Mixing
The nonsingular matrix closest to singular one
It is well known that given nonsingular matrix $A$, the distance to closest singular is given by $|A^{-1}|^{-1}$.
My question is given singular matrix $B$ what is the distance to closest nonsingular? Is it the same as first problem?
linear-algebra matrices matrix-calculus
edited Nov 21 '18 at 20:04
Surb
37.4k94375
asked Nov 21 '18 at 19:52
Studying OptimizationStudying Optimization
676
add a comment |
It is well known that given nonsingular matrix $A$, the distance to closest singular is given by $|A^{-1}|^{-1}$.
My question is given singular matrix $B$ what is the distance to closest nonsingular? Is it the same as first problem?
linear-algebra matrices matrix-calculus
edited Nov 21 '18 at 20:04
Surb
37.4k94375
asked Nov 21 '18 at 19:52
Studying OptimizationStudying Optimization
676
the distance is $epsilon$ for every $epsilon>0$.
– Surb
Nov 21 '18 at 19:57
@Jean-ClaudeArbaut, does it exist for certain norms or it doesnt exist at all? Where can I read about it?
– Studying Optimization
Nov 21 '18 at 20:01
@StudyingOptimization See my answer. And yes, it works for every norm as every norm on a finite dimensional space are equivalent.
– Surb
Nov 21 '18 at 20:03
add a comment |
It is well known that given nonsingular matrix $A$, the distance to closest singular is given by $|A^{-1}|^{-1}$.
My question is given singular matrix $B$ what is the distance to closest nonsingular? Is it the same as first problem?
linear-algebra matrices matrix-calculus
edited Nov 21 '18 at 20:04
Surb
37.4k94375
asked Nov 21 '18 at 19:52
Studying OptimizationStudying Optimization
676
It is well known that given nonsingular matrix $A$, the distance to closest singular is given by $|A^{-1}|^{-1}$.
My question is given singular matrix $B$ what is the distance to closest nonsingular? Is it the same as first problem?
linear-algebra matrices matrix-calculus
linear-algebra matrices matrix-calculus
edited Nov 21 '18 at 20:04
Surb
37.4k94375
asked Nov 21 '18 at 19:52
Studying OptimizationStudying Optimization
676
edited Nov 21 '18 at 20:04
Surb
37.4k94375
asked Nov 21 '18 at 19:52
Studying OptimizationStudying Optimization
676
edited Nov 21 '18 at 20:04
Surb
37.4k94375
edited Nov 21 '18 at 20:04
Surb
37.4k94375
edited Nov 21 '18 at 20:04
Surb
37.4k94375
37.4k94375
asked Nov 21 '18 at 19:52
Studying OptimizationStudying Optimization
676
asked Nov 21 '18 at 19:52
Studying OptimizationStudying Optimization
676
asked Nov 21 '18 at 19:52
Studying OptimizationStudying Optimization
676
676
the distance is $epsilon$ for every $epsilon>0$.
– Surb
Nov 21 '18 at 19:57
@Jean-ClaudeArbaut, does it exist for certain norms or it doesnt exist at all? Where can I read about it?
– Studying Optimization
Nov 21 '18 at 20:01
@StudyingOptimization See my answer. And yes, it works for every norm as every norm on a finite dimensional space are equivalent.
– Surb
Nov 21 '18 at 20:03
add a comment |
the distance is $epsilon$ for every $epsilon>0$.
– Surb
Nov 21 '18 at 19:57
@Jean-ClaudeArbaut, does it exist for certain norms or it doesnt exist at all? Where can I read about it?
– Studying Optimization
Nov 21 '18 at 20:01
@StudyingOptimization See my answer. And yes, it works for every norm as every norm on a finite dimensional space are equivalent.
– Surb
Nov 21 '18 at 20:03
the distance is $epsilon$ for every $epsilon>0$.
– Surb
Nov 21 '18 at 19:57
the distance is $epsilon$ for every $epsilon>0$.
– Surb
Nov 21 '18 at 19:57
@Jean-ClaudeArbaut, does it exist for certain norms or it doesnt exist at all? Where can I read about it?
– Studying Optimization
Nov 21 '18 at 20:01
@Jean-ClaudeArbaut, does it exist for certain norms or it doesnt exist at all? Where can I read about it?
– Studying Optimization
Nov 21 '18 at 20:01
@StudyingOptimization See my answer. And yes, it works for every norm as every norm on a finite dimensional space are equivalent.
– Surb
Nov 21 '18 at 20:03
@StudyingOptimization See my answer. And yes, it works for every norm as every norm on a finite dimensional space are equivalent.
– Surb
Nov 21 '18 at 20:03
add a comment |
1 Answer
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There is nonsingular matrix at distance at most $epsilon$ for every $epsilon >0$.
Indeed, let $B$ be any matrix, then $p(t)=det(B-t I)$ is polynomial of $t$ and has finitely many roots. In particular, there exists $tau>0$ such that $p(t)neq 0$ for all $tin (0,tau)$ and thus $B_t = B-t I$ is nonsingular for all $tin(0,tau)$. Now, for every $epsilon >0$, you can choose $tin (0,tau)$ small enough so that $|B-B_t|=t|I|< epsilon$.
answered Nov 21 '18 at 20:02
SurbSurb
37.4k94375
add a comment |
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There is nonsingular matrix at distance at most $epsilon$ for every $epsilon >0$.
Indeed, let $B$ be any matrix, then $p(t)=det(B-t I)$ is polynomial of $t$ and has finitely many roots. In particular, there exists $tau>0$ such that $p(t)neq 0$ for all $tin (0,tau)$ and thus $B_t = B-t I$ is nonsingular for all $tin(0,tau)$. Now, for every $epsilon >0$, you can choose $tin (0,tau)$ small enough so that $|B-B_t|=t|I|< epsilon$.
answered Nov 21 '18 at 20:02
SurbSurb
37.4k94375
add a comment |
There is nonsingular matrix at distance at most $epsilon$ for every $epsilon >0$.
Indeed, let $B$ be any matrix, then $p(t)=det(B-t I)$ is polynomial of $t$ and has finitely many roots. In particular, there exists $tau>0$ such that $p(t)neq 0$ for all $tin (0,tau)$ and thus $B_t = B-t I$ is nonsingular for all $tin(0,tau)$. Now, for every $epsilon >0$, you can choose $tin (0,tau)$ small enough so that $|B-B_t|=t|I|< epsilon$.
answered Nov 21 '18 at 20:02
SurbSurb
37.4k94375
add a comment |
There is nonsingular matrix at distance at most $epsilon$ for every $epsilon >0$.
Indeed, let $B$ be any matrix, then $p(t)=det(B-t I)$ is polynomial of $t$ and has finitely many roots. In particular, there exists $tau>0$ such that $p(t)neq 0$ for all $tin (0,tau)$ and thus $B_t = B-t I$ is nonsingular for all $tin(0,tau)$. Now, for every $epsilon >0$, you can choose $tin (0,tau)$ small enough so that $|B-B_t|=t|I|< epsilon$.
answered Nov 21 '18 at 20:02
SurbSurb
37.4k94375
There is nonsingular matrix at distance at most $epsilon$ for every $epsilon >0$.
Indeed, let $B$ be any matrix, then $p(t)=det(B-t I)$ is polynomial of $t$ and has finitely many roots. In particular, there exists $tau>0$ such that $p(t)neq 0$ for all $tin (0,tau)$ and thus $B_t = B-t I$ is nonsingular for all $tin(0,tau)$. Now, for every $epsilon >0$, you can choose $tin (0,tau)$ small enough so that $|B-B_t|=t|I|< epsilon$.
answered Nov 21 '18 at 20:02
SurbSurb
37.4k94375
answered Nov 21 '18 at 20:02
SurbSurb
37.4k94375
answered Nov 21 '18 at 20:02
SurbSurb
37.4k94375
answered Nov 21 '18 at 20:02
SurbSurb
37.4k94375
37.4k94375
add a comment |
add a comment |
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the distance is $epsilon$ for every $epsilon>0$.
– Surb
Nov 21 '18 at 19:57
@Jean-ClaudeArbaut, does it exist for certain norms or it doesnt exist at all? Where can I read about it?
– Studying Optimization
Nov 21 '18 at 20:01
@StudyingOptimization See my answer. And yes, it works for every norm as every norm on a finite dimensional space are equivalent.
– Surb
Nov 21 '18 at 20:03