Question

Question: Let \(\left\{ {{{\bf{v}}_{\bf{1}}},{{\bf{v}}_{\bf{2}}}} \right\}\) be an orthogonal set of nonzero vectors, and let \({c_{\bf{1}}}\), \({c_{\bf{2}}}\) be any nonzero scalars. Show that \(\left\{ {{c_{\bf{1}}}{{\bf{v}}_{\bf{1}}},{c_{\bf{2}}}{{\bf{v}}_{\bf{2}}}} \right\}\) is also an orthogonal set. Since orthgonality of a set is defined in terms of pairs of vectors, this shows that if the vectors in an orthogonal set are normalized, the new set will still be orthogonal.

Short answer

It is proved that the set \(\left\{ {{c_1}{{\bf{v}}_1},{c_2}{{\bf{v}}_2}} \right\}\) is an orthogonal set.

Step-by-step solution

Write the relation for the vectors \(\left\{ {{{\bf{v}}_{\bf{1}}},{{\bf{v}}_{\bf{2}}}} \right\}\)

Since the set \(\left\{ {{{\bf{v}}_1},{{\bf{v}}_2}} \right\}\) is an orthogonal set of nonzero vectors, so \({{\bf{v}}_1} \cdot {{\bf{v}}_2} = 0\).

Check the orthogonality for \(\left\{ {{c_{\bf{1}}}{{\bf{v}}_{\bf{1}}},{c_{\bf{2}}}{{\bf{v}}_{\bf{2}}}} \right\}\)

As \({c_1}\) and \({c_2}\) are non-zero scalars, then the dot product for the set of vectors \(\left\{ {{c_1}{{\bf{v}}_1},{c_2}{{\bf{v}}_2}} \right\}\) is shown below:

\(\begin{array}{c}\left( {{c_1}{{\bf{v}}_1}} \right) \cdot \left( {{c_2}{{\bf{v}}_2}} \right) = {c_1}\left( {{{\bf{v}}_1} \cdot \left( {{c_2}{{\bf{v}}_2}} \right)} \right)\\ = {c_1}{c_2}\left( {{{\bf{v}}_1} \cdot {{\bf{v}}_2}} \right)\\ = {c_1}{c_2}\left( 0 \right)\\ = 0\end{array}\)

Therefore, the set \(\left\{ {{c_1}{{\bf{v}}_1},{c_2}{{\bf{v}}_2}} \right\}\) is an orthogonal set.