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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

Expert verified

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

Step by step solution

01

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 \({\mathbb{R}^n}\).

02

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.

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