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

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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.




linear-algebra abstract-algebra general-topology connectedness




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  • 1




    How is the function $lambda _1$ defined?
    – Suzet
    Jul 22 at 6:07











  • In $Bbb C$, not every path from $-1$ to $1$ passes through $0$.
    – Lord Shark the Unknown
    Jul 22 at 7:02










  • I see. This solves my confusion. Thanks.
    – user9527
    Jul 22 at 7:10














up vote
0
down vote

favorite












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.




linear-algebra abstract-algebra general-topology connectedness




share|cite|improve this question

















  • 1




    How is the function $lambda _1$ defined?
    – Suzet
    Jul 22 at 6:07











  • In $Bbb C$, not every path from $-1$ to $1$ passes through $0$.
    – Lord Shark the Unknown
    Jul 22 at 7:02










  • I see. This solves my confusion. Thanks.
    – user9527
    Jul 22 at 7:10












up vote
0
down vote

favorite









up vote
0
down vote

favorite











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.




linear-algebra abstract-algebra general-topology connectedness




share|cite|improve this 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.





linear-algebra abstract-algebra general-topology connectedness





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edited Jul 22 at 6:17
























asked Jul 22 at 6:04









user9527

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  • 1




    How is the function $lambda _1$ defined?
    – Suzet
    Jul 22 at 6:07











  • In $Bbb C$, not every path from $-1$ to $1$ passes through $0$.
    – Lord Shark the Unknown
    Jul 22 at 7:02










  • I see. This solves my confusion. Thanks.
    – user9527
    Jul 22 at 7:10












  • 1




    How is the function $lambda _1$ defined?
    – Suzet
    Jul 22 at 6:07











  • In $Bbb C$, not every path from $-1$ to $1$ passes through $0$.
    – Lord Shark the Unknown
    Jul 22 at 7:02










  • I see. This solves my confusion. Thanks.
    – user9527
    Jul 22 at 7:10







1




1




How is the function $lambda _1$ defined?
– Suzet
Jul 22 at 6:07





How is the function $lambda _1$ defined?
– Suzet
Jul 22 at 6:07













In $Bbb C$, not every path from $-1$ to $1$ passes through $0$.
– Lord Shark the Unknown
Jul 22 at 7:02




In $Bbb C$, not every path from $-1$ to $1$ passes through $0$.
– Lord Shark the Unknown
Jul 22 at 7:02












I see. This solves my confusion. Thanks.
– user9527
Jul 22 at 7:10




I see. This solves my confusion. Thanks.
– user9527
Jul 22 at 7:10










2 Answers
2






active

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up vote
0
down vote



accepted










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?






share|cite|improve this answer





















  • Thanks. I understand we can define a path this way. But I could not see why the argument in question was wrong. Could you point out where the question was fundamentally wrong?
    – user9527
    Jul 22 at 6:18


















up vote
1
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$lambda_1$ 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.






share|cite|improve this answer





















  • I reformulated the question. Does it make sense now?
    – user9527
    Jul 22 at 6:17










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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
0
down vote



accepted










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?






share|cite|improve this answer





















  • Thanks. I understand we can define a path this way. But I could not see why the argument in question was wrong. Could you point out where the question was fundamentally wrong?
    – user9527
    Jul 22 at 6:18















up vote
0
down vote



accepted










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?






share|cite|improve this answer





















  • Thanks. I understand we can define a path this way. But I could not see why the argument in question was wrong. Could you point out where the question was fundamentally wrong?
    – user9527
    Jul 22 at 6:18













up vote
0
down vote



accepted







up vote
0
down vote



accepted






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?






share|cite|improve this answer













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?







share|cite|improve this answer













share|cite|improve this answer



share|cite|improve this answer











answered Jul 22 at 6:10









Lord Shark the Unknown

85.2k950111




85.2k950111











  • Thanks. I understand we can define a path this way. But I could not see why the argument in question was wrong. Could you point out where the question was fundamentally wrong?
    – user9527
    Jul 22 at 6:18

















  • Thanks. I understand we can define a path this way. But I could not see why the argument in question was wrong. Could you point out where the question was fundamentally wrong?
    – user9527
    Jul 22 at 6:18
















Thanks. I understand we can define a path this way. But I could not see why the argument in question was wrong. Could you point out where the question was fundamentally wrong?
– user9527
Jul 22 at 6:18





Thanks. I understand we can define a path this way. But I could not see why the argument in question was wrong. Could you point out where the question was fundamentally wrong?
– user9527
Jul 22 at 6:18











up vote
1
down vote













$lambda_1$ 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.






share|cite|improve this answer





















  • I reformulated the question. Does it make sense now?
    – user9527
    Jul 22 at 6:17














up vote
1
down vote













$lambda_1$ 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.






share|cite|improve this answer





















  • I reformulated the question. Does it make sense now?
    – user9527
    Jul 22 at 6:17












up vote
1
down vote










up vote
1
down vote









$lambda_1$ 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.






share|cite|improve this answer













$lambda_1$ 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.







share|cite|improve this answer













share|cite|improve this answer



share|cite|improve this answer











answered Jul 22 at 6:11









mathcounterexamples.net

24.1k21653




24.1k21653











  • I reformulated the question. Does it make sense now?
    – user9527
    Jul 22 at 6:17
















  • I reformulated the question. Does it make sense now?
    – user9527
    Jul 22 at 6:17















I reformulated the question. Does it make sense now?
– user9527
Jul 22 at 6:17




I reformulated the question. Does it make sense now?
– user9527
Jul 22 at 6:17












 

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