Question_S2

Question_S2

Note that we don’t have associativity here so we have to work from right to left.

For the first note that:

\begin{pmatrix}1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix}

is merely the identity matrix so:

\begin{pmatrix}1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix} \begin{pmatrix}x \\ y \\ z \\ w\end{pmatrix} = \begin{pmatrix}x \\ y \\ z \\ w\end{pmatrix}

We therefore have:

\begin{pmatrix}x&y&z&w\end{pmatrix}\begin{pmatrix}1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix} \begin{pmatrix}x \\ y \\ z \\ w\end{pmatrix} = \begin{pmatrix}x&y&z&w\end{pmatrix}\begin{pmatrix}x \\ y \\ z \\ w\end{pmatrix} = (x^2 + y^2 + z^2 + w^2)

As an aside, note that this is merely:

\begin{pmatrix}x\\y\\z\\w\end{pmatrix} \cdot \begin{pmatrix}x\\y\\z\\w\end{pmatrix}

We can hence rationalise that the identity:

\mathbf v^T \mathbf w = (\mathbf v \cdot \mathbf w)

holds for \mathbf v, \mathbf w \in \mathbb R^n say. (in practice we treat \mathbf v^T \mathbf w like a scalar instead of a 1 \times 1 matrix)

The second product follows similarly, we have:

\begin{pmatrix}1&0&0&0\\0&-1&0&0 \\ 0& 0 & -1 & 0 \\ 0& 0& 0 & -1\end{pmatrix} \begin{pmatrix}x \\y\\z\\w\end{pmatrix} = \begin{pmatrix}x + 0 \times y + 0\times z + 0 \times w\\0\times x -y + 0 \times z + 0 \times w\\0\times x + 0\times y - z + 0 \times w\\0\times x + 0 \times y + 0 \times z -w\end{pmatrix} =\begin{pmatrix}x\\-y\\-z\\-w\end{pmatrix}

And so:

\begin{pmatrix}x&y&z&w\end{pmatrix} \begin{pmatrix}1&0&0&0\\0&-1&0&0 \\ 0& 0 & -1 & 0 \\ 0& 0& 0 & -1\end{pmatrix} \begin{pmatrix}x \\y\\z\\w\end{pmatrix} = \begin{pmatrix}x & y & z & w\end{pmatrix}\begin{pmatrix}x\\-y\\-z\\-w\end{pmatrix} = (x^2 - y^2 - z^2 - w^2)

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