Chapter 2: Q17Q (page 93)
If \(A\) is invertible, then the columns of \({A^{ - {\bf{1}}}}\) are linearly independent. Explain why?
The columns of the inverse matrix are linearly independent.
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Explain why the columns of an \(n \times n\) matrix Aspan \({\mathbb{R}^{\bf{n}}}\) when
Ais invertible. (Hint:Review Theorem 4 in Section 1.4.)
A useful way to test new ideas in matrix algebra, or to make conjectures, is to make calculations with matrices selected at random. Checking a property for a few matrices does not prove that the property holds in general, but it makes the property more believable. Also, if the property is actually false, you may discover this when you make a few calculations.
38. Use at least three pairs of random \(4 \times 4\) matrices Aand Bto test the equalities \({\left( {A + B} \right)^T} = {A^T} + {B^T}\) and \({\left( {AB} \right)^T} = {A^T}{B^T}\). (See Exercise 37.) Report your conclusions. (Note:Most matrix programs use \(A'\) for \({A^{\bf{T}}}\).
Suppose the transfer function W(s) in Exercise 19 is invertible for some s. It can be showed that the inverse transfer function \(W{\left( s \right)^{ - {\bf{1}}}}\), which transforms outputs into inputs, is the Schur complement of \(A - BC - s{I_n}\) for the matrix below. Find the Sachur complement. See Exercise 15.
\(\left[ {\begin{array}{*{20}{c}}{A - BC - s{I_n}}&B\\{ - C}&{{I_m}}\end{array}} \right]\)
In exercises 11 and 12, mark each statement True or False. Justify each answer.
a. The definition of the matrix-vector product \(A{\bf{x}}\) is a special case of block multiplication.
b. If \({A_{\bf{1}}}\), \({A_{\bf{2}}}\), \({B_{\bf{1}}}\), and \({B_{\bf{2}}}\) are \(n \times n\) matrices, \[A = \left[ {\begin{array}{*{20}{c}}{{A_{\bf{1}}}}\\{{A_{\bf{2}}}}\end{array}} \right]\] and \(B = \left[ {\begin{array}{*{20}{c}}{{B_{\bf{1}}}}&{{B_{\bf{2}}}}\end{array}} \right]\), then the product \(BA\) is defined, but \(AB\) is not.
In Exercises 1–9, assume that the matrices are partitioned conformably for block multiplication. Compute the products shown in Exercises 1–4.
1. \(\left[ {\begin{array}{*{20}{c}}I&{\bf{0}}\\E&I\end{array}} \right]\left[ {\begin{array}{*{20}{c}}A&B\\C&D\end{array}} \right]\)
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