Hyperplanes of ##M_n(\mathbb{C})##

[geoffrey159]

Homework Statement

Show that any hyperplane of $M_n(\mathbb{C})$ contains at least an invertible matrix

Homework Equations

Let $H$ be an hyperplane of $M_n(\mathbb{C})$.

The Attempt at a Solution

By contradiction, assume that for any matrix $A\in H$, $A$ is not invertible.
Therefore 0 is an eigenvalue of A, and there exists a basis of $M_{n,1}(\mathbb{C})$ in which at least the first column of $A$ is 0.
This can be translated to : there is a surjection $\phi$ between the vector space $U = \{ M\in M_n(\mathbb{C}) : \forall i = 1...n\ m_{i1} = 0 \}$ and hyperplane $H$.

My problem to obtain a contradiction is that $\phi$ is not linear. If it was I would use the rank theorem and
$n^2 - 1 = \text{dim}(H) = \text{rk}(\phi) \le \text{dim}(U) = n^2 -n$
which is a contradiction as soon as $n\ge 2$.

How would you solve this ?

 

[S.G. Janssens]

By contradiction, assume that for any matrix $A\in H$, $A$ is not invertible.
Therefore 0 is an eigenvalue of A, and there exists a basis of $M_{n,1}(\mathbb{C})$ in which at least the first column of $A$ is 0

To me it is not immediately clear that there exists a basis that accomplishes this simultaneously for all $A \in H$.

How would you solve this ?

I would perhaps start by noting that there exists a nonzero $c \in \mathbb{C}^{n \times n}$ such that $H$ is linearly isomorphic to
$$
\left\{a \in \mathbb{C}^{n \times n}\,:\,\sum_{j = 1}^n\sum_{i=1}^nc_{ij}\overline{a_{ij}} = 0 \right\}
$$
(Every hyperplane can be written as the kernel of some non-zero functional.) Since $c$ is non-zero, you could then solve for one of the entries of $a$ while keeping the freedom to choose the other $n^2 - 1$ entries at will.

Does this help?

 

[geoffrey159]

We can write $M_n(\mathbb{C}) = H \bigoplus \mathbb{C} M_0$, where $H$ is the kernel of a linear form $\phi : M_n(\mathbb{C}) \to \mathbb{C}$, and $\phi(M_0) \neq 0$

Assuming that $H$ doesn't contain any invertible matrix, the family $(M_i)_{i\in I}$ of all invertible matrices has the form : $M_i = A_i + \lambda_i M_0$, with $\lambda_i \neq 0$.

But that means that for all $i\in I$, $\phi(\lambda_i^{-1} M_i) = \phi(M_0) \neq 0$. This implies that for all $i\in I$, $\lambda_i^{-1} M_i \in \mathbb{C}M_0$

But the restriction of $\phi$ to $\mathbb{C}M_0$ is a bijection onto $\mathbb{C}$. Since the $(\phi(\lambda_i ^{-1} M_i))_{i\in I}$ are all equal to $\phi(M_0)$, that means by injectivity that all invertible matrices should be proportional 2 by 2. This is absurd.

Is it correct?

 

[fresh_42]

But that means that for all $i\in I$, $\phi(\lambda_i^{-1} M_i) = \phi(M_0) \neq 0$. This implies that for all $i\in I$, $\lambda_i^{-1} M_i \in \mathbb{C}M_0$

It implies $\lambda_i^{-1} M_i = \lambda_i^{-1} A_i + M_0 ∉ H$, i.e. its projection on $\mathbb{C}M_0$ equals $1$ and not that it is included in $\mathbb{C}M_0$.
I think that you somehow need to use the fact that $GL_n(ℂ) ⊂ M_n(ℂ)$ is a dense subset, i.e. you probably need to wiggle a little bit on a point $M$ in $H$ to find an invertible point $M+εM'$ without leaving $H$.

Edit: IMO induction on n seems to be the easiest way.

 

[geoffrey159]

It implies $\lambda_i^{-1} M_i = \lambda_i^{-1} A_i + M_0 ∉ H$, i.e. its projection on $\mathbb{C}M_0$ equals $1$ and not that it is included in $\mathbb{C}M_0$.

Doomed ! Once again

 

[Samy_A]

Since $c$ is non-zero, you could then solve for one of the entries of $a$ while keeping the freedom to choose the other $n^2 - 1$ entries at will.

(bolding mine)
Without loss of generality, you can assume $a_{11}$ is that one entry.

 

[geoffrey159]

I think I have it :

Let $(E_i)_{i = 1...n^2-1}$ be a basis of $H$. Then the matrices $M_\lambda = E_i + \lambda E_j$ are in $H$.
But $P(\lambda) = \det M_\lambda$ is a polynomial of $\mathbb{C}_n[X]$.
Therefore it has at most $n$ zeros and there is a complex $\lambda_0$ for which $P(\lambda_0)\neq 0$.
Meaning that $M_{\lambda_0} \in H$ is invertible.

Is it good now?

 

[Samy_A]

I think I have it :

Let $(E_i)_{i = 1...n^2-1}$ be a basis of $H$. Then the matrices $M_\lambda = E_i + \lambda E_j$ are in $H$.
But $P(\lambda) = \det M_\lambda$ is a polynomial of $\mathbb{C}_n[X]$.
Therefore it has at most $n$ zeros and there is a complex $\lambda_0$ for which $P(\lambda_0)\neq 0$.
Meaning that $M_{\lambda_0} \in H$ is invertible.

Is it good now?

$E_1=\begin{pmatrix} 1 & 0\\ 0 & 0 \end{pmatrix}$, $E_2=\begin{pmatrix} 0& 1\\ 0 & 0 \end{pmatrix}$
What is the polynomial $P(\lambda)$?

 

[geoffrey159]

Hmmm I get it, but really, I'm starting to dry up with this problem !

 

[Samy_A]

Hmmm I get it, but really, I'm starting to dry up with this problem !

Look again at @Krylov 's tip and its implications.

 

[geoffrey159]

I did not understand what he said

 

[Samy_A]

I did not understand what he said

Since H is a hyperplane, the n² matrix elements of matrices in H satisfy a linear equation. Solve this equation for one the elements (without loss of generality, say $a_{11}$): that element is then a linear combination of all the others. To get a matrix in H, you then have n²-1 degrees of freedom, meaning you can chose n²-1 matrix elements freely. The last one will then be a linear combination of these freely chosen n²-1 elements.

 

[geoffrey159]

I'm sorry, but this is above my head, I don't follow you.
If you know how to solve it I'll be glad to read your solution. But it has already taken me too much time and I need to move on.
Really sorry.

Thread closed, unless someone posts a solution

 

[Samy_A]

Is $A=\begin{pmatrix} x & 0 & 0 & 0 & 1\\ 0 & 0 & 0 & 1 & 0\\ 0 & 0 & 1 & 0 & 0\\ 0 & 1 & 0 & 0 & 0\\ 1 & 0 & 0 & 0 & 0 \end{pmatrix}$ invertible?

 

[S.G. Janssens]

I did not understand what he said

Don't be afraid to ask me next time. @Samy_A, thank you for clarifying.

I'm sorry, but this is above my head, I don't follow you.
If you know how to solve it I'll be glad to read your solution. But it has already taken me too much time and I need to move on.
Really sorry.

Thread closed, unless someone posts a solution

I hope you are willing to let it rest a bit (maybe for a few days) and then reconsider with a fresh mind. Sometimes it takes a while to solve an exercise, don't give up on it.