Proof of Maximal Left Ideal Generation in the Ring Mn(F)

[xixi]

Let R = Mn(F) the ring consists of all n*n matrices over a field F and E = E11 + E22 + ... + En-1,n-1, where Eii is the elementary matrix( Eij is a matrix whose ij th element is 1 and the others are 0). Then the
following hold:
1.Every matrix of rank n-1 in any maximal left ideal generates the maximal left ideal itself . 2 .Moreover, the number of matrices in every maximal left ideal that can be a generator is the same as the number of matrices in the maximal left ideal RE11 +· · ·+REn−1,n−1 that can be a generator. 3.Furthermore why any maximal left ideal has a rank n-1 matrix.
what is the proof of above statements ? I really need help , it's so important for me.
Here is a hint for it but it seems incorrect as I will explain in the following.
(hint:
1. RE is a maximal left ideal.
2. If A is a rank n-1 matrix in R then A is equivalent to E, so that there are invertible matrices P and Q such that A = PEQ. Hence, RA=RPEQ=REQ=RE is a maximal left ideal. (note that if B is invertible and I is a left ideal (resp. a right ideal), then BI = I (resp. IB = I)).
From 1 and 2 we know that any rank n-1 matrix in R generates a maximal
left ideal. )
Since RE is a left ideal so for an invertible matrix Q , we have QRE=RE (as stated above) and not REQ=RE . so the red equivalence is incorrect .
So what is the correct proof?
THanks

 

[fresh_42]

$1.$ There cannot be a matrix of full rank, since these are units. Now given any matrix $M$ of rank $n-1$, then we can find a regular matrix $A$ such that $A\cdot M=A\cdot E \in RE$ (invert the regular block in $M$ and shift the redundant row at the end) and $M$ generates $RE$, which is the only maximal left ideal, and ...
$2.$ ... every matrix of rank $n-1$ is a generator, which are those which generates $RE_{1,1}+\ldots + RE_{n-1,n-1}\,.$
$3.$ Say we have an ideal $I$ with matrices of rank at most $n-2$. Then we can add a rank $n-1$ matrix $A$ such that $I \subsetneq I +RA \subsetneq R$ and $I$ cannot be maximal.