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Chapter 2: Q2.9-25E (page 93)
Let A be an \(n \times p\) matrix whose column space is p-dimensional. Explain why the column A must be linearly independent.
The p-columns form a basis for the column space of A if the vectors are linearly independent.
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In Exercises 1鈥�9, assume that the matrices are partitioned conformably for block multiplication. In Exercises 5鈥�8, find formulas for X, Y, and Zin terms of A, B, and C, and justify your calculations. In some cases, you may need to make assumptions about the size of a matrix in order to produce a formula. [Hint:Compute the product on the left, and set it equal to the right side.]
6. \[\left[ {\begin{array}{*{20}{c}}X&{\bf{0}}\\Y&Z\end{array}} \right]\left[ {\begin{array}{*{20}{c}}A&{\bf{0}}\\B&C\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}I&{\bf{0}}\\{\bf{0}}&I\end{array}} \right]\]
Show that if ABis invertible, so is B.
In Exercises 1鈥�9, assume that the matrices are partitioned conformably for block multiplication. Compute the products shown in Exercises 1鈥�4.
2. \[\left[ {\begin{array}{*{20}{c}}E&{\bf{0}}\\{\bf{0}}&F\end{array}} \right]\left[ {\begin{array}{*{20}{c}}A&B\\C&D\end{array}} \right]\]
In Exercises 1 and 2, compute each matrix sum or product if it is defined. If an expression is undefined, explain why. Let
\(A = \left( {\begin{aligned}{*{20}{c}}2&0&{ - 1}\\4&{ - 5}&2\end{aligned}} \right)\), \(B = \left( {\begin{aligned}{*{20}{c}}7&{ - 5}&1\\1&{ - 4}&{ - 3}\end{aligned}} \right)\), \(C = \left( {\begin{aligned}{*{20}{c}}1&2\\{ - 2}&1\end{aligned}} \right)\), \(D = \left( {\begin{aligned}{*{20}{c}}3&5\\{ - 1}&4\end{aligned}} \right)\) and \(E = \left( {\begin{aligned}{*{20}{c}}{ - 5}\\3\end{aligned}} \right)\)
\( - 2A\), \(B - 2A\), \(AC\), \(CD\).
Suppose Tand Ssatisfy the invertibility equations (1) and (2), where T is a linear transformation. Show directly that Sis a linear transformation. [Hint: Given u, v in \({\mathbb{R}^n}\), let \[{\mathop{\rm x}\nolimits} = S\left( {\mathop{\rm u}\nolimits} \right),{\mathop{\rm y}\nolimits} = S\left( {\mathop{\rm v}\nolimits} \right)\]. Then \(T\left( {\mathop{\rm x}\nolimits} \right) = {\mathop{\rm u}\nolimits} \), \[T\left( {\mathop{\rm y}\nolimits} \right) = {\mathop{\rm v}\nolimits} \]. Why? Apply Sto both sides of the equation \(T\left( {\mathop{\rm x}\nolimits} \right) + T\left( {\mathop{\rm y}\nolimits} \right) = T\left( {{\mathop{\rm x}\nolimits} + y} \right)\). Also, consider \(T\left( {cx} \right) = cT\left( x \right)\).]
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