The Transpose of a Nonsingular Matrix is Nonsingular

Linear algebra problems and solutions

Problem 558

Let $A$ be an $n\times n$ nonsingular matrix.

Prove that the transpose matrix $A^{\trans}$ is also nonsingular.

 
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Definition (Nonsingular Matrix).

By definition, $A^{\trans}$ is a nonsingular matrix if the only solution to
\[A^{\trans}\mathbf{x}=\mathbf{0}\] is the zero vector $\mathbf{x}=\mathbf{0}$ in $\R^n$.

Proof.

Suppose that
\[A^{\trans}\mathbf{v}=\mathbf{0} \tag{*}\] for an $n$-dimensional vector $\mathbf{v}$.
We prove that $\mathbf{v}=\mathbf{0}$.


Since $A$ is nonsingular, there exists a vector $\mathbf{u}\in \R^n$ such that
\[A\mathbf{u}=\mathbf{v}.\] (For those who know the inverse matrix, the vector $\mathbf{u}$ is given by $\mathbf{u}=A^{-1}\mathbf{v}$. The fact can be proved without using the inverse matrix, though.)

Hence we obtain from (*)
\[A^{\trans}A\mathbf{u}=\mathbf{0}.\] It follows that
\[\mathbf{u}^{\trans}A^{\trans}A\mathbf{u}=\mathbf{u}^{\trans}\mathbf{0}=0.\] By the property of the transpose, we have
\[\mathbf{u}^{\trans}A^{\trans}=(A\mathbf{u})^{\trans}.\] Thus we have
\[0=(A\mathbf{u})^{\trans}A\mathbf{u}=\|A\mathbf{u}\|^2.\] This yields the length $\|A\mathbf{u}\|=0$, and hence $A\mathbf{u}=\mathbf{0}$.
Since $\mathbf{v}=A\mathbf{u}$, we conclude that $\mathbf{v}=\mathbf{0}$.


Therefore if $A^{\trans}\mathbf{v}=\mathbf{0}$, then $\mathbf{v}=\mathbf{0}$.
This shows that the transpose $A^{\trans}$ is nonsingular.


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