Here we prove the following theorem. Theorem. For every linear operator f in Rn with standard matrix A the following conditions are equivalent:

1. f is invertible.
2. A is invertible.
3. f is surjective.
4. f is injective. Proof. Let us translate the third and the fourth conditions of this theorem into the language of matrices.

The third condition (f is surjective) means that for every vector b in Rn there exists a vector v in Rn such that Av=b, that is for every vector b the system of equations Av=b has a solution.

The fourth condition (f is injective) means that for every two vectors x and y

if Ax=Ay then x=y.

We can rewrite this condition in the following form:

if A(x-y)=0 then x-y=0.

Let us denote x-y by v. Since u and v were arbitrary vectors, v is also an arbitrary vector. Then we can write our condition in the following form:

if Av=0 then v=0

that is the homogeneous system of linear equations Av=0 has only the trivial solution.

Thus we can rewrite our theorem in as follows:

Theorem. For every linear operator f in Rn with standard matrix A the following conditions are equivalent:

1. f is invertible.
2. A is invertible.
3. For every vector b in Rn there exists a vector v in Rn such that Av=b.
4. For every vector v in Rn if Av=0 then v=0.

Now it is easy to see that the first two properties in this theorem are equivalent because of the theorem about a connection between invertible matrices and invertible operators. The last 3 conditions are equivalent by the second theorem about inverses. Thus all four conditions are equivalent. The theorem is proved. 