### Composantes connexes de \(GL_n(\mathbb{R})\)

In

on Euclidean norms, we needed to know that the set of matrices with determinant \(>0\) is connected. Here's a reminder of the proof:
**Lemma: **Let \(GL_n^+(\mathbb{R})=\{M\in M_n(\mathbb{R}),\det M >0\}\) and \(GL_n^-(\mathbb{R})=\{M\in M_n(\mathbb{R}),\det M <0\}\).

\(GL_n^+(\mathbb{R})\) and \(GL_n^-(\mathbb{R})\) are connected sets.

Find a path from \(I_n\) to \(T_{ij}(\lambda)=\)

Decompose an invertible matrix using the Gauss pivot method.

Conclude.

We can already see that \(GL_n(\mathbb{R})\) is not connected, since \(\det(.)\) is a continuous application, and \(\det(GL_n(\mathbb{R}))=\mathbb{R}^*\) which is not a connected subset of \(\mathbb{R}\).

Consider matrices:

\(T_{ij}(\lambda) = I_n + \lambda E_{ij}=\)

where \(\lambda\in\mathbb{R}\)

and

\( D_{i}(\lambda)=I_n + (-1+\lambda)E_{ii} =\)

where \(\lambda\in\mathbb{R}^*\).

Multiplying on the left by \(T_{ij}(\lambda)\) gives the same operation on the rows: \[ L_i\leftarrow L_i+\lambda L_j \] Multiplying on the right gives the corresponding operation on the columns.

Multiplying on the left by \(D_{ij}(\lambda)\) gives the same operation on the rows: \[ L_i\leftarrow \lambda L_i \] Multiplying on the right gives the corresponding operation on the columns.

Two lines can be interchanged by multiplying on the left by: \[ T_{ij}(-1)T_{ji}(1)T_{ij}(-1) \]

###### Decomposition of a matrix of \(GL_n^+(\mathbb{R})\) into a product of transvection matrices.

Let \(M\in GL^+_n(\mathbb{R})\). Let \(k\in\lBrack 1,n\rBrack\) be such that \(m_{k1}\neq0\).

We use the Gauss pivot method. We multiply by on the right by \(T_{ik}(\frac{m_{11}-1}{m_{k1}})\) so as to have a 1 in the upperleft position. Then for \(l\in\lBrack 2,n\rBrack\), multiply by \(T_{l1}(-m_{l1})\) to place 0's throughout the first column.

Then multiply on the right by \(T_{k1}(-m_{1l})\) to cancel the first line except for the pivot.

We can therefore find matrices \(T\) such that:

\[T_nT_{n-1}\dots T_1MT'_1\dots T'_{n-1}=\]

Repeat until a the following matrix is obtained:

\(\prod_i T_i\times M \times\prod_j T'_j=\)
\(=D_n(\delta)\)

where \(\delta=\det M>0\)

Since the matrices \(T_{ij}\) are invertible (\(T_{ij}(\lambda)^{-1}=T_{ij}(-\lambda)\)), \(M\) decomposes as: \[M=\prod_i T_i\times D_n(\delta)\times\prod_j T_j\]

###### Continuous path from \(I_n\) to a matrix of \(GL_n^+(\mathbb{R})\).

Now we notice that the following matrix, of determinant 1, provides a continuous path from \(I_n\) to \(T_{ij}(\lambda)\):

The following matrix \(D_n(1-t+t\delta)\) allows you to go continuously from \(I_n\) to \(D_n(\lambda)\):

We have \(\det(D_n(1-t+t\delta))\in |1,\delta|\subset \mathbb{R}^{+*}\).

Thus, given the decomposition obtained, by multiplying the above matrices we can construct a continuous path from \(I_n\) to \(M\), included in \(GL_n^+(\mathbb{R})\).

###### General case.

Let \(M_1,M_2\in GL_n(\mathbb{R})\) be two matrices whose determinants have the same sign.

The matrix \(M_2M_1^{-1}\) is therefore in \(GL_n^+(\mathbb{R})\) and we can find a matrix \(M(t)\) such that:

- \(t\mapsto M(t)\) continues.
- \(\forall t\in[0,1],\quad M(t)\in GL_n^+(\mathbb{R})\).
- \(M(0)=I_n\) and \(M(1)=M_2M_1^{-1}\)

We then see that \(t\mapsto M(t)M_1\) is a continuous path from \(M_1\) to \(M_2\).