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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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Most popular questions from this chapter

Suppose \({{\bf{p}}_{\bf{1}}}\), \({{\bf{p}}_{\bf{2}}}\), \({{\bf{p}}_{\bf{3}}}\), and \({{\bf{p}}_{\bf{4}}}\) are specific polynomials that span a two-dimensional subspace H of \({P_{\bf{5}}}\). Describe how one can find a basis for H by examining the four polynomials and making almost no computations.

Question: Exercises 12-17 develop properties of rank that are sometimes needed in applications. Assume the matrix \(A\) is \(m \times n\).

16. If \(A\) is an \(m \times n\) matrix of rank\(r\), then a rank factorization of \(A\) is an equation of the form \(A = CR\), where \(C\) is an \(m \times r\) matrix of rank\(r\) and \(R\) is an \(r \times n\) matrix of rank \(r\). Such a factorization always exists (Exercise 38 in Section 4.6). Given any two \(m \times n\) matrices \(A\) and \(B\), use rank factorizations of \(A\) and \(B\) to prove that rank\(\left( {A + B} \right) \le {\mathop{\rm rank}\nolimits} A + {\mathop{\rm rank}\nolimits} B\).

(Hint: Write \(A + B\) as the product of two partitioned matrices.)

Question 11: Let\(S\)be a finite minimal spanning set of a vector space\(V\). That is,\(S\)has the property that if a vector is removed from\(S\), then the new set will no longer span\(V\). Prove that\(S\)must be a basis for\(V\).

Define by \(T\left( {\mathop{\rm p}\nolimits} \right) = \left( {\begin{array}{*{20}{c}}{{\mathop{\rm p}\nolimits} \left( 0 \right)}\\{{\mathop{\rm p}\nolimits} \left( 1 \right)}\end{array}} \right)\). For instance, if \({\mathop{\rm p}\nolimits} \left( t \right) = 3 + 5t + 7{t^2}\), then \(T\left( {\mathop{\rm p}\nolimits} \right) = \left( {\begin{array}{*{20}{c}}3\\{15}\end{array}} \right)\).

  1. Show that \(T\) is a linear transformation. (Hint: For arbitrary polynomials p, q in \({{\mathop{\rm P}\nolimits} _2}\), compute \(T\left( {{\mathop{\rm p}\nolimits} + {\mathop{\rm q}\nolimits} } \right)\) and \(T\left( {c{\mathop{\rm p}\nolimits} } \right)\).)
  2. Find a polynomial p in \({{\mathop{\rm P}\nolimits} _2}\) that spans the kernel of \(T\), and describe the range of \(T\).

In Exercise 1, find the vector x determined by the given coordinate vector \({\left( x \right)_{\rm B}}\) and the given basis \({\rm B}\).

1. \({\rm B} = \left\{ {\left( {\begin{array}{*{20}{c}}{\bf{3}}\\{ - {\bf{5}}}\end{array}} \right),\left( {\begin{array}{*{20}{c}}{ - {\bf{4}}}\\{\bf{6}}\end{array}} \right)} \right\},{\left( x \right)_{\rm B}} = \left( {\begin{array}{*{20}{c}}{\bf{5}}\\{\bf{3}}\end{array}} \right)\)

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