Question

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

Short answer

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

Step-by-step solution

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 \(\left( {\begin{array}{*{20}{c}}I&{{A^{ - 1}}}\end{array}} \right)\).

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\({\mathbb{R}^n}\).