# How to show path-connectedness of $GL(n,\mathbb{C})$

Well, I am not getting any hint how to show $GL_n(\mathbb{C})$ is path connected. So far I have thought that let $A$ be any invertible complex matrix and $I$ be the idenity matrix, I was trying to show a path from $A$ to $I$ then define $f(t)=At+(1-t)I$ for $t\in[0,1]$ which is possible continous except where the $\operatorname{det}{f(t)}=0$ i.e. which has $n$ roots and I can choose a path in $\mathbb{C}\setminus\{t_1,\dots,t_n\}$ where $t_1,\dots,t_n$ are roots of $\operatorname{det}{f(t)}=0$, is my thinking was correct? Could anyone tell me the solution?

• This path can fail in liying on $GL_n(\mathbb{C})$. For example, if you take $A=-I$, at the middle point $t=1/2$ you will get $f(t)=0$ May 1, 2012 at 19:45
• +1 Looks good to me. @matgaio, the idea seems to be that you go around that troublesome $t=1/2$ - there is room for that, when you let $t$ be a complex number! It might have been clearer to define $f(t)$ for all complex $t$, exclude the finite set of points, and then select a path from $0$ to $1$. May 1, 2012 at 19:47
• Uhm, ok. I was understanding he was trying to go straight from $A$ to $I$. May 1, 2012 at 19:50
• This thread is slightly relevant. May 1, 2012 at 20:02

• If $P$ is a polynomial of degree $n$, the set $\{\lambda,P(\lambda)\neq 0\}$ is path connected (because its complement is finite, so you can pick a polygonal path).
• Let $P(t):=\det(A+t(I-A))$. We have that $P(0)=\det A\neq 0$, and $P(1)=\det I=1\neq 0$, so we can find a path $\gamma\colon[0,1]\to\mathbb C$ such that $\gamma(0)=0$, $\gamma(1)=1$, and $P(\gamma(t))\neq 0$ for all $t$. Finally, put $\Gamma(t):=A+\gamma(t)(I-A)$.
• If $B_1$ and $B_2$ are two invertible matrices, consider $\gamma(t):=B_2\cdot\gamma(t)$, where we chose $\gamma$ for $A:=B_2^{-1}B_1$.
• Your argument prove that if $f$ is any polynomial (even holomorphic) function on $\bf C^n$, the $f\not=0$ is path -connected. Feb 21, 2016 at 7:14
• Hi Davide, $\gamma(t)$ is in $\mathbb C$, but $B_2$ is a matrix. What do you mean by $\gamma(t):=B_2\cdot\gamma(t)$? And I am also confused because $\gamma$ appears on both left and right sides. Thanks. Apr 28, 2021 at 15:14
• @SamWong This must be a typo. What I think Davide meant to write is $\gamma(t) = B_2 \cdot \Gamma(t)$, where $\Gamma$ is a path between $B_2^{-1}B_1$ and $I$. Note that here $\gamma$ is not as it was in Davide's second bullet point - it would probably be better to use another Greek letter, but there you go. Jan 4, 2022 at 17:57

Use $\Gamma(t) = e^{t \log A + (1-t) \log B}$. This is well defined since $A,B$ are invertible. $\Gamma(t)$ is clearly invertible for all $t$, $\Gamma(0) = B$, $\Gamma(1) = A$.

• This idea is not mine, I just can't remember where I saw it, it was in a control theory context. May 1, 2012 at 20:20
• what is $log A$? Feb 21, 2016 at 7:11
• @Thomas: Any matrix $B$ such that $e^B = A$ will do, to show the existence, pick a branch of $\log$ (call it $l$) that does not pass through any eigenvalue and take a curve $\gamma$ that encircles each eigenvalue once while remaining in the domain of $l$, then let $\log A = {1 \over 2 \pi i} \int_\gamma l(z) (zI-A)^{-1} dz$. Feb 21, 2016 at 17:28
• Now I remember, a professor of mine, Shankar Sastry, showed this to me. Apr 8, 2018 at 5:57
• How would you connect matrix with determinant +1 to one of -1? Apr 13, 2018 at 14:04

Since any matrix $A\in GL_n(\mathbb C)$ has only finitely many eigenvalues, and 0 isn't one of them, there is a point $z\in S^1$ such that the line through the origin containing $z$ doesn't intersect any of the eigenvalues of $A$. Now, consider the path $f(t)=At+z(1-t)I$. This has determinant 0 iff $z(t-1)$ is an eigenvalue of $At$, which happens iff $z(1-1/t)$ is an eigenvalue of $A$ (this doesn't work when $t=0$, but then it is clear that the determinant is non-zero). By construction, it isn't for any $t\in[0,1]$ so this defines a path form $A$ to $zI$. now there is a path not passing through 0 from $z$ to 1, and this gives rise to a path from $zI$ to $I$, and so concatenating the two paths, we get a path from $A$ to $I$, showing that $GL_n(\mathbb C)$ is path connected.

Following idea shows connectedness of $Gl(n,\mathbb{C})$ ..

As $A\in Gl(n,\mathbb{C})$ you do have an upper triangular matrix which is similar to $A$.

See subgroup of Invertible, Upper triangular matrices as $$\underbrace{\mathbb{C}^*\times\mathbb{C}^*\times \cdots\times\mathbb{C}^*}_{n- times}\times \underbrace{\mathbb{C}\times\mathbb{C}\times \mathbb{C}\times \cdots \times\mathbb{C}}_{\dfrac{n(n-1)}{2}times}$$ As $\mathbb{C}^*\times\mathbb{C}^*\times \cdots\times\mathbb{C}^*$ is connected and $\mathbb{C}$ is connected so is the above product.

See that conjugation is continuous so preserves connectedness.

By which i mean that $\{BUB^{-1} : U - \text{Upper Triangular}\}$ is connected.

$Gl(n,\mathbb{C})$ being union of connected sets is also connected.

See that all these conjugates have Identity matrix in common.

• But $\Bbb{C}^*$ is connected.
– user297008
Apr 14, 2017 at 9:14
• @iota yes, you are correct. I don't know why I have said so before.thanks
– user87543
Apr 14, 2017 at 12:22

I think the answer by Praphulla Koushik has the (to me) most elementary approach, but it is awkwardly formulated, so I will try to restate it more simply. It suffices to check two statements:

1. The group $B$ of invertible upper triangular matrices is path-connected.

2. The union of conjugates of $B$ (each of which contains $I$) equals $GL(n,\Bbb C)$.

Then every $A\in GL(n,\Bbb C)$ can be joined to $I$ by a path that stays within a conjugate of $B$ that contains$~A$.

For point 1. it suffices to observe that $B$ is homeomorphic to the Cartesian product of the group $T$ of invertible diagonal matrices and the set $U$ of strictly upper triangular matrices; the former is a power of $\Bbb C^*$, which is path-connected, and the latter to a power of$~\Bbb C$.

Point 2. follows from the well known fact that any complex matrix is trigonalisable. This is equivalent to saying that every linear operator$~T$ on $\Bbb C^n$ admits a complete flag of $T$-stable subspaces. To find one, one chooses a left-eigenvector (a linear form on$~\Bbb C^n$) of$~T$, whose kernel is a $T$-stable hyperplane$~H$, and applies induction on the dimension to the restriction of $T$ to$~H$ to find the remaining $T$-stable subspaces for the flag.

By singular value decomposition and the fact that every unitary matrix is unitarily diagonalisable, we can write every invertible matrix $A$ as a product of the form $(US_1U^\ast)S_2(VS_3V^\ast)$, where $U,V$ are unitary matrices and $S_1,S_2,S_3$ are invertible diagonal matrices. So, it suffices to show that each such diagonal matrix is path-connected to $I$, and this boils down to the scalar case, which is trivial.