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Chapter 2: Q26Q (page 93)

Explain why the columns of \({A^{\bf{2}}}\) span \({\mathbb{R}^n}\) whenever the columns of Aare linearly independent.

Short Answer

Expert verified

The columns of \({A^2}\) are invertible, and they span \({\mathbb{R}^n}\).

Step by step solution

01

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

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

02

Determine the linear independence of the matrix

The columns of the\(n \times n\)matrix Aare linearly independent, so it must have n pivot columns or pivots (no free variable). This matrix can be row reduced to an identity matrix.

Matrix A and the identity matrix I are row equivalent. This shows that matrix A is invertible.

Thus, the columns of\({A^2}\)are also invertible, and they span\({\mathbb{R}^n}\).

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Most popular questions from this chapter

Find the inverse of the matrix \(\left( {\begin{aligned}{*{20}{c}}{\bf{3}}&{ - {\bf{4}}}\\{\bf{7}}&{ - {\bf{8}}}\end{aligned}} \right)\).

Let \(A = \left( {\begin{aligned}{*{20}{c}}1&1&1\\1&2&3\\1&4&5\end{aligned}} \right)\), and \(D = \left( {\begin{aligned}{*{20}{c}}2&0&0\\0&3&0\\0&0&5\end{aligned}} \right)\). Compute \(AD\) and \(DA\). Explain how the columns or rows of A change when A is multiplied by D on the right or on the left. Find a \(3 \times 3\) matrix B, not the identity matrix or the zero matrix, such that \(AB = BA\).

(M) Read the documentation for your matrix program, and write the commands that will produce the following matrices (without keying in each entry of the matrix).

  1. A \({\bf{5}} \times {\bf{6}}\) matrix of zeros
  2. A \({\bf{3}} \times {\bf{5}}\) matrix of ones
  3. The \({\bf{6}} \times {\bf{6}}\) identity matrix
  4. A \({\bf{5}} \times {\bf{5}}\) diagonal matrix, with diagonal entries 3, 5, 7, 2, 4

When a deep space probe launched, corrections may be necessary to place the probe on a precisely calculated trajectory. Radio elementary provides a stream of vectors, \({{\bf{x}}_{\bf{1}}},....,{{\bf{x}}_k}\), giving information at different times about how the probe鈥檚 position compares with its planned trajectory. Let \({X_k}\) be the matrix \(\left[ {{x_{\bf{1}}}.....{x_k}} \right]\). The matrix \({G_k} = {X_k}X_k^T\) is computed as the radar data are analyzed. When \({x_{k + {\bf{1}}}}\) arrives, a new \({G_{k + {\bf{1}}}}\) must be computed. Since the data vector arrive at high speed, the computational burden could be serve. But partitioned matrix multiplication helps tremendously. Compute the column-row expansions of \({G_k}\) and \({G_{k + {\bf{1}}}}\) and describe what must be computed in order to update \({G_k}\) to \({G_{k + {\bf{1}}}}\).

1. Find the inverse of the matrix \(\left( {\begin{aligned}{*{20}{c}}{\bf{8}}&{\bf{6}}\\{\bf{5}}&{\bf{4}}\end{aligned}} \right)\).
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