Question

Explain why the columns of an $n\times n$ matrix Aspan ${\mathbb{R}}^{\mathbf{n}}$ when

Ais invertible. (Hint:Review Theorem 4 in Section 1.4.)

Short answer

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

Step-by-step solution

Write the algorithm for obtaining $A^{-1}$

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

Explain the invertible and matrix span ${\mathbb{R}}^{\mathbf{n}}$

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

Thus, the columns of an$n\times n$matrix Aspan${\mathbb{R}}^{\mathbf{n}}$when Aisinvertible.