I am asked to do the following:
Let $\vec y$ be the vector
$$ \vec y = \begin{bmatrix} 6&-3&9&8\\ \end{bmatrix} $$
Compute the following composition: $$||\vec y||^2 = ||\bar y.\vec 1||^2 + ||\vec y- \bar y.\vec 1||^2$$ where $\vec 1$ stands for the vector $\vec 1 = \begin{bmatrix} 1&1&1&1\\ \end{bmatrix}.$ Also, $\bar y$ is the arithmetic mean of the components of $\vec y$.
To start off, I would say to first solve for $\bar y. \vec 1$ and computing $\bar y. \vec 1$ yields $\begin {bmatrix} 5&5&5&5\\ \end {bmatrix}$ since the arithmetic mean of $\bar y = 5.$ How do I go about solving for the square length $|| \bar y. \vec 1||^2?$ Also, I know that $\vec y - \bar y. \vec 1 = \begin {bmatrix} 1&-8&4&3\\ \end {bmatrix}.$ We need to keep in mind that $||\vec y||^2$ is a scalar!
$$|| \bar y. \vec 1||^2=\begin {bmatrix} 5&5&5&5\end {bmatrix}\begin {bmatrix} 5\\5\\5\\5 \end {bmatrix} = 5^2+5^2+5^2+5^2=100$$ and in the same way $$\|\vec y - \bar y. \vec 1 \|^2= \begin {bmatrix} 1&-8&4&3 \end {bmatrix}\begin {bmatrix} 1\\-8\\4\\3 \end {bmatrix}= 1^2+(-8)^2+4^2+3^2=90$$