1. ## Quaternion

Let $q=a+bi+cj+dk$ be a nonzero quaternion.
(a) Prove that the left multiplication $L_q:\mathbb{H}\to\mathbb{H}$ given by $L_q(x)=qx$ is an isomorphism.

$L_q(x_1)=L_q(x_2)\Rightarrow qx_1=qx_2$
By def., the Quaternion Algebra is a vector space equipped with a ring. Therefore, $q^{-1}$ exist.
$q^{-1}qx_1=q^{-1}qx_2\Rightarrow x_1=x_2$
Thus, $L_q(x)$ is monic.

$y=L_q(x)=qx\Rightarrow q^{-1}y=x$
$L_q(q^{-1}y)=qq^{-1}y=y$
Therefore, f is an epimorphism and f is an isomorphism.

(b) Find the matrix of $L_q$ relative to the basis $1,i,j,k\in\mathbb{H}$

Can someone walk step by step through this piece? I don't remember what to do.

2. ## Re: Quaternion

vector spaces equipped with a ring do not necessarily have inverses. for example, what is x^-1 in R[x]?

however, quaternions are special, they form a division ring, which is what we need for (multiplicative) inverses to exist.

it is this same property (existence of inverses) that allows you to prove surjectivity.

but those are minor points, easily fixed by the addition of a single word ("divison" in front of ring) in your proof.

you also need to show that $L_q$ is a linear map. i recommend invoking the distributive law, and the fact that real numbers

commute with quaternions.

for part (b) you are looking for a 4x4 (real) matrix that corresponds to (left) mulitplication by q. fix an arbitrary quaternion q' in H,

and calculate qq', collecting terms by basis elements. for example, if q' = a' + b'i + c'j +d'k, the real term will be:

aa' - bb' - cc' - dd', so the first row should be [a -b -c -d].

3. ## Re: Quaternion

Originally Posted by Deveno
aa' - bb' - cc' - dd', so the first row should be [a -b -c -d].
Why is the first row a -b -c -d and not aa' -bb' -cc' dd'?

$\begin{bmatrix}aa'&-bb'&-cc'&-dd'\\ab'&a'b&cd'&-dc'\\ac'&a'c&-bd'&db'\\ad'&a'd&bc'&-cb'\end{bmatrix}$

This is what I have obtained.

4. ## Re: Quaternion

if q' = a' + b'i + c'j + d'k, in the basis {1,i,j,k} q' has coordinates (a',b',c',d'), so your matrix multiplication should look like this:

$\begin{bmatrix}a&-b&-c&-d\\b&a&d&-c\\c&d&a&-b\\d&-c&b&a \end{bmatrix} \begin{bmatrix}a'\\b'\\c'\\d' \end{bmatrix} = \begin{bmatrix}aa'-bb'-cc'-dd'\\ab'+a'b+cd'-c'd\\ac'+a'c-bd'+b'd\\ad'+a'd+bc'-b'c \end{bmatrix}$

the 4x4 matrix is the matrix for $L_q$, the result is qq' (as a quaternial product).

5. ## Re: Quaternion

Originally Posted by Deveno
if q' = a' + b'i + c'j + d'k, in the basis {1,i,j,k} q' has coordinates (a',b',c',d'), so your matrix multiplication should look like this:

$\begin{bmatrix}a&-b&-c&-d\\b&a&d&-c\\c&-d&a&b\\d&c&-b&a \end{bmatrix} \begin{bmatrix}a'\\b'\\c'\\d' \end{bmatrix} = \begin{bmatrix}aa'-bb'-cc'-dd'\\ab'+a'b+cd'-c'd\\ac'+a'c-bd'+b'd\\ad'+a'd+bc'-b'c \end{bmatrix}$

the 4x4 matrix is the matrix for $L_q$, the result is qq' (as a quaternial product).
I don't understand the first matrix. Why is it that?

6. ## Re: Quaternion

um, there were some errors in it. i hope they're fixed now.

look, $L_q$ is a linear transformation from H to H. we are regarding H as a 4-dimensional real vector space. what does a matrix from R^4 to R^4 look like?

we are looking for a 4x4 matrix $L_q$ such that $L_q(q') = qq'$.

that is, this matrix should take q'-->qq'. what kind of thing takes a 4-vector to a 4-vector?

7. ## Re: Quaternion

Originally Posted by Deveno
um, there were some errors in it. i hope they're fixed now.

look, $L_q$ is a linear transformation from H to H. we are regarding H as a 4-dimensional real vector space. what does a matrix from R^4 to R^4 look like?
I still don't understand how you came up with that matrix.

8. ## Re: Quaternion

if q = a + bi + cj + dk and q' = a' + b'i + c'j + d'k, then qq' = (aa' - bb' - cc' - dd') + (ab' + a'b + cd' - c'd)i + (ac' + a'c - bd' + b'd)j + (ad' + a'd + bc' - b'c)k, right?

in the basis {1,i,j,k} this is the 4-vector (aa'-bb'-cc'-dd',ab'+a'b+cd'-c'd,ac'+a'c-b'd+b'd,ad'+a'd+bc'-b'c).

now, the first entry of that 4-vector is the inner product of the first row-vector of the matrix for Lq with (a',b',c',d').

the only part of the real number aa'-bb'-cc'-dd' involving a' is the term aa', hence the first entry of the first row for Lq is a.

similarly, the only term involving b' is the term -bb', so b' must get multplied by -b, c' must get multiplied by -c, d' must get multiplied by -d.

so the first row for Lq is [a -b -c -d].

to find the second row for Lq, we look at the "i" coefficient in the product qq'. what is a' multiplied by in the i coefficient? well the only a' term in the i coefficient is ab' (=ba', since these are real numbers).

so the first entry of the 2nd row for Lq is b. note that each "primed" entry of q' appears just once in every coefficient of qq'. so the "other part" of the term

involving that coefficient of q' (whether it is a',b',c' or d') must be the entry for Lq in that row of Lq. it's just "fill in the blanks" using matrix multiplication rules.