In linear algebra, a complement to a subspace of a vector space is another subspace which forms an internal direct sum. Two such spaces are mutually complementary.
Formally, if $U$ is a subspace of $V$, then $W$ is a complement of $U$ if and only if $V$ is the direct sum of $U$ and $W$, $V=U\oplus W$, that is:
$V=U+W$
$U\cap W=\emptyset$
I am solving an exercise in which $V=\mathbb{R}^8$ and $U$ has the following basis
\begin{equation} B_U=\left\{\left(\begin{array}{c} 0 \\ 3 \\ 0 \\ 0 \\ 1 \\ 0 \\ 0 \\ 0 \end{array}\right), \left(\begin{array}{c} -1 \\ 0 \\ 2 \\ 0 \\ 0 \\ 1 \\ 0 \\ 0 \end{array}\right), \left(\begin{array}{c} 0 \\ -2 \\ 0 \\ 1 \\ 0 \\ 0 \\ 1 \\ 0 \end{array}\right), \left(\begin{array}{c} 0 \\ 0 \\ -3 \\ 0 \\ 0 \\ 0 \\ 0 \\ 1 \end{array}\right), \left(\begin{array}{c} -1 \\ 0 \\ 0 \\ 0 \\ 0 \\ 3 \\ 0 \\ 0 \end{array}\right), \left(\begin{array}{c} 0 \\ 0 \\ 0 \\ -1 \\ 0 \\ 0 \\ -3 \\ 0 \end{array}\right)\right\} \end{equation} The exercise asks me to calculate a basis for complement $W$. Due to the definition of complement space, I have to find two elements of $\mathbb{R}^8$ which added to $B_U$ do not change the linear independence of the elements of $B_U$.
From the operational point of view, as I find these two elements?
Thank you very much