Prove that $cond(A)ge frac{||A||}{||A-B||}$ for any induced matrix norm
$begingroup$
Prove that for any induced matrix norm:
$cond(A)ge frac{leftlVert A rightrVert}{leftlVert A-B rightrVert}$
Where $A$ is an invertible matrix, and $B$ is a singular matrix.
The condition number is: $cond(A) := leftlVert A rightrVert leftlVert A^{-1} rightrVert$
I have tried to prove that $leftlVert A rightrVert leftlVert A-B rightrVert ge 1$ ,but I'm not sure how to use the fact that $B$ is singular.
matrices norm condition-number matrix-norms
edited Jan 18 at 21:16
Rodrigo de Azevedo
13.1k41960
asked Jan 8 at 18:54
useruser
52
$endgroup$
add a comment |
$begingroup$
Prove that for any induced matrix norm:
$cond(A)ge frac{leftlVert A rightrVert}{leftlVert A-B rightrVert}$
Where $A$ is an invertible matrix, and $B$ is a singular matrix.
The condition number is: $cond(A) := leftlVert A rightrVert leftlVert A^{-1} rightrVert$
I have tried to prove that $leftlVert A rightrVert leftlVert A-B rightrVert ge 1$ ,but I'm not sure how to use the fact that $B$ is singular.
matrices norm condition-number matrix-norms
edited Jan 18 at 21:16
Rodrigo de Azevedo
13.1k41960
asked Jan 8 at 18:54
useruser
52
$endgroup$
add a comment |
$begingroup$
Prove that for any induced matrix norm:
$cond(A)ge frac{leftlVert A rightrVert}{leftlVert A-B rightrVert}$
Where $A$ is an invertible matrix, and $B$ is a singular matrix.
The condition number is: $cond(A) := leftlVert A rightrVert leftlVert A^{-1} rightrVert$
I have tried to prove that $leftlVert A rightrVert leftlVert A-B rightrVert ge 1$ ,but I'm not sure how to use the fact that $B$ is singular.
matrices norm condition-number matrix-norms
edited Jan 18 at 21:16
Rodrigo de Azevedo
13.1k41960
asked Jan 8 at 18:54
useruser
52
$endgroup$
Prove that for any induced matrix norm:
$cond(A)ge frac{leftlVert A rightrVert}{leftlVert A-B rightrVert}$
Where $A$ is an invertible matrix, and $B$ is a singular matrix.
The condition number is: $cond(A) := leftlVert A rightrVert leftlVert A^{-1} rightrVert$
I have tried to prove that $leftlVert A rightrVert leftlVert A-B rightrVert ge 1$ ,but I'm not sure how to use the fact that $B$ is singular.
matrices norm condition-number matrix-norms
matrices norm condition-number matrix-norms
edited Jan 18 at 21:16
Rodrigo de Azevedo
13.1k41960
asked Jan 8 at 18:54
useruser
52
edited Jan 18 at 21:16
Rodrigo de Azevedo
13.1k41960
asked Jan 8 at 18:54
useruser
52
edited Jan 18 at 21:16
Rodrigo de Azevedo
13.1k41960
edited Jan 18 at 21:16
Rodrigo de Azevedo
13.1k41960
edited Jan 18 at 21:16
Rodrigo de Azevedo
13.1k41960
13.1k41960
asked Jan 8 at 18:54
useruser
52
asked Jan 8 at 18:54
useruser
52
asked Jan 8 at 18:54
useruser
52
52
add a comment |
add a comment |
1 Answer
1
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$begingroup$
Let $zin ker B$ be a unit vector. It is easy to see that
$$
|Az|=|Az-Bz|le |A-B|.
$$ On the other hand, we have
$$
1 = |z|=|A^{-1}Az|le |A^{-1}||Az|.
$$ Therefore, we have
$$
|A^{-1}|^{-1}le |Az|le |A-B|,
$$ and
$$
frac{|A|}{|A-B|}le|A||A^{-1}|=text{cond}(A)
$$ follows.
answered Jan 8 at 19:12
SongSong
18.6k21651
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$begingroup$
Let $zin ker B$ be a unit vector. It is easy to see that
$$
|Az|=|Az-Bz|le |A-B|.
$$ On the other hand, we have
$$
1 = |z|=|A^{-1}Az|le |A^{-1}||Az|.
$$ Therefore, we have
$$
|A^{-1}|^{-1}le |Az|le |A-B|,
$$ and
$$
frac{|A|}{|A-B|}le|A||A^{-1}|=text{cond}(A)
$$ follows.
answered Jan 8 at 19:12
SongSong
18.6k21651
$endgroup$
add a comment |
$begingroup$
Let $zin ker B$ be a unit vector. It is easy to see that
$$
|Az|=|Az-Bz|le |A-B|.
$$ On the other hand, we have
$$
1 = |z|=|A^{-1}Az|le |A^{-1}||Az|.
$$ Therefore, we have
$$
|A^{-1}|^{-1}le |Az|le |A-B|,
$$ and
$$
frac{|A|}{|A-B|}le|A||A^{-1}|=text{cond}(A)
$$ follows.
answered Jan 8 at 19:12
SongSong
18.6k21651
$endgroup$
add a comment |
$begingroup$
Let $zin ker B$ be a unit vector. It is easy to see that
$$
|Az|=|Az-Bz|le |A-B|.
$$ On the other hand, we have
$$
1 = |z|=|A^{-1}Az|le |A^{-1}||Az|.
$$ Therefore, we have
$$
|A^{-1}|^{-1}le |Az|le |A-B|,
$$ and
$$
frac{|A|}{|A-B|}le|A||A^{-1}|=text{cond}(A)
$$ follows.
answered Jan 8 at 19:12
SongSong
18.6k21651
$endgroup$
Let $zin ker B$ be a unit vector. It is easy to see that
$$
|Az|=|Az-Bz|le |A-B|.
$$ On the other hand, we have
$$
1 = |z|=|A^{-1}Az|le |A^{-1}||Az|.
$$ Therefore, we have
$$
|A^{-1}|^{-1}le |Az|le |A-B|,
$$ and
$$
frac{|A|}{|A-B|}le|A||A^{-1}|=text{cond}(A)
$$ follows.
answered Jan 8 at 19:12
SongSong
18.6k21651
answered Jan 8 at 19:12
SongSong
18.6k21651
answered Jan 8 at 19:12
SongSong
18.6k21651
answered Jan 8 at 19:12
SongSong
18.6k21651
18.6k21651
add a comment |
add a comment |
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