# Le A be an n x n matrix. Prove that A is non-singular if and only if rank(A) = n

My intuition tells me that i have to use the rank nullity theorem

Rank(A) + dim(ker(A)) = n

if Rank(A) = n, then that implies that dim(ker(A)) = 0 which means that the ker(A) only has the zero vector... but how can i use this information to prove that A is non-singular/invertible?

This is just a guess on my part, but if Rank(A) is n, then that must mean that the row reduced echelon form(RREF) of A is the identity matrix because the rank of A is the number of NON ZERO rows in the RREF(A), and Rank(A) = n implies that there are no zero rows in RREF(A)

This is as much as i can do on my own

(Please note, i have looked at other posts in Math Exchange that ask the same question, but the answers are too abstract and confusing for me, i just need someone to explain this in steps and ,if possible, to explain this in a more simple way)

• Do you know the defintion of a non-singular matrix? – Qudit Dec 15 '17 at 21:45
• a matrix that is invertible and has a determinant of 0 – Soon_to_be_code_master Dec 15 '17 at 21:47
• @Soon_to_be_code_master Let's do it without determinant, which involves to many calculations as a multilinear function.Note that in a matrix, row operations doesn't affect the linear (in)dependence of the columns.(you're just operating on two entries,leaving the rest fixed.) Having full (row/column) rank in a square matrix means that you have all columns are linearly independent.So $Ax=0$ can have only trivial solutions.Now, to find the right inverse $A^{-1}$, we do it column by column.In the augmented matrix $[A\mid e_i]$ with $n+1$ column vectors $\Bbb{F}^n$, can it be linearly independent? – GNUSupporter 8964民主女神 地下教會 Dec 15 '17 at 21:49
• @Soon_to_be_code_master "A matrix that is invertible and has determinant of 0" Such a matrix does not exist. A matrix that is invertbile has nonzero determinant and a matrix with determinant zero is non-invertible. Your question has to do with a part of the invertible matrix theorem. It is unclear which results you have already proven and what tools you have available to you unless you tell us yourself. – JMoravitz Dec 15 '17 at 22:03
• You cite the rank-nullity theorem. How about $dim(\ker(A))=0\iff A$ is one to one (injective)? How about $Rank(A)=n\iff A$ is surjective (for $A$ a linear transformation from $\Bbb R^m$ to $\Bbb R^n$)? How do you have "non-singular matrix" defined? (your earlier definition is incorrect). How do you have "invertible matrix" defined? – JMoravitz Dec 15 '17 at 22:05

If $\ker A=\{0\}$, then the matrix is injective, and since its image is $n$ dimensional, it is surjective, so it is invertible.

Then $$1=\det(I)=\det (AA^{-1})=\det(A)\det(A^{-1})$$

so $\det(A) \neq 0$.

• My professor did not talk about matrices being either injective or surjective. But the way that you explained this, it seems to me that a matrix must be both surjective and injective to be invertible. Can you elaborate on why a matrix is injective if its kernel only has the zero vector and on why the matrix is surjective if its image is n dimensional? – Soon_to_be_code_master Dec 16 '17 at 0:37
• What is your definition of rank? Are you familiar with linear transformations? – Andres Mejia Dec 16 '17 at 0:46
• my definition of a rank is the number of non zero rows in a row reduced echelon form of a matrix, i'm not too familiar with linear transformations – Soon_to_be_code_master Dec 16 '17 at 1:02

$A$ is non-singular if it is invertable.

If $A$ is invertable it is a bijection.

That is $A\mathbf u = A\mathbf v \iff \bf u=v$ (otherwise how does the matrix inverse map a vector in the image back to the original vector?)

Suppose $A$ has a non-trivial kernel.

Then there is a vector $\bf x \ne 0$ such that $A\bf x = 0$ But $A\bf 0 = 0,$ which means that $A$ is not a bijection, and $A$ is not invertable.

This property, singular matrices have a non-trivial kernel, is frequently very definition of singular, with it being non-invertable and having determinant $= 0$ as a consequence.