Question

Let \({{\bf{v}}_1} = \left[ {\begin{array}{*{20}{c}}1\\0\\1\end{array}} \right]\), \({{\bf{v}}_{\bf{2}}} = \left[ {\begin{array}{*{20}{c}}{\bf{0}}\\{\bf{1}}\\1\end{array}} \right]\), \({{\bf{v}}_{\bf{3}}} = \left[ {\begin{array}{*{20}{c}}{\bf{0}}\\{\bf{1}}\\{\bf{0}}\end{array}} \right]\), and let H be the set of vectors in \({\mathbb{R}^{\bf{3}}}\) whose second and third entries are equal. Then every vector in H has a unique expansion as a linear combination of \({{\bf{v}}_{\bf{1}}}\), \({{\bf{v}}_{\bf{2}}}\), \({{\bf{v}}_{\bf{3}}}\), because

\(\left[ {\begin{array}{*{20}{c}}s\\t\\t\end{array}} \right] = s\left[ {\begin{array}{*{20}{c}}{\bf{1}}\\{\bf{0}}\\{\bf{1}}\end{array}} \right] + \left( {t - s} \right)\left[ {\begin{array}{*{20}{c}}{\bf{0}}\\{\bf{1}}\\{\bf{1}}\end{array}} \right] + s\left[ {\begin{array}{*{20}{c}}{\bf{0}}\\{\bf{1}}\\{\bf{1}}\end{array}} \right]\)for any \(s\), and \(t\). Is \(\left\{ {{v_1},{v_2},{v_3}} \right\}\) a basis for H? Why or why not?

Short answer

The set\(\left\{ {{v_1},{v_2},{v_3}} \right\}\)is not a basis for H.

Step-by-step solution

State the invertible matrix theorem

Recall the invertible matrix theorem, which states that if the square matrix is invertible, then the columns are linearly independent, and they form a basis for \({\mathbb{R}^n}\).

State the set of vectors in Span

The vectors\(\left\{ {{v_1},{v_2},{v_3}} \right\}\)must span H in order to provide a basis for H. So,\(H = {\rm{Span}}\left\{ {{v_1},{v_2},{v_3}} \right\}\).

The vectors in set H have identical second and third entries, and H is a subset of\(\left\{ {{v_1},{v_2},{v_3}} \right\}\).But there are some vectors that are not in H.

So,\(H\)is not in\({\rm{Span}}\left\{ {{v_1},{v_2},{v_3}} \right\}\).

Therefore,\(\left\{ {{v_1},{v_2},{v_3}} \right\}\)is not a basis for H.