An injective or surjective square matrix is bijective

Question

Is the argument below valid?

Let $A$ a square matrix $n \times n$. Suppose $A$ is injective, ie $Ker(A) = {0}$. Therefore the columns of $A$ are linearly independent. We have $n$ vectors that are linearly independent, which means the set of vectors (the columns of A) are a basis of $R^n$. Therefore, all vectors $\in R^n$ are a (unique) linear combination of the columns of $A$. Therefore, $A$ has a unique solution to all vectors in $R^n$. Therefore, $Im(A) = R^n$. Therefore $A$ is surjective. Therefore $A$ is bijective.

Let $A$ a square matrix $n \times n$. Suppose $A$ is surjective, ie $Im(A) = R^n$. Therefore every element of $R^n$ is a linear combination of the columns of $A$. We have $n$ columns, meaning a generating set of $R^n$ of $n$ vectors. Therefore the columns of $A$ must be linearly independent. Therefore $Ker(A) = {0}$. Therefore $A$ is injective. Therefore $A$ is bijective.

Answer 1

Your argument is correct. Here are a few others.

If $A$ is injective:

• you have $\ker A=0$, so $A$ has all nonzero eigenvalues. Thus $\det A\ne0$ and so $A$ is invertible.

• if $e_1,\ldots,e_n$ is a basis, then using that $A$ is injective it is easy to show that $Ae_1,\ldots,Ae_n$ are linearly independent, and so $A$ is surjective.

If $A$ is surjective:

• Given a basis $f_1,\ldots,f_n$, there exist elements $e_1,\ldots,e_n$ with $Ae_k=f_k$ for all $k$. The linearity of $A$ implies then that $e_1,\ldots,e_n$ are linearly independent. The linear map $B$ induced by $f_k\longmapsto e_k$ is then an inverse for $A$.