Question

This exercise shows one way to define the quaternions,discovered in 1843 by the Irish mathematician Sir W.R. Hamilton (1805-1865).Consider the set H of all $\mathbf{4}\mathbf{\times }\mathbf{4}$matrices M of the form

$\mathit{M}\mathbf{=}\left[\begin{array}{cccc}p& -q& -r& -s\\ q& p& s& -r\\ r& -s& p& q\\ s& r& -q& p\end{array}\right]$

where p,q,r,s are arbitrary real numbers.We can write M more sufficiently in partitioned form as

$\mathit{M}\mathbf{=}\left(\begin{array}{cc}A& -B^T\\ B& A^T\end{array}\right)$

where A and B are rotation-scaling matrices.

a.Show that H is closed under addition:If M and N are in H then so is$\mathit{M}\mathbf{+}\mathit{N}$

M+Nb.Show that H is closed under scalar multiplication .If M is in H and K is an arbitrary scalar then kM is in H.

c.Parts (a) and (b) Show that H is a subspace of the linear space ${\mathit{R}}^{\mathbf{4}\mathbf{\times }\mathbf{4}}$ .Find a basis of H and thus determine the dimension of H.

d.Show that H is closed under multiplication If M and N are in H then so is MN.

e.Show that if M is in H,then so is ${\mathit{M}}^{\mathbf{T}}$.

f.For a matrix M in H compute ${\mathit{M}}^{\mathbf{T}}\mathit{M}$.

g.Which matrices M in H are invertible.If a matrix M in H is invertible is${\mathit{M}}^{\mathbf{-}\mathbf{1}}$ necessarily in H as well?

h. If M and N are in H,does the equation$\mathit{M}\mathit{N}\mathbf{=}\mathit{N}\mathit{M}$always hold?

Short answer

(a) It is proved that M+N is also closed under addition.

(b) kM also closed under scalar multiplication.

(c) $\dim \left(H\right)=2$

(d) MN is also closed under addition.

(e) It is proved thatMis inHand $M^T$is also in H.

(f) $M^{T}M=\left(\begin{array}{cc}{A}^2+{B^T}^2& AB-A^T{B}^T\\ AB-A^T{B}^T& B^2+A^{T2}\end{array}\right)$

(g) M is invertible then M^-1be also in the invertible as well.

(h) Yes,$MN=NM$ .

Step-by-step solution

Explanation for the dimension of the space

If a linear space has a basis with nelements, then all other bases of V, consists of n elements as well.

Also, we say that n is the dimension of V:$dim\left(V\right)=n$

(a)Step2:Matrix under addition

Consider the set H of all $4\times 4$matrices M of the form.

$M=\left[\begin{array}{l}p-q-r-s\\ qps-r\\ r-spq\\ sr-qp\end{array}\right]$

where p,q,r,s are arbitrary real numbers.We can write M more sufficiently in partitioned form as:

$M=\left(\begin{array}{cc}A& -B^T\\ B& A^T\end{array}\right)$

where A and B are rotation-scaling matrices.

Assume that H is the subspace which is closed under addition.

We know that M has the partitioned form and N also be considered in the same partitioned form.

Hence M and N are in the subspace H.

Since we know that the subspace H is closed under addition,then M+N is also closed under addition.

(b)Step:3 Matrix under scalar multiplication

Consider the set H of all $4\times 4$matrices M of the form.

$M=\left[\begin{array}{l}p-q-r-s\\ qps-r\\ r-spq\\ sr-qp\end{array}\right]$

where p,q,r,s are arbitrary real numbers .We can write M more sufficiently in partitioned form as

$M=\left(\begin{array}{cc}A& -B^T\\ B& A^T\end{array}\right)$

where A and B are rotation-scaling matrices.

Assume that H is the subspace which is closed under scalar multiplication.

We know that M has the partitioned form and k also be considered in the arbitrary scalar

Hence M is in the subspace H.

Since we know that the subspace H is closed under scalar multiplication,then the arbitrary k is kM also closed under scalar multiplication.

(c)Step 4: Dimension of the subspace H

Assume that a linear space has a basis with n elements then all other bases of H, consists of n elements as well.

Also we say that n is the dimension of H: $\dim \left(H\right)=n$

Here n is two then the dimension in the subspace of the linear space H is 2.

$\dim \left(H\right)=2$

Hence the solution

(d)Step 6: Matrix under multiplication

Consider the set H of all$4\times 4$ matrices M of the form.

$M=\left[\begin{array}{l}p-q-r-s\\ qps-r\\ r-spq\\ sr-qp\end{array}\right]$

where p,q,r,s are arbitrary real numbers .We can write M more sufficiently in partitioned form as:

$M=\left(\begin{array}{cc}A& -B^T\\ B& A^T\end{array}\right)$

where A and B are rotation-scaling matrices.

Assume that H is the subspace which is closed under multiplication.

We know that M has the partitioned form and N also be considered in the same partitioned form.

Hence M and N are in the subspace H.

Since we know that the subspace H is closed under multiplication,then MN is also closed under addition.

(e)Step5: Transpose of a matrix

Consider the set H of all $4\times 4$matrices M of the form.

$M=\left[\begin{array}{l}p-q-r-s\\ qps-r\\ r-spq\\ sr-qp\end{array}\right]$

where p,q,r,s are arbitrary real numbers .We can write M more sufficiently in partitioned form as:

$M=\left(\begin{array}{cc}A& -B^T\\ B& A^T\end{array}\right)$

where A andB are rotation-scaling matrices.

Assume that H is the subspace

We know that M has the partitioned form.

Now we have to find the transpose of the matrix M.

$M^T=\left(\begin{array}{cc}A& B\\ -B^T& A^T\end{array}\right)$

(f)

The matrix and the transpose of the matrix be as follows

$\begin{array}{c}{M}^{T}M=\left(\begin{array}{cc}A& B\\ -B^T& A^T\end{array}\right)\left(\begin{array}{cc}A& -B^T\\ B& A^T\end{array}\right)\\ =\left(\begin{array}{cc}{A}^2+{B^T}^2& AB-A^T{B}^T\\ AB-A^T{B}^T& B^2+A^{T2}\end{array}\right)\end{array}$

Thus, the solution.

(g)Step6: Invertible of a matrix

Consider the set H of all $4\times 4$matrices M of the form.

$M=\left[\begin{array}{l}p-q-r-s\\ qps-r\\ r-spq\\ sr-qp\end{array}\right]$

where p,q,r,s are arbitrary real numbers .We can write M more sufficiently in partitioned form as:

$M=\left(\begin{array}{cc}A& -B^T\\ B& A^T\end{array}\right)$

where A and B are rotation-scaling matrices.

Assume that H is the subspace

We know that M has the partitioned form and M^-1 be the invertible matrix

Hence M and M^-1be the matrix in the subspace H

Thus, the M is invertible then M^-1be also in the invertible as well

(h)

Using the matrix M and N as the partitioned form,then $MN-NM=0$be the equation.

Hence the equation$MN=NM$ holds