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Chapter 2: Q22Q (page 93)

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.)

Short Answer

Expert verified

The columns of an\(n \times n\)matrix Aspan\({\mathbb{R}^{\bf{n}}}\)when Ais invertible because the equation\(A{\bf{x}} = {\bf{b}}\)has a unique solution for each b.

Step by step solution

01

Write the algorithm for obtaining \({A^{ - 1}}\)

The inverse of an\(m \times m\)matrix A can be computed using the augmented matrix\(\left( {\begin{aligned}{*{20}{c}}A&I\end{aligned}} \right)\), where\(I\)is theidentity matrix. Matrix Ahas inverse only if \(\left( {\begin{aligned}{*{20}{c}}A&I\end{aligned}} \right)\) is row equivalent to \(\left( {\begin{aligned}{*{20}{c}}I&{{A^{ - 1}}}\end{aligned}} \right)\).

02

Explain the invertible and matrix span \({\mathbb{R}^{\bf{n}}}\)

For each value of vector bin the equation,\(A{\bf{x}} = {\bf{b}}\)has aunique solution if the matrix A is invertible. So, the columns of an\(n \times n\)matrix Amust span\({\mathbb{R}^{\bf{n}}}\).

Thus, the columns of an\(n \times n\)matrix Aspan\({\mathbb{R}^{\bf{n}}}\)when Aisinvertible.

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Most popular questions from this chapter

Prove Theorem 2(d). (Hint: The \(\left( {i,j} \right)\)- entry in \(\left( {rA} \right)B\) is \(\left( {r{a_{i1}}} \right){b_{1j}} + ... + \left( {r{a_{in}}} \right){b_{nj}}\).)

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 \(E = \left( {\begin{aligned}{*{20}{c}}{ - 5}\\3\end{aligned}} \right)\)

\(A + 2B\), \(3C - E\), \(CB\), \(EB\).

Describe in words what happens when you compute \({A^{\bf{5}}}\), \({A^{{\bf{10}}}}\), \({A^{{\bf{20}}}}\), and \({A^{{\bf{30}}}}\) for \(A = \left( {\begin{aligned}{*{20}{c}}{1/6}&{1/2}&{1/3}\\{1/2}&{1/4}&{1/4}\\{1/3}&{1/4}&{5/12}\end{aligned}} \right)\).

If A, B, and X are \(n \times n\) invertible matrices, does the equation \({C^{ - 1}}\left( {A + X} \right){B^{ - 1}} = {I_n}\) have a solution, X? If so, find it.

Suppose \(\left( {B - C} \right)D = 0\), where Band Care \(m \times n\) matrices and \(D\) is invertible. Show that B = C.
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