1
$\begingroup$

Suppose that $A$ is a $n \times n$ matrix over $\mathbb R$ such that $\mathrm {rank} (A)=n$ then I know that $A$ is of full row rank and consequently has full column rank since row rank of $A$ is same as the column rank of $A$. Which is same as saying that the column vectors of $A$ are linearly independent and cosequently the row vectors of $A$ are also so. From here how can I formally prove that $A$ is non-singular i.e. $\det (A) \neq 0$? A trick just came into my mind which is suppose $T : \mathbb R^n \longrightarrow \mathbb R^n$ be a map defined by $T(X)=AX$, $X \in \mathbb R^n$. Then $T$ can be easily proved to be a linear operator on $\mathbb R^n$. Now using rank-nullity-dimension theorem we have $\mathrm{rank} (T) + \mathrm{nullity} (T) = \dim \mathbb R^n$. Now $\mathrm{rank} (T) = \mathrm{column\ rank} (A) =\mathrm{rank} (A)=n$ and $\dim \mathbb R^n=n$. So we have $\mathrm{nullity} (T) = 0$. But $\mathrm{nullity} (T) = \mathrm{nullity} (A)$. So $\mathrm{nullity} (A) = 0$. This implies that the homogeneous system of equations $AX=0$ has only one solution which is precisely the zero solution.

From here how can I prove that $A$ is non-singular? Please help me.

Thank you in advance.

EDIT $:$

There is an another way of looking into it. We know that row equivalent matrices have the same row space. So if $A$ is row equivalent to the row reduced echelon matrix $R$. Then from the above fact it follows that $\mathrm{rank} (R)= \mathrm{rank} (A)=n$. Now if the row reduced echelon matrix have the full rank does it have any choice other than the identity matrix? I'm thinking a bit. Probably not because if leading non-zero element in the $i$-th non-zero row occurs in the $k_i$-th column for $1 \leq i \leq n$. Then we must have $k_1<k_2<k_3<\cdots<k_n$. Since there are exactly $n$ columns of $A$ we should have $k_i=i$ for each $1 \leq i \leq n$. But then all other entries in the $i$-th column should be equal to zero since $R$ is the row reduced echelon matrix. Since leading non-zero entries are all $1$ so this gives rise to a identity matrix. So $A$ is row equivalent to a identity matrix. Hence there exist a invertible matrix $P$ such that $I=PA$ or in other words $A=P^{-1}$ proving that $A$ is invertible and consequently $A$ is non-singular.

$\endgroup$
  • 1
    $\begingroup$ Since the nullity of $A$ is $0$, you know that $A$ is one-to-one. Since $A$ has rank $n$, it is onto. Together these give that $A$ is invertible. $\endgroup$ – John Griffin Oct 4 '17 at 6:36
  • $\begingroup$ What's your definition of "non-singular"? To me it looks like you're pretty much done, but that may depend on that I use another definition - that singular means having non-trivial null-space. $\endgroup$ – skyking Oct 4 '17 at 6:59
  • $\begingroup$ I have assumed that non-singular means $\det (A) \neq 0$. $\endgroup$ – user251057 Oct 4 '17 at 7:01
1
$\begingroup$

Your map $T$ is surjectve, since its image is generated by the columns of $A$. Since $T$ is an endomorphism of $\mathbb{R}^n$, which is finite-dimensional, $T$ is injective too (I know you proved this, by my way is far shorter) and therefore an isomorphism. And an endomorphism of $\mathbb{R}^n$ is an isomorphism if and only if the determinant of its matrix with respect to the canonical basis (that is, the matrix $A$) is not $0$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy