Let \({\mathop{\rm u}\nolimits} = \left[ {\begin{array}{*{20}{c}}1\\2\end{array}} \right]\). Find \({\mathop{\rm v}\nolimits} \) in \({\mathbb{R}^3}\) such that \(\left[ {\begin{array}{*{20}{c}}1&{ - 3}&4\\2&{ - 6}&8\end{array}} \right] = {{\mathop{\rm uv}\nolimits} ^T}\) .

Short Answer

Expert verified

\({\mathop{\rm v}\nolimits} = \left[ {\begin{array}{*{20}{c}}1\\{ - 3}\\4\end{array}} \right]\)

Step by step solution

01

Determine vector \({\mathop{\rm v}\nolimits} \) in \({\mathbb{R}^3}\)

It is noted that the second row of the matrix is twice the first row. Therefore, when \({\mathop{\rm v}\nolimits} = \left( {1, - 3,4} \right)\), which is the first row of the matrix.

\(\begin{array}{c}{{\mathop{\rm uv}\nolimits} ^T} = \left[ {\begin{array}{*{20}{c}}1\\2\end{array}} \right]\left[ {\begin{array}{*{20}{c}}1&{ - 3}&4\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}1&{ - 3}&4\\2&{ - 6}&8\end{array}} \right]\end{array}\)

Thus, \({\mathop{\rm v}\nolimits} = \left[ {\begin{array}{*{20}{c}}1\\{ - 3}\\4\end{array}} \right]\).

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Most popular questions from this chapter

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

  1. Show that if \(B\) is \(n \times p\), then rank\(AB \le {\mathop{\rm rank}\nolimits} A\). (Hint: Explain why every vector in the column space of \(AB\) is in the column space of \(A\).
  2. Show that if \(B\) is \(n \times p\), then rank\(AB \le {\mathop{\rm rank}\nolimits} B\). (Hint: Use part (a) to study rank\({\left( {AB} \right)^T}\).)

Let \({M_{2 \times 2}}\) be the vector space of all \(2 \times 2\) matrices, and define \(T:{M_{2 \times 2}} \to {M_{2 \times 2}}\) by \(T\left( A \right) = A + {A^T}\), where \(A = \left( {\begin{array}{*{20}{c}}a&b\\c&d\end{array}} \right)\).

  1. Show that \(T\)is a linear transformation.
  2. Let \(B\) be any element of \({M_{2 \times 2}}\) such that \({B^T} = B\). Find an \(A\) in \({M_{2 \times 2}}\) such that \(T\left( A \right) = B\).
  3. Show that the range of \(T\) is the set of \(B\) in \({M_{2 \times 2}}\) with the property that \({B^T} = B\).
  4. Describe the kernel of \(T\).

If a\({\bf{6}} \times {\bf{3}}\)matrix A has a rank 3, find dim Nul A, dim Row A, and rank\({A^T}\).

(M) Determine whether w is in the column space of \(A\), the null space of \(A\), or both, where

\({\mathop{\rm w}\nolimits} = \left( {\begin{array}{*{20}{c}}1\\1\\{ - 1}\\{ - 3}\end{array}} \right),A = \left( {\begin{array}{*{20}{c}}7&6&{ - 4}&1\\{ - 5}&{ - 1}&0&{ - 2}\\9&{ - 11}&7&{ - 3}\\{19}&{ - 9}&7&1\end{array}} \right)\)

In Exercise 7, find the coordinate vector \({\left( x \right)_{\rm B}}\) of x relative to the given basis \({\rm B} = \left\{ {{b_{\bf{1}}},...,{b_n}} \right\}\).

7. \({b_{\bf{1}}} = \left( {\begin{array}{*{20}{c}}{\bf{1}}\\{ - {\bf{1}}}\\{ - {\bf{3}}}\end{array}} \right),{b_{\bf{2}}} = \left( {\begin{array}{*{20}{c}}{ - {\bf{3}}}\\{\bf{4}}\\{\bf{9}}\end{array}} \right),{b_{\bf{3}}} = \left( {\begin{array}{*{20}{c}}{\bf{2}}\\{ - {\bf{2}}}\\{\bf{4}}\end{array}} \right),x = \left( {\begin{array}{*{20}{c}}{\bf{8}}\\{ - {\bf{9}}}\\{\bf{6}}\end{array}} \right)\)

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