Is it true that given a matrix $A_{m\times n}$, $A$ is regular / invertible if and only if $m=n$ and $A$ is a basis in $\mathbb{R}^n$?
Seems so to me, but I haven't seen anything in my book yet that says it directly.
Is it true that given a matrix $A_{m\times n}$, $A$ is regular / invertible if and only if $m=n$ and $A$ is a basis in $\mathbb{R}^n$?
Seems so to me, but I haven't seen anything in my book yet that says it directly.
$\mathrm{rk}(A_{m \times n}) \leq \min(m, n)$, from which follows $\mathrm{rk}(A_{m \times n}B_{n \times m}) \leq \min(m, n)$, while $\mathrm{rk}(E_{m \times m}) = m$, so, obviously, $m$ should be equal to $n$ in order for $A$ to be invertible.
Also, if the strings (or columns) of $A_{n \times n}$ do not form the basis in ${\mathbb R}^n$, it means that some of the strings / columns are linearly dependent, and thus \mathrm{rk}(A) < n, $|A| = 0$, so that A is not invertible.
From the other hand, if the strings (or columns) of $A_{n \times n}$ do form the basis, it means that they are linearly independent, and $|A| \ne 0$. In that case, you can find $A^{-1}$ by the Cramer's rule.
Aren't both statements simply saying that A has full rank?