Chapter 4: Q8E (page 191)
Suppose a \({\bf{5}} \times {\bf{6}}\) matrix A has four pivot columns. What is dim Nul A? Is \({\bf{Col}}\,A = {\mathbb{R}^{\bf{3}}}\)? Why or why not?
dim Nul A=2, \({\rm{Col}}\,A \ne {\mathbb{R}^4}\)
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Justify the following equalities:
a.\({\rm{dim Row }}A{\rm{ + dim Nul }}A = n{\rm{ }}\)
b.\({\rm{dim Col }}A{\rm{ + dim Nul }}{A^T} = m\)
Define a linear transformation by \(T\left( {\mathop{\rm p}\nolimits} \right) = \left( {\begin{array}{*{20}{c}}{{\mathop{\rm p}\nolimits} \left( 0 \right)}\\{{\mathop{\rm p}\nolimits} \left( 0 \right)}\end{array}} \right)\). Find \(T:{{\mathop{\rm P}\nolimits} _2} \to {\mathbb{R}^2}\)polynomials \({{\mathop{\rm p}\nolimits} _1}\) and \({{\mathop{\rm p}\nolimits} _2}\) in \({{\mathop{\rm P}\nolimits} _2}\) that span the kernel of T, and describe the range of 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?
If a \({\bf{3}} \times {\bf{8}}\) matrix A has a rank 3, find dim Nul A, dim Row A, and rank \({A^T}\).
Let \(B = \left\{ {\left( {\begin{array}{*{20}{c}}{\bf{1}}\\{ - {\bf{4}}}\end{array}} \right),\,\left( {\begin{array}{*{20}{c}}{ - {\bf{2}}}\\{\bf{9}}\end{array}} \right)\,} \right\}\). Since the coordinate mapping determined by B is a linear transformation from \({\mathbb{R}^{\bf{2}}}\) into \({\mathbb{R}^{\bf{2}}}\), this mapping must be implemented by some \({\bf{2}} \times {\bf{2}}\) matrix A. Find it. (Hint: Multiplication by A should transform a vector x into its coordinate vector \({\left( {\bf{x}} \right)_B}\)).
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