(1) [math]\Rightarrow[/math] (10): Obvious from the definition of invertible matrices.[br][br](10) [math]\Rightarrow[/math] (4): Suppose [math]CA=I[/math]. Let [math]x[/math] be a vector such that [math]Ax=0[/math]. Then we have [br][math](CA)x=Ix\Rightarrow C(Ax)=x \Rightarrow C0=x \Rightarrow x=0[/math] [br]That is, [math]Ax=0[/math] has only the trivial solution.[br][br](4) [math]\Rightarrow[/math] (3): (4) implies that there are no free variables when [math]A[/math] is row reduced to the one in echelon form, which means that [math]A[/math] has n pivot positions.[br][br](3) [math]\Rightarrow[/math] (2): Suppose [math]A[/math] has n pivot positions. The matrix in reduced echelon form obtained by the row reduction algorithm must be the n x n identity matrix.[br][br](2) [math]\Rightarrow[/math] (1): Assume (2), there exists a sequence of elementary matrices [math]E_1, E_2, \ldots, E_p[/math] such that [math]E_p\cdots E_1A=I[/math]. Hence [math](E_1^{-1}\cdots E_p^{-1})(E_p\cdots E_1)A=E_1^{-1}\cdots E_p^{-1}I[/math], which implies that [math]A=E_1^{-1}\cdots E_p^{-1}[/math]. Since the product of any elementary matrices is invertible, [math]A[/math] is an invertible matrix.

(1) [math]\Rightarrow[/math] (11): Obvious from the definition of invertible matrices.[br][br](11) [math]\Rightarrow[/math] (7): Assume (11), [math]AD=I[/math]. Then, for any [math]b[/math] in [math]\mathbb{R}^n[/math], we have[br][math](AD)b=Ib=b \Rightarrow A(Db)=b[/math][br]That is, [math]Db[/math] is a solution to the equation [math]Ax=b[/math].[br][br](7) [math]\Rightarrow [/math] (1): Assume (7). That is to say, the linear system corresponding to [math]Ax=b[/math] is consistent for any [math]b[/math]. Use the row reduction algorithm to transform the augmented matrix [math]\left( A \ | \ b \right)[/math] to the one in reduced echelon form. Assume there are fewer than n pivot positions, the bottom row of the augmented matrix in reduced echelon form must look like [math]\left( 0 \cdots 0 \ | \ \ast \right)[/math], where [math]\ast[/math] is a real number depending on the choice of [math]b[/math]. Since [math]b[/math] can be any column vector, we can choose it such that [math]\ast[/math] is a nonzero number i.e. contradicting the fact that the linear system is always consistant for any [math]b[/math]. Hence, [math]A[/math] must have n pivot positions i.e. (2) is true, which implies (1) by the previous result.

(7) [math]\Leftrightarrow[/math] (9): Obvious[br][br](7) [math]\Leftrightarrow[/math] (8): Let [math]A=\left( \mathbf{a_1} \ \mathbf{a_2} \ \cdots \ \mathbf{a_n} \right)[/math]. Then [math]Ax=b \Leftrightarrow x_1\mathbf{a_1}+x_2\mathbf{a_2}+\cdots +x_n\mathbf{a_n}=b[/math] i.e [math]b[/math] is a linear combination of the column vectors of [math]A[/math]. Therefore, (7) [math]\Leftrightarrow \text{Span}\left\{\mathbf{a_1},\mathbf{a_2}, \ldots, \mathbf{a_n}\right\}=\mathbb{R}^n \Leftrightarrow[/math] (8).[br][br](4) [math]\Leftrightarrow[/math] (5) [math]\Leftrightarrow[/math] (6): See [url=https://www.geogebra.org/m/mjvfnzmb]here[/url][br][br](1) [math]\Leftrightarrow[/math] (12): If [math]A[/math] is invertible, we already proved that [math]A^T[/math] is also invertible and [math](A^T)^{-1}=(A^{-1})^T[/math]. Conversely, if [math]A^T[/math] is invertible, by what we have just proved, [math](A^T)^T[/math] is invertible. But [math](A^T)^T=A[/math]. Hence [math]A[/math] is invertible.[br]