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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

Show that block upper triangular matrix \(A\) in Example 5is invertible if and only if both \({A_{{\bf{11}}}}\) and \({A_{{\bf{12}}}}\) are invertible. [Hint: If \({A_{{\bf{11}}}}\) and \({A_{{\bf{12}}}}\) are invertible, the formula for \({A^{ - {\bf{1}}}}\) given in Example 5 actually works as the inverse of \(A\).] This fact about \(A\) is an important part of several computer algorithims that estimates eigenvalues of matrices. Eigenvalues are discussed in chapter 5.

In Exercises 1鈥�9, assume that the matrices are partitioned conformably for block multiplication. Compute the products shown in Exercises 1鈥�4.

3. \[\left[ {\begin{array}{*{20}{c}}{\bf{0}}&I\\I&{\bf{0}}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}W&X\\Y&Z\end{array}} \right]\]

In Exercises 27 and 28, view vectors in \({\mathbb{R}^n}\)as\(n \times 1\)matrices. For \({\mathop{\rm u}\nolimits} \) and \({\mathop{\rm v}\nolimits} \) in \({\mathbb{R}^n}\), the matrix product \({{\mathop{\rm u}\nolimits} ^T}v\) is a \(1 \times 1\) matrix, called the scalar product, or inner product, of u and v. It is usually written as a single real number without brackets. The matrix product \({{\mathop{\rm uv}\nolimits} ^T}\) is a \(n \times n\) matrix, called the outer product of u and v. The products \({{\mathop{\rm u}\nolimits} ^T}{\mathop{\rm v}\nolimits} \) and \({{\mathop{\rm uv}\nolimits} ^T}\) will appear later in the text.

28. If u and v are in \({\mathbb{R}^n}\), how are \({{\mathop{\rm u}\nolimits} ^T}{\mathop{\rm v}\nolimits} \) and \({{\mathop{\rm v}\nolimits} ^T}{\mathop{\rm u}\nolimits} \) related? How are \({{\mathop{\rm uv}\nolimits} ^T}\) and \({\mathop{\rm v}\nolimits} {{\mathop{\rm u}\nolimits} ^T}\) related?

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

Show that if the columns of Bare linearly dependent, then so are the columns of AB.
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