Why this argument does not make sense on connected components of $GL_n(\mathbb R)$?

Question

Let $S \in GL_n(\mathbb R)$ with $\det(S) > 0$. Suppose $\gamma: [0,1] \to GL_n(\mathbb R)$ is a continuous path from $S$ to $I$. $\gamma$ exists since $GL_n(\mathbb R)_+$ is connected. Suppose $S$ has a negative eigenvalue. Since $t \mapsto \gamma(t)$ is continuous, there are $n$ continuous functions $\lambda_j: [0,1] \to \mathbb C$ such that for each $t$, the spectrum of $\gamma(t)$ is equal to $\{\lambda_j(t)\}$. I am thinking: for some $j$, we should have $\lambda_j(\gamma(0)) = \lambda_j(S) < 0$ and $\lambda_j(\gamma(1) ) = 1 > 0$. This says at some $t \in [0,1]$, the eigenvalue is $0$ and so there should not exist a path. But this is a clearly wrong argument since we know $GL_n(\mathbb R)_+$ is connected. However, I could not figure out why this is wrong.

Answer 1

Let's consider a concrete example: $n=2$, $S=-I$, and $\gamma(t) =\pmatrix{-\cos \pi t&\sin \pi t\\-\sin \pi t&-\cos \pi t}$. Where is there a matrix with determinant zero here?

Answer 2

$\lambda_1$ is a real number. Moreover, $\lambda_1 \circ \gamma$ is as you mentioned a map from $[0,1]$ to $GL_n(\mathbb R)$. So saying that $\lambda_1(S)<0$ doesn’t make sense.