Hyperplane of Matrices Let $M(n \times n, \mathbb{C})$ be vector space of complex matrices $n \times n$. Show that every hyperplane in this space has a invertible matrix.
For to solve this problem I use the fact that the invertible matrices are dense in $M(n \times n, \mathbb{C})$. Given any matrix $A$, if we get any neighborhood $V$ containing A there is at least a intertible matrix in $V$. But, I hope a solution using only basic Linear Algebra. Someone can help me? Thanks. 
 A: Let $e_1,\ldots,e_n$ be the canonical basis of $\mathbb{C}^n$.
Define $N_{kl}=e_ke_l^t$ and $B_{kl}=ie_ke_l^t+ie_le_k^t+e_ke_k^t-e_le_l^t$, where $i$ is the imaginary unit. Notice that $tr(N_{kl})=tr(B_{kl})=0$ and $N_{kl}^2=B_{kl}^2=0$, if $k\neq l$.
Consider the set $N=\{N_{kl}\ |\ k\neq l\}\cup\{B_{12},\ldots,B_{1n}\}$. Notice that $N$ is a linear independent set. Therefore $N$ is a basis for the subspace of matrices with trace zero.
If a $(n^2-1)-$ dimensional subspace contains $N$  then it contains the subspace of matrices with trace zero. Now, we can always find an invertible complex matrix with trace zero.
If a $(n^2-1)-$ dimensional subspace $H$ does not contain $N$. Let $A\in N\setminus H$. Define $W$ the vector space spanned by $A$ and $Id$. Since $W\cap H \neq 0$ (by a dimensionality argument) then there is $aId+bA\in H\setminus \{0\}$. If $a=0$ then $bA\in H$ and $A\in H$. Absurd.  Thus, $a\neq 0$ and $(aId+bA)(\frac{1}{a}Id-\frac{b}{a^2}A)=Id$. 
Now, let $H_1$ be a hyperplane. There is a $(n^2-1)-$ dimensional subspace $H_0$ and a matrix $A$ such that $H_1=\{A+X,\ X \in H_0 \}$. We have just proved that there is an invertible matrix $B\in H_0$. Now, $\det(A-\lambda B)=\det(B)\det(B^{-1}A-\lambda Id).$ If $\lambda$ is not an eigenvalue of $B^{-1}A$ then $\det(A-\lambda B)\neq 0$. Notice that $A-\lambda B\in H_1$.
