(i) Each of <em>u</em>, <em>v</em>, and <em>w</em> are vectors in R<em>ⁿ</em>, so they each have size <em>n</em> × 1 (i.e. <em>n</em> rows and 1 column). So <em>u </em>and <em>v</em> both have size <em>n</em> × 1, while <em>w</em>ᵀ has size 1 × <em>n</em>.
<em>M</em> is an <em>n</em> × <em>n</em> matrix, so the matrix <em>A</em> has been partitioned into the blocks
where <em>α</em> is a scalar with size 1 × 1. So <em>A</em> has size (<em>n</em> + 1) × (<em>n</em> + 1).
(ii) Multiplying both sides (on the left is the only sensible way) by the given matrix gives
and of course <em>M</em> ⁻¹ <em>M</em> = <em>I</em> (the identity matrix), so
-<em>w</em>ᵀ <em>M</em> ⁻¹ <em>M</em> + <em>w</em>ᵀ = -<em>w</em>ᵀ + <em>w</em>ᵀ = 0ᵀ (the zero vector transposed)
(iii) Simplifying the system further gives
So now, setting <em>y</em> = <em>M</em> ⁻¹<em>u</em> and <em>z</em> = <em>M</em> ⁻¹ <em>v</em> gives
Given that <em>α</em> - <em>w</em>ᵀ<em>y</em> ≠ 0, it follows that
(iv) Combining the result from (iii) with the first row gives