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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

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Write your given vectors in a matrix, fill up with the standard unit vectors $$ \begin{pmatrix} 0 & -1 & 0 & 0 & -1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 3 & 0 & -2 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 2 & 0 &-3 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 1 &0 & 0 & -1& 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0\\ 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 & 3 & 0& 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0 & 0 & -3& 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \end{pmatrix} $$ Now compute a row echelon form using row operations. $$ \begin{pmatrix} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 & 1 & 0 & -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 1 &0 & 0 & -1& 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1\\ 0 & 0 & 0 & 0 & 2 & 0 & 1 & 0 & 0 & 0 & 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & -2& 0 & 1 & 0 & 2 &-3 & 0 & 0 & 0\\ 0 & 0 & 0 &0 & 0 & 0 & 3 & 0 & 1 & 0 & 0 & 1 & 0 & 3\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 &-1 & 0 &-3 & 3 & 0 & 1 & 0\\ \end{pmatrix} $$ So, if I didn't make a computation error, $e_1$ and $e_2$ will span a complement for your given $U$, as their columns correspond to pivot columns in the row echelon form.

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Thanks. But: who are $e_1, e_2$? –  Mark Nov 16 '12 at 9:02
1  
@Mark: $e_i$ refers to the vector with a $1$ in the $i$th position and $0$s everywhere else. –  wj32 Nov 16 '12 at 9:14
    
This is not very clear as an answer. I suppose that "their columns" refers to the columns that originally held $e_1$ and $e_2$, in other words columns $7$ and $8$. Also, I think there should be a few words on why the row echelon form says anything about the original problem to begin with, and more specifically why the pivot columns among the added columns identify a set of standard basis vectors that span a complementary subspace. –  Marc van Leeuwen Nov 27 at 10:55

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