\end{bmatrix} = \begin{bmatrix}1&0\\0&1\end{bmatrix} $$

while, as indicated above, $$e_1 = (1,0)$$ ?

Can you clarify?

Peter

 

[Deveno]

Hi Deveno … thanks for your extensive help ...

Sorry to be slow, but I need your further assistance ...

In your example of $$\mathbb{R}^2$$ as an $$\mathbb{R}$$-algebra, you write:

" … … Given the basis $$\{ e_1, e_2 \} = \{ (1,0), (0,1) \} $$ … … "

So we have $$e_1 = (1,0)$$ and $$e_2 = (0,1)$$ … …

Then you write:

" … … $$e_1e_1 = ( E_{11} + E_{22} ) ( E_{11} + E_{22} )$$ … … "

so, that is $$e_1 = ( E_{11} + E_{22} )$$

BUT … …

$$ E_{11} + E_{22} = \begin{bmatrix} 1&0 \\ 0&0 \end{bmatrix} + \begin{bmatrix}0&0\\0&1\end{bmatrix} = \begin{bmatrix}1&0\\0&1\end{bmatrix} $$

while, as indicated above, $$e_1 = (1,0)$$ ?

Can you clarify?

Peter

Given a vector space ($F$-module) $V$, one can realize this vector space many different ways. For example, one can embed an isomorph of $\Bbb R^2$ in $\Bbb R^4$ a LOT of different ways, just send any basis $\{v_1,v_2\}$ to two linearly independent vectors of $\Bbb R^4$.

Here, that isomorphism, of $\Bbb C$ as a 2-dimensional $\Bbb R$-algebra into $\text{Mat}_4(\Bbb R)$ is given by:

$e_1 = (1,0) \mapsto E_{11} + E_{22} = \begin{bmatrix}1&0\\0&1\end{bmatrix}$

$e_2 = (0,1) \mapsto E_{21} - E_{12} = \begin{bmatrix}0&-1\\1&0\end{bmatrix}$.

Of course, it might not be obvious that $\{E_{11}+E_{22},E_{21}-E_{12}\}$ is linearly independent, but if:

$c_1(E_{11}+E_{22}) + c_2(E_{21}-E_{12}) = 0$ (the 0-matrix)

then we have a linear combination of the $E_{ij}$ that sums to 0, and since the $E_{ij}$ are linearly independent, it follows that $c_1 = c_2 = 0$.

The important thing about this embedding is not only is it a VECTOR-SPACE isomorphism, but it preserves the RING multiplication as well.

So we have "the same algebraic structure" (complex numbers), as on one hand "two-dimensional things" (2-vectors in the plane), and (equivalently) as "four-dimensional things" (2x2 matrices).

If we extend by linearity, we obtain the algebra monomorphism:

$\phi: \Bbb C \to \text{Mat}_2(\Bbb R)$ given by:

$\phi(a+bi) = \begin{bmatrix}a&-b\\b&a\end{bmatrix}$

Again, it is instructive to work out that $\phi$ is both a ring-homomorphism and a vector space homomorphism with trivial kernel.

In particular, since $\Bbb C$ is isomorphic to $\phi(\Bbb C)$, it follows that for these two basis choices, we have the same multiplication constants, so it really doesn't matter if we talk about $e_j$ or $\phi(e_j)$, they "act the same".