Given data:

The given space vector is: \(\displaystyle{V}={\left\lbrace{v}-{u},{w}-{v},{u}-{w}\right\rbrace}\).

The expression to check whether the space vector is linearly dependent or independent is

\(\displaystyle{c}_{{{1}}}{\left({v}-{u}\right)}+{c}_{{{2}}}{\left({w}-{v}\right)}+{c}_{{{3}}}{\left({u}-{w}\right)}={\left({0},{0},{0}\right)}\)

Further, solve the above expression.

\(\displaystyle{c}_{{{1}}}{v}-{c}_{{{1}}}{u}+{c}_{{{2}}}{w}-{c}_{{{2}}}{v}+{c}_{{{3}}}{u}-{c}_{{{3}}}{w}={\left({0},{0},{0}\right)}\)

\(\displaystyle{\left({c}_{{{1}}}-{c}_{{{2}}}\right)}{v}-{\left({c}_{{{1}}}-{c}_{{{3}}}\right)}{u}+{\left({c}_{{{2}}}-{c}_{{{3}}}\right)}{w}={\left({0},{0},{0}\right)}\)

\(\displaystyle{c}_{{{1}}}-{c}_{{{2}}}={0}\)

\(\displaystyle{c}_{{{1}}}-{c}_{{{3}}}={0}\)

Step 2

The above equation gives the value of the scalar multiples.

\(\displaystyle{c}_{{{1}}}={c}_{{{2}}}={c}_{{{3}}}\)

As the \(\displaystyle{c}_{{{1}}},{c}_{{{2}}}\), and \(\displaystyle{c}_{{{3}}}\) are the scalar multiples of the given space vector, and they are equal, so the space vector is linearly dependent.

Thus, the given space vector is linearly dependent.