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Can someone help me to understand the last two steps in Hadamard transpose times Hardamard in the given image? I don’t understand how/why the $8I$’s got multiplied by $Q_8$.

Gilbert Strang, page 242:

Then $Q=H/2$ has orthonormal columns. Dividing by $2$ gives unit vectors in $Q$. A $5$-by-$5$ Hadamard matrix is impossible because the dot product of columns would have five $1$’s and/or $-1$’s and could not add to zero. $H_8$ has orthogonal columns of length $\sqrt 8$. $$\def\T{\mathrm T} H^\T_8 H_8 = \left[ \begin{array}{lr} H^\T & H^\T \\ H^\T & -H^\T \end{array} \right] \left[ \begin{array}{lr} H & H \\ H & -H \end{array} \right] = \begin{bmatrix} 2H^\T H & 0 \\ 0 & 2H^\T H \end{bmatrix} = \begin{bmatrix} 8I & 0 \\ 0 & 8I \end{bmatrix} . Q_8 = \frac {H_8} {\sqrt 8}$$

Rócherz
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user82039
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    I don’t own that book, and I may need to know how the inner $H$ is defined, but are you sure that dot is a product and not a sentence full-stop? – Rócherz Feb 04 '24 at 01:04
  • Thank you for formatting it @Rócherz. The question is as below and above was the provided answer. $H_8$=\begin{bmatrix} {H_4}& {H_4} \ {H_4}& {-H_4} \ \end{bmatrix}

    is the next Hadamard matrix with 1's -1's. What is the product of $H^T_8H_8$? Here $H_4$ is defined in terms of $H_2$.

    $H_4$=\begin{bmatrix} {H_2}& {H_2} \ {H_2}& {-H_2} \ \end{bmatrix} and $H_2$=\begin{bmatrix} 1& 1 \ 1& -1 \ \end{bmatrix}

     – user82039 Feb 04 '24 at 04:31

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