Question envolving linear algebra, Hyperplane and $GL(n,F)$ I need to prove that $H \cap GL(n,F) \neq {\emptyset}$, where $H$ is a hyperplane of $F^{n \times n}$. My professor gave me some tips. 
a)Suppose that $H \cap GL(n,F) = \emptyset$, then prove that $\forall N \notin H$, there are $M\in H$ and $\lambda \in F$ s.t $I = M + \lambda N$.
b)Prove that every nilpotent matriz in $H$.
Ok, item a):
Let $N \notin H$, since $H$ is a hyperplane, we have $F^{n \times n} = H \oplus <N>$, every matrix $A \in F^{n \times n }$ can be written by $A = M + \lambda N$ where $M \in H$ and $\lambda N \in <N>$. In particular $A = I$.
b) For this item, suppose that there is a nilpotent matrix $A \notin H$. Hence, by a) $I = M+ \lambda A$, for some $M \in H$ and $\lambda \in F $. 
Since $M \in H$ and $H \cap Gl(n,F) = \emptyset$, we have $0=\det(M) = \det(I- \lambda A)$, that is $\lambda $ is eigenvalue of $A$. But $A$ is nilpotent, then $\lambda = 0$. So we have a contradiction.
Therefore, if $A$ is nilpotent, then $A \in H$.
Ok, I've proved this itens, but how can I conclude that $H \cap Gl(n,F) \neq \emptyset$??
 A: I don't know exactly what your professor had in mind, so let me propose a possible solution. First of all, notice that the statement is false for $n = 1$ (since then $H  =\{0\}$). Let thus $n \geq 2$. Consider the matrices
$$A_1 := \begin{pmatrix}0 & 1 \\ 1 & 0\end{pmatrix} \text{ and } A_2 := \begin{pmatrix}0 & 1 & 1 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix}.$$
You can easily check that these two matrices are invertible, since their respective determinants are $-1$ and $1$. Now, define a matrix $A$ as follows. If $n$ is even, let $A$ be the block diagonal matrix $$A = \mathrm{diag}(\underbrace{A_1,...,A_1}_{n/2 \text{ times}}),$$ and if $n$ is odd, then $n \geq 3$ and define $$A := \mathrm{diag}(A_2,\underbrace{A_1,...,A_1}_{(n-3)/2 \text{ times}}).$$ In any case, $A$ is invertible, since it is block diagonal and its blocks are invertible.
To conclude, as you showed, $H$ contains every nilpotent matrix. Thus, observe that $H$ contains the $n^2-n$ matrices $E_{ij}$ for $i \neq j$ whose $(i,j)$-th entry is $1$ and the other entries are $0$ (all of the matrices $E_{ij}$ are nilpotent). Of course, the matrix $A$ can be written as a linear combination of the $E_{ij}$'s (since the diagonal entries of $A$ are $0$). Hence, since $H$ is a linear subspace, $A \in H$.
