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The original question: Let $\vec{q}$ be a unit vector $( \vec{q} \in \mathbb{R}^{n}, \left \| q \right \|=1 )$ and suppose that $\vec{q} \ne \vec{e}_{1}$ . Let $\vec{a} = \vec{q}-\vec{e}_{1}$ ​ and $H_{\vec{a}}$ be the Householder matrix generated by the vector $\vec{a}$. Prove that the first column of $H_{\vec{a}}$ is the vector $\vec{q}$.

I understand that I have some vector $a$, which is the difference between two unit vectors, generating a Householder reflection. I need to show that $a$ plus $e_{1}$ is equal to the first column of $H_{a}$, which I can output by multiplication of the reflector and $e_{1}$. I feel like I am missing a basic principle here but I have wrestled with this for a few days and need some help.

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    I have gotten farther with this problem now. $\vec{a}$ is the difference of q and $e_{1}$ so $a^{t}a$ will always be 2. I'm not sure how to express it mathematically, but $a^{t}a$ is the dot product of $a$ with itself or the sum of each term squared, so the first element of $a$ will be $q_{1} - 1$ and sum of every other term will be $q - q_{1}$ and I know that in $a^{t}a$ those along with the first term will sum to 2. – 2017-02-20
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    Also, I see that $aa^{t}$ multiplied by the $e_{1}$ vector produces the first column of $aa^{t}$ and the first column is a reflection of $a$, $-a$. I'm sure that is because the first element, again, is $q - 1$, but need to write that mathematically. – 2017-02-20

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