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Suppose \({A^n} = {\bf{0}}\) for some \({\bf{n}} > {\bf{1}}\). Find an inverse for \(I - A\).

Short Answer

Expert verified

The inverse of \(I - A\) is \(I + A + {A^2} + ... + {A^{n - 1}}\).

Step by step solution

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01

Use the matrix computation in Exercise 3

That is \(\left( {I - A} \right)\left( {I + A + {A^2}} \right) = I - {A^3}\). In general,

\(\left( {I - A} \right)\left( {I + A + {A^2} + ... + {A^{n - 1}}} \right) = I - {A^n}\)

02

Suppose \({A^n} = {\bf{0}}\)

Substitute \({A^n} = 0\) in the above equation to obtain:

\(\left( {I - A} \right)\left( {I + A + {A^2} + ... + {A^{n - 1}}} \right) = I\)

This implies \(I - A\) is invertibleand \({\left( {I - A} \right)^{ - 1}} = I + A + {A^2} + ... + {A^{n - 1}}\)

03

Conclusion

Hence, theinverse of \(I - A\) is \(I + A + {A^2} + ... + {A^{n - 1}}\).

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Most popular questions from this chapter

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 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?

In Exercise 9 mark each statement True or False. Justify each answer.

9. a. In order for a matrix B to be the inverse of A, both equations \(AB = I\) and \(BA = I\) must be true.

b. If A and B are \(n \times n\) and invertible, then \({A^{ - {\bf{1}}}}{B^{ - {\bf{1}}}}\) is the inverse of \(AB\).

c. If \(A = \left( {\begin{aligned}{*{20}{c}}a&b\\c&d\end{aligned}} \right)\) and \(ab - cd \ne {\bf{0}}\), then A is invertible.

d. If A is an invertible \(n \times n\) matrix, then the equation \(Ax = b\) is consistent for each b in \({\mathbb{R}^{\bf{n}}}\).

e. Each elementary matrix is invertible.

If \(A = \left( {\begin{aligned}{*{20}{c}}1&{ - 2}\\{ - 2}&5\end{aligned}} \right)\) and \(AB = \left( {\begin{aligned}{*{20}{c}}{ - 1}&2&{ - 1}\\6&{ - 9}&3\end{aligned}} \right)\), determine the first and second column of B.

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

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