Why is the norm of a matrix larger than its eigenvalue?

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I know there are different definitions of Matrix Norm, but I want to use the definition on WolframMathWorld, and Wikipedia also gives a similar definition.

The definition states as below:

Given a square complex or real $ntimes n$ matrix $A$, a matrix norm $|A|$ is a nonnegative number associated with $A$ having the properties

1.$|A|>0$ when $Aneq0$ and $|A|=0$ iff $A=0$,

2.$|kA|=|k||A|$ for any scalar $k$,

3.$|A+B|leq|A|+|B|$, for $n times n$ matrix $B$

4.$|AB|leq|A||B|$.

Then, as the website states, we have $|A|geq|lambda|$, here $lambda$ is an eigenvalue of $A$. I don't know how to prove it, by using just these four properties.

edited Jul 18 at 0:26

mechanodroid

22.2k52041

asked Jul 18 at 0:00

Eric Yewen Sun

442216

2

@mechanodroid Is it? Can you verify point 4? (The submultiplicative property of matrix norms)
â€“Â Clement C.
Jul 18 at 0:05

@ClementC. Sorry, my bad.
â€“Â mechanodroid
Jul 18 at 0:07

add a commentÂ |Â

up vote
21
down vote

favorite

I know there are different definitions of Matrix Norm, but I want to use the definition on WolframMathWorld, and Wikipedia also gives a similar definition.

The definition states as below:

Given a square complex or real $ntimes n$ matrix $A$, a matrix norm $|A|$ is a nonnegative number associated with $A$ having the properties

1.$|A|>0$ when $Aneq0$ and $|A|=0$ iff $A=0$,

2.$|kA|=|k||A|$ for any scalar $k$,

3.$|A+B|leq|A|+|B|$, for $n times n$ matrix $B$

4.$|AB|leq|A||B|$.

Then, as the website states, we have $|A|geq|lambda|$, here $lambda$ is an eigenvalue of $A$. I don't know how to prove it, by using just these four properties.

edited Jul 18 at 0:26

mechanodroid

22.2k52041

asked Jul 18 at 0:00

Eric Yewen Sun

442216

2

@mechanodroid Is it? Can you verify point 4? (The submultiplicative property of matrix norms)
â€“Â Clement C.
Jul 18 at 0:05

@ClementC. Sorry, my bad.
â€“Â mechanodroid
Jul 18 at 0:07

add a commentÂ |Â

up vote
21
down vote

favorite

I know there are different definitions of Matrix Norm, but I want to use the definition on WolframMathWorld, and Wikipedia also gives a similar definition.

The definition states as below:

Given a square complex or real $ntimes n$ matrix $A$, a matrix norm $|A|$ is a nonnegative number associated with $A$ having the properties

1.$|A|>0$ when $Aneq0$ and $|A|=0$ iff $A=0$,

2.$|kA|=|k||A|$ for any scalar $k$,

3.$|A+B|leq|A|+|B|$, for $n times n$ matrix $B$

4.$|AB|leq|A||B|$.

Then, as the website states, we have $|A|geq|lambda|$, here $lambda$ is an eigenvalue of $A$. I don't know how to prove it, by using just these four properties.

edited Jul 18 at 0:26

mechanodroid

22.2k52041

asked Jul 18 at 0:00

Eric Yewen Sun

442216

I know there are different definitions of Matrix Norm, but I want to use the definition on WolframMathWorld, and Wikipedia also gives a similar definition.

The definition states as below:

Given a square complex or real $ntimes n$ matrix $A$, a matrix norm $|A|$ is a nonnegative number associated with $A$ having the properties

1.$|A|>0$ when $Aneq0$ and $|A|=0$ iff $A=0$,

2.$|kA|=|k||A|$ for any scalar $k$,

3.$|A+B|leq|A|+|B|$, for $n times n$ matrix $B$

4.$|AB|leq|A||B|$.

Then, as the website states, we have $|A|geq|lambda|$, here $lambda$ is an eigenvalue of $A$. I don't know how to prove it, by using just these four properties.

edited Jul 18 at 0:26

mechanodroid

22.2k52041

asked Jul 18 at 0:00

Eric Yewen Sun

442216

edited Jul 18 at 0:26

mechanodroid

22.2k52041

edited Jul 18 at 0:26

mechanodroid

22.2k52041

edited Jul 18 at 0:26

mechanodroid

22.2k52041

asked Jul 18 at 0:00

Eric Yewen Sun

442216

asked Jul 18 at 0:00

Eric Yewen Sun

442216

asked Jul 18 at 0:00

Eric Yewen Sun

442216

2

@mechanodroid Is it? Can you verify point 4? (The submultiplicative property of matrix norms)
â€“Â Clement C.
Jul 18 at 0:05

@ClementC. Sorry, my bad.
â€“Â mechanodroid
Jul 18 at 0:07

add a commentÂ |Â

2

@mechanodroid Is it? Can you verify point 4? (The submultiplicative property of matrix norms)
â€“Â Clement C.
Jul 18 at 0:05

@ClementC. Sorry, my bad.
â€“Â mechanodroid
Jul 18 at 0:07

@mechanodroid Is it? Can you verify point 4? (The submultiplicative property of matrix norms)
â€“Â Clement C.
Jul 18 at 0:05

@ClementC. Sorry, my bad.
â€“Â mechanodroid
Jul 18 at 0:07

add a commentÂ |Â

2 Answers
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up vote
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Suppose $v$ is an eigenvector for $A$ corresponding to $lambda$. Form the "eigenmatrix" $B$ by putting $v$ in all the columns. Then $AB = lambda B$. So, by properties $2$ and $4$ (and $1$, to make sure $|B| > 0$),
$$|lambda| |B| = |lambda B| = |AB| le |A| |B|.$$
Hence, $|A| ge |lambda|$ for all eigenvalues $lambda$.

answered Jul 18 at 0:09

Theo Bendit

12.1k1844

Nice trick. Thank you!
â€“Â Eric Yewen Sun
Jul 18 at 0:12

add a commentÂ |Â

up vote
3
down vote

Let $|cdot|$ be a matrix norm.

It is known that the spectral radius $r(A) = lim_ntoinfty |A^n|^frac1n$ has the property $|lambda| le r(A)$ for all $lambdain sigma(A)$.

Indeed, let $lambda in mathbbC$ such that $|lambda| > r(A)$.

Then $I - frac1lambda A$ is invertible. Namely, check that the inverse is given by $sum_n=0^inftyfrac1lambda^nA^n$.

This series converges absolutely because $frac1$ is less than the radius of convergence of the power series $sum_n=1^infty |A|^nx^n$, which is $frac1A^n = frac1r(A)$.

Hence $$lambda I - A = lambdaleft(I - frac1lambda Aright)$$

is also invertible so $lambda notin sigma(A)$.

Now using submultiplicativity we get $|A^n| le |A|^n$ so

$$|lambda| le r(A) = lim_ntoinfty |A^n|^frac1n le lim_ntoinfty |A|^ncdotfrac1n = |A|$$

answered Jul 18 at 0:23

mechanodroid

22.2k52041

I had a look at the Wikipedia page. It seems that the identity $r(A) = lim_ntoinfty |A^n|^frac1n$ holds for natural matrix norms, i.e. operator norms induced by norms on $mathbbR^n$. I don't have a counterexample, but I suspect it doesn't hold in general.
â€“Â Theo Bendit
Jul 18 at 0:27

2

@TheoBendit I used the abstract spectral radius defined in Banach algebras simply as $lim_ntoinfty |A^n|^frac1n$. Perhaps the name is not appropriate for matrices. Secondly, every two matrix norms are equivalent so if you take a natural matrix norm $|cdot|_1$ we have $m|cdot|_1 le |cdot| le M|cdot|_1$ so $$m^1/n|A^n|_1^1/n le |A^n|^1/n le M^1/n|A^n|_1^1/n$$ Letting $ntoinfty$ gives $lim_ntoinfty |A^n|_1^frac1n = lim_ntoinfty |A^n|^frac1n$. So it should hold for every matrix norm.
â€“Â mechanodroid
Jul 18 at 0:32

1

@TheoBendit Actually it is there in the Wikipedia page under the name Gelfand's Formula. I guess the only nontrivial thing in my answer is to show that the sequence $(|A^n|^1/n)_n$ indeed converges.
â€“Â mechanodroid
Jul 18 at 0:36

add a commentÂ |Â

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2 Answers
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active

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2 Answers
2

active

oldest

votes

up vote
41
down vote

accepted

answered Jul 18 at 0:09

Theo Bendit

12.1k1844

Nice trick. Thank you!
â€“Â Eric Yewen Sun
Jul 18 at 0:12

add a commentÂ |Â

up vote
41
down vote

accepted

answered Jul 18 at 0:09

Theo Bendit

12.1k1844

Nice trick. Thank you!
â€“Â Eric Yewen Sun
Jul 18 at 0:12

add a commentÂ |Â

up vote
41
down vote

accepted

answered Jul 18 at 0:09

Theo Bendit

12.1k1844

answered Jul 18 at 0:09

Theo Bendit

12.1k1844

answered Jul 18 at 0:09

Theo Bendit

12.1k1844

answered Jul 18 at 0:09

Theo Bendit

12.1k1844

answered Jul 18 at 0:09

Theo Bendit

12.1k1844

Nice trick. Thank you!
â€“Â Eric Yewen Sun
Jul 18 at 0:12

add a commentÂ |Â

Nice trick. Thank you!
â€“Â Eric Yewen Sun
Jul 18 at 0:12

Nice trick. Thank you!
â€“Â Eric Yewen Sun
Jul 18 at 0:12

add a commentÂ |Â

up vote
3
down vote

Let $|cdot|$ be a matrix norm.

It is known that the spectral radius $r(A) = lim_ntoinfty |A^n|^frac1n$ has the property $|lambda| le r(A)$ for all $lambdain sigma(A)$.

Indeed, let $lambda in mathbbC$ such that $|lambda| > r(A)$.

Then $I - frac1lambda A$ is invertible. Namely, check that the inverse is given by $sum_n=0^inftyfrac1lambda^nA^n$.

This series converges absolutely because $frac1$ is less than the radius of convergence of the power series $sum_n=1^infty |A|^nx^n$, which is $frac1A^n = frac1r(A)$.

Hence $$lambda I - A = lambdaleft(I - frac1lambda Aright)$$

is also invertible so $lambda notin sigma(A)$.

Now using submultiplicativity we get $|A^n| le |A|^n$ so

$$|lambda| le r(A) = lim_ntoinfty |A^n|^frac1n le lim_ntoinfty |A|^ncdotfrac1n = |A|$$

answered Jul 18 at 0:23

mechanodroid

22.2k52041

I had a look at the Wikipedia page. It seems that the identity $r(A) = lim_ntoinfty |A^n|^frac1n$ holds for natural matrix norms, i.e. operator norms induced by norms on $mathbbR^n$. I don't have a counterexample, but I suspect it doesn't hold in general.
â€“Â Theo Bendit
Jul 18 at 0:27

2

@TheoBendit I used the abstract spectral radius defined in Banach algebras simply as $lim_ntoinfty |A^n|^frac1n$. Perhaps the name is not appropriate for matrices. Secondly, every two matrix norms are equivalent so if you take a natural matrix norm $|cdot|_1$ we have $m|cdot|_1 le |cdot| le M|cdot|_1$ so $$m^1/n|A^n|_1^1/n le |A^n|^1/n le M^1/n|A^n|_1^1/n$$ Letting $ntoinfty$ gives $lim_ntoinfty |A^n|_1^frac1n = lim_ntoinfty |A^n|^frac1n$. So it should hold for every matrix norm.
â€“Â mechanodroid
Jul 18 at 0:32

1

@TheoBendit Actually it is there in the Wikipedia page under the name Gelfand's Formula. I guess the only nontrivial thing in my answer is to show that the sequence $(|A^n|^1/n)_n$ indeed converges.
â€“Â mechanodroid
Jul 18 at 0:36

add a commentÂ |Â

up vote
3
down vote

Let $|cdot|$ be a matrix norm.

It is known that the spectral radius $r(A) = lim_ntoinfty |A^n|^frac1n$ has the property $|lambda| le r(A)$ for all $lambdain sigma(A)$.

Indeed, let $lambda in mathbbC$ such that $|lambda| > r(A)$.

Then $I - frac1lambda A$ is invertible. Namely, check that the inverse is given by $sum_n=0^inftyfrac1lambda^nA^n$.

This series converges absolutely because $frac1$ is less than the radius of convergence of the power series $sum_n=1^infty |A|^nx^n$, which is $frac1A^n = frac1r(A)$.

Hence $$lambda I - A = lambdaleft(I - frac1lambda Aright)$$

is also invertible so $lambda notin sigma(A)$.

Now using submultiplicativity we get $|A^n| le |A|^n$ so

$$|lambda| le r(A) = lim_ntoinfty |A^n|^frac1n le lim_ntoinfty |A|^ncdotfrac1n = |A|$$

answered Jul 18 at 0:23

mechanodroid

22.2k52041

I had a look at the Wikipedia page. It seems that the identity $r(A) = lim_ntoinfty |A^n|^frac1n$ holds for natural matrix norms, i.e. operator norms induced by norms on $mathbbR^n$. I don't have a counterexample, but I suspect it doesn't hold in general.
â€“Â Theo Bendit
Jul 18 at 0:27

2

@TheoBendit I used the abstract spectral radius defined in Banach algebras simply as $lim_ntoinfty |A^n|^frac1n$. Perhaps the name is not appropriate for matrices. Secondly, every two matrix norms are equivalent so if you take a natural matrix norm $|cdot|_1$ we have $m|cdot|_1 le |cdot| le M|cdot|_1$ so $$m^1/n|A^n|_1^1/n le |A^n|^1/n le M^1/n|A^n|_1^1/n$$ Letting $ntoinfty$ gives $lim_ntoinfty |A^n|_1^frac1n = lim_ntoinfty |A^n|^frac1n$. So it should hold for every matrix norm.
â€“Â mechanodroid
Jul 18 at 0:32

1

@TheoBendit Actually it is there in the Wikipedia page under the name Gelfand's Formula. I guess the only nontrivial thing in my answer is to show that the sequence $(|A^n|^1/n)_n$ indeed converges.
â€“Â mechanodroid
Jul 18 at 0:36

add a commentÂ |Â

up vote
3
down vote

Let $|cdot|$ be a matrix norm.

It is known that the spectral radius $r(A) = lim_ntoinfty |A^n|^frac1n$ has the property $|lambda| le r(A)$ for all $lambdain sigma(A)$.

Indeed, let $lambda in mathbbC$ such that $|lambda| > r(A)$.

Then $I - frac1lambda A$ is invertible. Namely, check that the inverse is given by $sum_n=0^inftyfrac1lambda^nA^n$.

This series converges absolutely because $frac1$ is less than the radius of convergence of the power series $sum_n=1^infty |A|^nx^n$, which is $frac1A^n = frac1r(A)$.

Hence $$lambda I - A = lambdaleft(I - frac1lambda Aright)$$

is also invertible so $lambda notin sigma(A)$.

Now using submultiplicativity we get $|A^n| le |A|^n$ so

$$|lambda| le r(A) = lim_ntoinfty |A^n|^frac1n le lim_ntoinfty |A|^ncdotfrac1n = |A|$$

answered Jul 18 at 0:23

mechanodroid

22.2k52041

Let $|cdot|$ be a matrix norm.

It is known that the spectral radius $r(A) = lim_ntoinfty |A^n|^frac1n$ has the property $|lambda| le r(A)$ for all $lambdain sigma(A)$.

Indeed, let $lambda in mathbbC$ such that $|lambda| > r(A)$.

Then $I - frac1lambda A$ is invertible. Namely, check that the inverse is given by $sum_n=0^inftyfrac1lambda^nA^n$.

This series converges absolutely because $frac1$ is less than the radius of convergence of the power series $sum_n=1^infty |A|^nx^n$, which is $frac1A^n = frac1r(A)$.

Hence $$lambda I - A = lambdaleft(I - frac1lambda Aright)$$

is also invertible so $lambda notin sigma(A)$.

Now using submultiplicativity we get $|A^n| le |A|^n$ so

$$|lambda| le r(A) = lim_ntoinfty |A^n|^frac1n le lim_ntoinfty |A|^ncdotfrac1n = |A|$$

answered Jul 18 at 0:23

mechanodroid

22.2k52041

answered Jul 18 at 0:23

mechanodroid

22.2k52041

answered Jul 18 at 0:23

mechanodroid

22.2k52041

answered Jul 18 at 0:23

mechanodroid

22.2k52041

I had a look at the Wikipedia page. It seems that the identity $r(A) = lim_ntoinfty |A^n|^frac1n$ holds for natural matrix norms, i.e. operator norms induced by norms on $mathbbR^n$. I don't have a counterexample, but I suspect it doesn't hold in general.
â€“Â Theo Bendit
Jul 18 at 0:27

2

@TheoBendit I used the abstract spectral radius defined in Banach algebras simply as $lim_ntoinfty |A^n|^frac1n$. Perhaps the name is not appropriate for matrices. Secondly, every two matrix norms are equivalent so if you take a natural matrix norm $|cdot|_1$ we have $m|cdot|_1 le |cdot| le M|cdot|_1$ so $$m^1/n|A^n|_1^1/n le |A^n|^1/n le M^1/n|A^n|_1^1/n$$ Letting $ntoinfty$ gives $lim_ntoinfty |A^n|_1^frac1n = lim_ntoinfty |A^n|^frac1n$. So it should hold for every matrix norm.
â€“Â mechanodroid
Jul 18 at 0:32

1

@TheoBendit Actually it is there in the Wikipedia page under the name Gelfand's Formula. I guess the only nontrivial thing in my answer is to show that the sequence $(|A^n|^1/n)_n$ indeed converges.
â€“Â mechanodroid
Jul 18 at 0:36

add a commentÂ |Â

I had a look at the Wikipedia page. It seems that the identity $r(A) = lim_ntoinfty |A^n|^frac1n$ holds for natural matrix norms, i.e. operator norms induced by norms on $mathbbR^n$. I don't have a counterexample, but I suspect it doesn't hold in general.
â€“Â Theo Bendit
Jul 18 at 0:27

2

@TheoBendit I used the abstract spectral radius defined in Banach algebras simply as $lim_ntoinfty |A^n|^frac1n$. Perhaps the name is not appropriate for matrices. Secondly, every two matrix norms are equivalent so if you take a natural matrix norm $|cdot|_1$ we have $m|cdot|_1 le |cdot| le M|cdot|_1$ so $$m^1/n|A^n|_1^1/n le |A^n|^1/n le M^1/n|A^n|_1^1/n$$ Letting $ntoinfty$ gives $lim_ntoinfty |A^n|_1^frac1n = lim_ntoinfty |A^n|^frac1n$. So it should hold for every matrix norm.
â€“Â mechanodroid
Jul 18 at 0:32

1

@TheoBendit Actually it is there in the Wikipedia page under the name Gelfand's Formula. I guess the only nontrivial thing in my answer is to show that the sequence $(|A^n|^1/n)_n$ indeed converges.
â€“Â mechanodroid
Jul 18 at 0:36

I had a look at the Wikipedia page. It seems that the identity $r(A) = lim_ntoinfty |A^n|^frac1n$ holds for natural matrix norms, i.e. operator norms induced by norms on $mathbbR^n$. I don't have a counterexample, but I suspect it doesn't hold in general.
â€“Â Theo Bendit
Jul 18 at 0:27

@TheoBendit I used the abstract spectral radius defined in Banach algebras simply as $lim_ntoinfty |A^n|^frac1n$. Perhaps the name is not appropriate for matrices. Secondly, every two matrix norms are equivalent so if you take a natural matrix norm $|cdot|_1$ we have $m|cdot|_1 le |cdot| le M|cdot|_1$ so $$m^1/n|A^n|_1^1/n le |A^n|^1/n le M^1/n|A^n|_1^1/n$$ Letting $ntoinfty$ gives $lim_ntoinfty |A^n|_1^frac1n = lim_ntoinfty |A^n|^frac1n$. So it should hold for every matrix norm.
â€“Â mechanodroid
Jul 18 at 0:32

@TheoBendit Actually it is there in the Wikipedia page under the name Gelfand's Formula. I guess the only nontrivial thing in my answer is to show that the sequence $(|A^n|^1/n)_n$ indeed converges.
â€“Â mechanodroid
Jul 18 at 0:36

add a commentÂ |Â

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