Mixing
How to prove the set of all symmetric matrices with eigenvalues in $(0,2)$ is connected?
$begingroup$
Prove that space $X$ of all symmetric matrices in $GL_2(mathbb R)$ with both the eigenvalues belonging to the interval $(0,2),$ with the topology inherited from $M_2(mathbb R) $ is connected.
Space of all symmetric matrices in $M_2(mathbb R)$ is path -connected.
I was not able to show why $det(lambda A+(1-lambda)B)neq 0$ where $lambda in (0,1)$ and $A,Bin GL_2(mathbb R), A=A^T,B=B^T.$
Also how to use the eigenvalues from $(0,2)$ to prove the connectedness.
I hope my doubts are clear to you.
Any help is appreciated. Thank you.
general-topology matrices connectedness symmetric-matrices
edited Jan 22 at 10:08
José Carlos Santos
166k22132235
asked Jan 22 at 9:56
nurun neshanurun nesha
1,0362623
$endgroup$
add a comment |
$begingroup$
Prove that space $X$ of all symmetric matrices in $GL_2(mathbb R)$ with both the eigenvalues belonging to the interval $(0,2),$ with the topology inherited from $M_2(mathbb R) $ is connected.
Space of all symmetric matrices in $M_2(mathbb R)$ is path -connected.
I was not able to show why $det(lambda A+(1-lambda)B)neq 0$ where $lambda in (0,1)$ and $A,Bin GL_2(mathbb R), A=A^T,B=B^T.$
Also how to use the eigenvalues from $(0,2)$ to prove the connectedness.
I hope my doubts are clear to you.
Any help is appreciated. Thank you.
general-topology matrices connectedness symmetric-matrices
edited Jan 22 at 10:08
José Carlos Santos
166k22132235
asked Jan 22 at 9:56
nurun neshanurun nesha
1,0362623
$endgroup$
add a comment |
$begingroup$
Prove that space $X$ of all symmetric matrices in $GL_2(mathbb R)$ with both the eigenvalues belonging to the interval $(0,2),$ with the topology inherited from $M_2(mathbb R) $ is connected.
Space of all symmetric matrices in $M_2(mathbb R)$ is path -connected.
I was not able to show why $det(lambda A+(1-lambda)B)neq 0$ where $lambda in (0,1)$ and $A,Bin GL_2(mathbb R), A=A^T,B=B^T.$
Also how to use the eigenvalues from $(0,2)$ to prove the connectedness.
I hope my doubts are clear to you.
Any help is appreciated. Thank you.
general-topology matrices connectedness symmetric-matrices
edited Jan 22 at 10:08
José Carlos Santos
166k22132235
asked Jan 22 at 9:56
nurun neshanurun nesha
1,0362623
$endgroup$
Prove that space $X$ of all symmetric matrices in $GL_2(mathbb R)$ with both the eigenvalues belonging to the interval $(0,2),$ with the topology inherited from $M_2(mathbb R) $ is connected.
Space of all symmetric matrices in $M_2(mathbb R)$ is path -connected.
I was not able to show why $det(lambda A+(1-lambda)B)neq 0$ where $lambda in (0,1)$ and $A,Bin GL_2(mathbb R), A=A^T,B=B^T.$
Also how to use the eigenvalues from $(0,2)$ to prove the connectedness.
I hope my doubts are clear to you.
Any help is appreciated. Thank you.
general-topology matrices connectedness symmetric-matrices
general-topology matrices connectedness symmetric-matrices
edited Jan 22 at 10:08
José Carlos Santos
166k22132235
asked Jan 22 at 9:56
nurun neshanurun nesha
1,0362623
edited Jan 22 at 10:08
José Carlos Santos
166k22132235
asked Jan 22 at 9:56
nurun neshanurun nesha
1,0362623
edited Jan 22 at 10:08
José Carlos Santos
166k22132235
edited Jan 22 at 10:08
José Carlos Santos
166k22132235
edited Jan 22 at 10:08
José Carlos Santos
166k22132235
166k22132235
asked Jan 22 at 9:56
nurun neshanurun nesha
1,0362623
asked Jan 22 at 9:56
nurun neshanurun nesha
1,0362623
asked Jan 22 at 9:56
nurun neshanurun nesha
1,0362623
1,0362623
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
The space $SO_2(mathbb{R})$ is connected. Now let$$Lambda=left{begin{bmatrix}lambda_1&0\0&lambda_2end{bmatrix},middle|,lambda_1,lambda_2in(0,2)right}.$$The set $Lambda$ is connected too. So, the range of the map$$begin{array}{ccc}SO_2(mathbb{R})timesLambda&longrightarrow&GL_2(mathbb{R})\(P,D)&mapsto&P^{-1}DPend{array}$$is connected too. But this range is $X$.
answered Jan 22 at 10:07
José Carlos SantosJosé Carlos Santos
166k22132235
$endgroup$
$begingroup$
I had another question too: whether $det(lambda A+(1-lambda ) B)$ equal to $0$ or not? where $lambda in (0,1)$ and $A,B$ are symmetric matrices with determinant non-zero.
$endgroup$
– nurun nesha
Jan 22 at 10:53
1
$begingroup$
Yes, it can be $0$. Suppose that $A=operatorname{Id}$, the $B=-operatorname{Id}$, and that $lambda=frac12$.
$endgroup$
– José Carlos Santos
Jan 22 at 11:01
add a comment |
$begingroup$
We can prove much more (in $S_n(mathbb{R})$ , the set of $ntimes n$ real symmetric matrices).
$textbf{Proposition 1}$. The set $Z_I={Ain S_n(mathbb{R});spectrum(A)subset I}$ is convex (where $I$ is some interval of $mathbb{R}$).
$textbf{Proof}$. We use the fact that if $Ain S_n(mathbb{R})$, then $sup_{||x||_2=1}x^TAx=sup(spectrum(A)),inf_{||x||_2=1}x^TAx=inf(spectrum(A))$.
Then $lambda x^TAx+(1-lambda)x^TBxin I$ when $A,Bin Z_I,||x||_2=1,lambdain [0,1]$. $square$
$textbf{Proposition 2}$. If $I=[a,b],a<b$, then $Z_I$ is homeomorphic to the compact ball $B_{n(n+1)/2}$.
$textbf{Proof}.$ $Z_I$ is also closed (the eigenvalues are continuous functions of the entries) and bounded ($||A||_2=sup_{lambdain spectrum(A)}(|lambda|))$.
Finally, $Z_I$ is convex compact and, therefore, homeomorphic to the compact ball in $mathbb{R}^k$, where $k$ is the number of degrees of freedom of $S_n(mathbb{R})$.
answered Jan 28 at 20:19
loup blancloup blanc
23.6k21851
$endgroup$
add a comment |
$begingroup$
One-line proof: every $Ain X$ is path-connected to $I$ by the line segment ${tA+(1-t)I: 0leq tleq 1}subseteq X$.
Your original approach also works. Let $A,Bin X$ and $0leq tleq 1$. Then $A,B$ and in turn $tA+(1-t)B$ are positive definite and hence all eigenvalues of $tA+(1-t)B$ are positive. Also,
$$
|tA+(1-t)B|_2
leq t|A|_2+(1-t)|B|_2
< 2t+2(1-t)=2.
$$
Therefore all eigenvalues of $tA+(1-t)B$ lie inside the interval $(0,2)$. Consequently, $X$ is path-connected because ${tA+(1-t)B: 0leq tleq 1}subseteq X$ for any $A,Bin X$.
answered Jan 30 at 11:45
user1551user1551
73.5k566129
$endgroup$
add a comment |
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3 Answers
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active
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3 Answers
3
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$begingroup$
The space $SO_2(mathbb{R})$ is connected. Now let$$Lambda=left{begin{bmatrix}lambda_1&0\0&lambda_2end{bmatrix},middle|,lambda_1,lambda_2in(0,2)right}.$$The set $Lambda$ is connected too. So, the range of the map$$begin{array}{ccc}SO_2(mathbb{R})timesLambda&longrightarrow&GL_2(mathbb{R})\(P,D)&mapsto&P^{-1}DPend{array}$$is connected too. But this range is $X$.
answered Jan 22 at 10:07
José Carlos SantosJosé Carlos Santos
166k22132235
$endgroup$
$begingroup$
I had another question too: whether $det(lambda A+(1-lambda ) B)$ equal to $0$ or not? where $lambda in (0,1)$ and $A,B$ are symmetric matrices with determinant non-zero.
$endgroup$
– nurun nesha
Jan 22 at 10:53
1
$begingroup$
Yes, it can be $0$. Suppose that $A=operatorname{Id}$, the $B=-operatorname{Id}$, and that $lambda=frac12$.
$endgroup$
– José Carlos Santos
Jan 22 at 11:01
add a comment |
$begingroup$
The space $SO_2(mathbb{R})$ is connected. Now let$$Lambda=left{begin{bmatrix}lambda_1&0\0&lambda_2end{bmatrix},middle|,lambda_1,lambda_2in(0,2)right}.$$The set $Lambda$ is connected too. So, the range of the map$$begin{array}{ccc}SO_2(mathbb{R})timesLambda&longrightarrow&GL_2(mathbb{R})\(P,D)&mapsto&P^{-1}DPend{array}$$is connected too. But this range is $X$.
answered Jan 22 at 10:07
José Carlos SantosJosé Carlos Santos
166k22132235
$endgroup$
$begingroup$
I had another question too: whether $det(lambda A+(1-lambda ) B)$ equal to $0$ or not? where $lambda in (0,1)$ and $A,B$ are symmetric matrices with determinant non-zero.
$endgroup$
– nurun nesha
Jan 22 at 10:53
1
$begingroup$
Yes, it can be $0$. Suppose that $A=operatorname{Id}$, the $B=-operatorname{Id}$, and that $lambda=frac12$.
$endgroup$
– José Carlos Santos
Jan 22 at 11:01
add a comment |
$begingroup$
The space $SO_2(mathbb{R})$ is connected. Now let$$Lambda=left{begin{bmatrix}lambda_1&0\0&lambda_2end{bmatrix},middle|,lambda_1,lambda_2in(0,2)right}.$$The set $Lambda$ is connected too. So, the range of the map$$begin{array}{ccc}SO_2(mathbb{R})timesLambda&longrightarrow&GL_2(mathbb{R})\(P,D)&mapsto&P^{-1}DPend{array}$$is connected too. But this range is $X$.
answered Jan 22 at 10:07
José Carlos SantosJosé Carlos Santos
166k22132235
$endgroup$
The space $SO_2(mathbb{R})$ is connected. Now let$$Lambda=left{begin{bmatrix}lambda_1&0\0&lambda_2end{bmatrix},middle|,lambda_1,lambda_2in(0,2)right}.$$The set $Lambda$ is connected too. So, the range of the map$$begin{array}{ccc}SO_2(mathbb{R})timesLambda&longrightarrow&GL_2(mathbb{R})\(P,D)&mapsto&P^{-1}DPend{array}$$is connected too. But this range is $X$.
answered Jan 22 at 10:07
José Carlos SantosJosé Carlos Santos
166k22132235
answered Jan 22 at 10:07
José Carlos SantosJosé Carlos Santos
166k22132235
answered Jan 22 at 10:07
José Carlos SantosJosé Carlos Santos
166k22132235
answered Jan 22 at 10:07
José Carlos SantosJosé Carlos Santos
166k22132235
166k22132235
$begingroup$
I had another question too: whether $det(lambda A+(1-lambda ) B)$ equal to $0$ or not? where $lambda in (0,1)$ and $A,B$ are symmetric matrices with determinant non-zero.
$endgroup$
– nurun nesha
Jan 22 at 10:53
1
$begingroup$
Yes, it can be $0$. Suppose that $A=operatorname{Id}$, the $B=-operatorname{Id}$, and that $lambda=frac12$.
$endgroup$
– José Carlos Santos
Jan 22 at 11:01
add a comment |
$begingroup$
I had another question too: whether $det(lambda A+(1-lambda ) B)$ equal to $0$ or not? where $lambda in (0,1)$ and $A,B$ are symmetric matrices with determinant non-zero.
$endgroup$
– nurun nesha
Jan 22 at 10:53
1
$begingroup$
Yes, it can be $0$. Suppose that $A=operatorname{Id}$, the $B=-operatorname{Id}$, and that $lambda=frac12$.
$endgroup$
– José Carlos Santos
Jan 22 at 11:01
$begingroup$
I had another question too: whether $det(lambda A+(1-lambda ) B)$ equal to $0$ or not? where $lambda in (0,1)$ and $A,B$ are symmetric matrices with determinant non-zero.
$endgroup$
– nurun nesha
Jan 22 at 10:53
$begingroup$
I had another question too: whether $det(lambda A+(1-lambda ) B)$ equal to $0$ or not? where $lambda in (0,1)$ and $A,B$ are symmetric matrices with determinant non-zero.
$endgroup$
– nurun nesha
Jan 22 at 10:53
1
1
$begingroup$
Yes, it can be $0$. Suppose that $A=operatorname{Id}$, the $B=-operatorname{Id}$, and that $lambda=frac12$.
$endgroup$
– José Carlos Santos
Jan 22 at 11:01
$begingroup$
Yes, it can be $0$. Suppose that $A=operatorname{Id}$, the $B=-operatorname{Id}$, and that $lambda=frac12$.
$endgroup$
– José Carlos Santos
Jan 22 at 11:01
add a comment |
$begingroup$
We can prove much more (in $S_n(mathbb{R})$ , the set of $ntimes n$ real symmetric matrices).
$textbf{Proposition 1}$. The set $Z_I={Ain S_n(mathbb{R});spectrum(A)subset I}$ is convex (where $I$ is some interval of $mathbb{R}$).
$textbf{Proof}$. We use the fact that if $Ain S_n(mathbb{R})$, then $sup_{||x||_2=1}x^TAx=sup(spectrum(A)),inf_{||x||_2=1}x^TAx=inf(spectrum(A))$.
Then $lambda x^TAx+(1-lambda)x^TBxin I$ when $A,Bin Z_I,||x||_2=1,lambdain [0,1]$. $square$
$textbf{Proposition 2}$. If $I=[a,b],a<b$, then $Z_I$ is homeomorphic to the compact ball $B_{n(n+1)/2}$.
$textbf{Proof}.$ $Z_I$ is also closed (the eigenvalues are continuous functions of the entries) and bounded ($||A||_2=sup_{lambdain spectrum(A)}(|lambda|))$.
Finally, $Z_I$ is convex compact and, therefore, homeomorphic to the compact ball in $mathbb{R}^k$, where $k$ is the number of degrees of freedom of $S_n(mathbb{R})$.
answered Jan 28 at 20:19
loup blancloup blanc
23.6k21851
$endgroup$
add a comment |
$begingroup$
We can prove much more (in $S_n(mathbb{R})$ , the set of $ntimes n$ real symmetric matrices).
$textbf{Proposition 1}$. The set $Z_I={Ain S_n(mathbb{R});spectrum(A)subset I}$ is convex (where $I$ is some interval of $mathbb{R}$).
$textbf{Proof}$. We use the fact that if $Ain S_n(mathbb{R})$, then $sup_{||x||_2=1}x^TAx=sup(spectrum(A)),inf_{||x||_2=1}x^TAx=inf(spectrum(A))$.
Then $lambda x^TAx+(1-lambda)x^TBxin I$ when $A,Bin Z_I,||x||_2=1,lambdain [0,1]$. $square$
$textbf{Proposition 2}$. If $I=[a,b],a<b$, then $Z_I$ is homeomorphic to the compact ball $B_{n(n+1)/2}$.
$textbf{Proof}.$ $Z_I$ is also closed (the eigenvalues are continuous functions of the entries) and bounded ($||A||_2=sup_{lambdain spectrum(A)}(|lambda|))$.
Finally, $Z_I$ is convex compact and, therefore, homeomorphic to the compact ball in $mathbb{R}^k$, where $k$ is the number of degrees of freedom of $S_n(mathbb{R})$.
answered Jan 28 at 20:19
loup blancloup blanc
23.6k21851
$endgroup$
add a comment |
$begingroup$
We can prove much more (in $S_n(mathbb{R})$ , the set of $ntimes n$ real symmetric matrices).
$textbf{Proposition 1}$. The set $Z_I={Ain S_n(mathbb{R});spectrum(A)subset I}$ is convex (where $I$ is some interval of $mathbb{R}$).
$textbf{Proof}$. We use the fact that if $Ain S_n(mathbb{R})$, then $sup_{||x||_2=1}x^TAx=sup(spectrum(A)),inf_{||x||_2=1}x^TAx=inf(spectrum(A))$.
Then $lambda x^TAx+(1-lambda)x^TBxin I$ when $A,Bin Z_I,||x||_2=1,lambdain [0,1]$. $square$
$textbf{Proposition 2}$. If $I=[a,b],a<b$, then $Z_I$ is homeomorphic to the compact ball $B_{n(n+1)/2}$.
$textbf{Proof}.$ $Z_I$ is also closed (the eigenvalues are continuous functions of the entries) and bounded ($||A||_2=sup_{lambdain spectrum(A)}(|lambda|))$.
Finally, $Z_I$ is convex compact and, therefore, homeomorphic to the compact ball in $mathbb{R}^k$, where $k$ is the number of degrees of freedom of $S_n(mathbb{R})$.
answered Jan 28 at 20:19
loup blancloup blanc
23.6k21851
$endgroup$
We can prove much more (in $S_n(mathbb{R})$ , the set of $ntimes n$ real symmetric matrices).
$textbf{Proposition 1}$. The set $Z_I={Ain S_n(mathbb{R});spectrum(A)subset I}$ is convex (where $I$ is some interval of $mathbb{R}$).
$textbf{Proof}$. We use the fact that if $Ain S_n(mathbb{R})$, then $sup_{||x||_2=1}x^TAx=sup(spectrum(A)),inf_{||x||_2=1}x^TAx=inf(spectrum(A))$.
Then $lambda x^TAx+(1-lambda)x^TBxin I$ when $A,Bin Z_I,||x||_2=1,lambdain [0,1]$. $square$
$textbf{Proposition 2}$. If $I=[a,b],a<b$, then $Z_I$ is homeomorphic to the compact ball $B_{n(n+1)/2}$.
$textbf{Proof}.$ $Z_I$ is also closed (the eigenvalues are continuous functions of the entries) and bounded ($||A||_2=sup_{lambdain spectrum(A)}(|lambda|))$.
Finally, $Z_I$ is convex compact and, therefore, homeomorphic to the compact ball in $mathbb{R}^k$, where $k$ is the number of degrees of freedom of $S_n(mathbb{R})$.
answered Jan 28 at 20:19
loup blancloup blanc
23.6k21851
answered Jan 28 at 20:19
loup blancloup blanc
23.6k21851
answered Jan 28 at 20:19
loup blancloup blanc
23.6k21851
answered Jan 28 at 20:19
loup blancloup blanc
23.6k21851
23.6k21851
add a comment |
add a comment |
$begingroup$
One-line proof: every $Ain X$ is path-connected to $I$ by the line segment ${tA+(1-t)I: 0leq tleq 1}subseteq X$.
Your original approach also works. Let $A,Bin X$ and $0leq tleq 1$. Then $A,B$ and in turn $tA+(1-t)B$ are positive definite and hence all eigenvalues of $tA+(1-t)B$ are positive. Also,
$$
|tA+(1-t)B|_2
leq t|A|_2+(1-t)|B|_2
< 2t+2(1-t)=2.
$$
Therefore all eigenvalues of $tA+(1-t)B$ lie inside the interval $(0,2)$. Consequently, $X$ is path-connected because ${tA+(1-t)B: 0leq tleq 1}subseteq X$ for any $A,Bin X$.
answered Jan 30 at 11:45
user1551user1551
73.5k566129
$endgroup$
add a comment |
$begingroup$
One-line proof: every $Ain X$ is path-connected to $I$ by the line segment ${tA+(1-t)I: 0leq tleq 1}subseteq X$.
Your original approach also works. Let $A,Bin X$ and $0leq tleq 1$. Then $A,B$ and in turn $tA+(1-t)B$ are positive definite and hence all eigenvalues of $tA+(1-t)B$ are positive. Also,
$$
|tA+(1-t)B|_2
leq t|A|_2+(1-t)|B|_2
< 2t+2(1-t)=2.
$$
Therefore all eigenvalues of $tA+(1-t)B$ lie inside the interval $(0,2)$. Consequently, $X$ is path-connected because ${tA+(1-t)B: 0leq tleq 1}subseteq X$ for any $A,Bin X$.
answered Jan 30 at 11:45
user1551user1551
73.5k566129
$endgroup$
add a comment |
$begingroup$
One-line proof: every $Ain X$ is path-connected to $I$ by the line segment ${tA+(1-t)I: 0leq tleq 1}subseteq X$.
Your original approach also works. Let $A,Bin X$ and $0leq tleq 1$. Then $A,B$ and in turn $tA+(1-t)B$ are positive definite and hence all eigenvalues of $tA+(1-t)B$ are positive. Also,
$$
|tA+(1-t)B|_2
leq t|A|_2+(1-t)|B|_2
< 2t+2(1-t)=2.
$$
Therefore all eigenvalues of $tA+(1-t)B$ lie inside the interval $(0,2)$. Consequently, $X$ is path-connected because ${tA+(1-t)B: 0leq tleq 1}subseteq X$ for any $A,Bin X$.
answered Jan 30 at 11:45
user1551user1551
73.5k566129
$endgroup$
One-line proof: every $Ain X$ is path-connected to $I$ by the line segment ${tA+(1-t)I: 0leq tleq 1}subseteq X$.
Your original approach also works. Let $A,Bin X$ and $0leq tleq 1$. Then $A,B$ and in turn $tA+(1-t)B$ are positive definite and hence all eigenvalues of $tA+(1-t)B$ are positive. Also,
$$
|tA+(1-t)B|_2
leq t|A|_2+(1-t)|B|_2
< 2t+2(1-t)=2.
$$
Therefore all eigenvalues of $tA+(1-t)B$ lie inside the interval $(0,2)$. Consequently, $X$ is path-connected because ${tA+(1-t)B: 0leq tleq 1}subseteq X$ for any $A,Bin X$.
answered Jan 30 at 11:45
user1551user1551
73.5k566129
answered Jan 30 at 11:45
user1551user1551
73.5k566129
answered Jan 30 at 11:45
user1551user1551
73.5k566129
answered Jan 30 at 11:45
user1551user1551
73.5k566129
73.5k566129
add a comment |
add a comment |
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