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Which of the subspaces \({\rm{Row }}A\) , \({\rm{Col }}A\), \({\rm{Nul }}A\), \({\rm{Row}}\,{A^T}\) , \({\rm{Col}}\,{A^T}\) , and \({\rm{Nul}}\,{A^T}\) are in \({\mathbb{R}^m}\) and which are in \({\mathbb{R}^n}\) ? How many distinct subspaces are in this list?.

Short Answer

Expert verified

There are four distinct subspaces in the list.

Step by step solution

01

Assume an arbitrary system and relate the given statement with it

Assume that the system has matrix A of the size \(m \times n\). So, The null space in \(A\) must be the subspace of \({\mathbb{R}^n}\); the rows in \(A\) must be the subspace of \({\mathbb{R}^n}\); thecolumns in \(A\) must be the subspace of \({\mathbb{R}^m}\).

As the matrix \({A^T}\) has the size \(n \times m\), the null space in \({A^T}\) is a subspace of \({\mathbb{R}^m}\); the rows in \({A^T}\) are the subspace of \({\mathbb{R}^m}\); the columns in \({A^T}\) are the subspace of \({\mathbb{R}^n}\).

02

Draw a conclusion

As \({\rm{Row}}\,{A^T} = {\rm{Col}}\,A\) and \({\rm{Row}}\,A = {\rm{Col}}\,{A^T}\) , there are four possible distinct subspaces— \({\rm{Row}}\,A\),\({\rm{Col}}\,A\), \({\rm{Nul}}\,A\), \({\rm{Nul}}\,{A^T}\).

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Most popular questions from this chapter

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

If the null space of A \({\bf{7}} \times {\bf{6}}\) matrix A is 4-dimensional, what is the dimension of the column space of A?

Question: Determine if the matrix pairs in Exercises 19-22 are controllable.

20. \(A = \left( {\begin{array}{*{20}{c}}{.8}&{ - .3}&0\\{.2}&{.5}&1\\0&0&{ - .5}\end{array}} \right),B = \left( {\begin{array}{*{20}{c}}1\\1\\0\end{array}} \right)\).

(M) Let \(H = {\mathop{\rm Span}\nolimits} \left\{ {{{\mathop{\rm v}\nolimits} _1},{{\mathop{\rm v}\nolimits} _2}} \right\}\) and \(K = {\mathop{\rm Span}\nolimits} \left\{ {{{\mathop{\rm v}\nolimits} _3},{{\mathop{\rm v}\nolimits} _4}} \right\}\), where

\({{\mathop{\rm v}\nolimits} _1} = \left( {\begin{array}{*{20}{c}}5\\3\\8\end{array}} \right),{{\mathop{\rm v}\nolimits} _2} = \left( {\begin{array}{*{20}{c}}1\\3\\4\end{array}} \right),{{\mathop{\rm v}\nolimits} _3} = \left( {\begin{array}{*{20}{c}}2\\{ - 1}\\5\end{array}} \right),{{\mathop{\rm v}\nolimits} _4} = \left( {\begin{array}{*{20}{c}}0\\{ - 12}\\{ - 28}\end{array}} \right)\)

Then \(H\) and \(K\) are subspaces of \({\mathbb{R}^3}\). In fact, \(H\) and \(K\) are planes in \({\mathbb{R}^3}\) through the origin, and they intersect in a line through 0. Find a nonzero vector w that generates that line. (Hint: w can be written as \({c_1}{{\mathop{\rm v}\nolimits} _1} + {c_2}{{\mathop{\rm v}\nolimits} _2}\) and also as \({c_3}{{\mathop{\rm v}\nolimits} _3} + {c_4}{{\mathop{\rm v}\nolimits} _4}\). To build w, solve the equation \({c_1}{{\mathop{\rm v}\nolimits} _1} + {c_2}{{\mathop{\rm v}\nolimits} _2} = {c_3}{{\mathop{\rm v}\nolimits} _3} + {c_4}{{\mathop{\rm v}\nolimits} _4}\) for the unknown \({c_j}'{\mathop{\rm s}\nolimits} \).)

[M] Let \(A = \left[ {\begin{array}{*{20}{c}}7&{ - 9}&{ - 4}&5&3&{ - 3}&{ - 7}\\{ - 4}&6&7&{ - 2}&{ - 6}&{ - 5}&5\\5&{ - 7}&{ - 6}&5&{ - 6}&2&8\\{ - 3}&5&8&{ - 1}&{ - 7}&{ - 4}&8\\6&{ - 8}&{ - 5}&4&4&9&3\end{array}} \right]\).

  1. Construct matrices \(C\) and \(N\) whose columns are bases for \({\mathop{\rm Col}\nolimits} A\) and \({\mathop{\rm Nul}\nolimits} A\), respectively, and construct a matrix \(R\) whose rows form a basis for Row\(A\).
  2. Construct a matrix \(M\) whose columns form a basis for \({\mathop{\rm Nul}\nolimits} {A^T}\), form the matrices \(S = \left[ {\begin{array}{*{20}{c}}{{R^T}}&N\end{array}} \right]\) and \(T = \left[ {\begin{array}{*{20}{c}}C&M\end{array}} \right]\), and explain why \(S\) and \(T\) should be square. Verify that both \(S\) and \(T\) are invertible.
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