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small spectral radius

sarrah

Member
Jun 20, 2014
60
Hello Colleagues

We know that for any square matrix | Lambda(A) | < ||A|| . I was looking for a matrix whether real symmetric with a large norm but small spectral radius or a general real matrix again with a large norm but small singular values. Any example will do, n up to 5
grateful thanks
sarrah
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,736
Hello Colleagues

We know that for any square matrix | Lambda(A) | < ||A|| . I was looking for a matrix whether real symmetric with a large norm but small spectral radius or a general real matrix again with a large norm but small singular values. Any example will do, n up to 5
grateful thanks
sarrah
Hi sarrah,

If a matrix is real symmetric, then its singular values are the same as the absolute values of its eigenvalues.
Consequently the norm (largest singular value) is the same as the largest absolute eigenvalue, which is the spectral radius.

We can get a general real matrix with a large norm and a small spectral radius by taking eigenvectors that are almost colinear but that have opposite eigenvalues.
As an example, we can pick \begin{pmatrix}0&-10\\-0.1&0\end{pmatrix} which has spectral radius $1$ and norm $10$.
 
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sarrah

Member
Jun 20, 2014
60
Dear Klaas

I am grateful for taking my question
yet the matrix you gave is not real symmetric. My question is whether one knows of a real symmetric matrix with big norm and small eigenvalues OR a real matrix only like the one you wrote but having small singular values. Your matrix has big singular values
I am sorry
Sarah
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,736
Dear sarrah,

I'm sorry, it appears I misread the second part of your question.

However, it seems the first part of my answer was ignored, so let me try to clarify.
If $A$ is a symmetric matrix, then the Spectral Theorem tells us that it can be written as $A=BDB^T$ where $D$ is a real diagonal matrix with the eigenvalues on its diagonal, and $B$ is an orthogonal matrix with eigenvectors.
Consequently we have $\|A\| = \sigma_{max}(A)= \sqrt{\lambda_{max}(A^TA)} = \sqrt{\lambda_{max}((BDB^T)^TBDB^T)} = \sqrt{\lambda_{max}(BD^2B^T)}=|\lambda_{max}(A)|=\rho(A)$.
That is, the norm and the spectral radius of a real symmetric matrix are the same.
Put otherwise, it is not possible for the norm to be large and the spectral radius to be small.

As for the second part of your question, the norm of a matrix is equal to its largest singular value.
It means that a matrix with a large norm also has a large singular value. All other singular values can be small though.
The matrix I gave, has norm $10$ and singular values $10$ and $0.1$, so I think it still provides a correct answer.
 
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sarrah

Member
Jun 20, 2014
60
Hello Klaas
Sorry to have wasted your time Klaas. I know that the spectral norm of a real symmetric matrix is equal to its spectral radius. Meaning that the spectral norm is the smallest of all norms since the spectral radius is less than ||A|| (one or infinity or frobenius, etc..) So perhaps I can find a matrix with small spectral norm yet with large ||A|| (I mean here norm one or infinity say).
Again for any real matrix A, true the spectral norm is equal to its larger singular value, but consider this matrix

2 10^9 -2x10^9
-10^-9 5 -3
2x10^-9 -3 2

this matrix has as eigenvalues 1 , 2 , 6 and singular values 0 , 2.23 , 3.6 and norm infinity 3x10^9
another pathological but bad example for me is

-149 -50 -154
537 180 546
-27 -9 -25

its spectral radius is 3 , but its spectral norm is 817. ||A|| infinity is 1263 (spectral norm is big and norm infinity is comparable so bad example for me).
What I was looking for is something like the first matrix but to be real symmetric having a large norm one or infinity (preferrably a positive matrix) but with a small spectral radius
I apologize once more to you
grateful
Sarrah
 

sarrah

Member
Jun 20, 2014
60
No I am extremely embarassed, I recalculated the singular values of the first matrix it is huge. Because I doubted it since I know that the spectral radius must be less than the spectral norm. What I was really looking for is
either
1. A real matrix with large norm 1 or infinity but small singular value, or
2. A real symmetric matrix with large norm 1 or infinity but small spectral radius.
sincerely
sarrah
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,694
I'm not sure if this answers your question, but any two norms on the space of $n\times n$ matrices are equivalent. Given norms $\|.\|$ and $\|.\|^*$, there will always be a positive number $c$ (depending only on $n$) such that $\|A\| \leqslant c\|A\|^*$ and $\|A\|^* \leqslant c\|A\|$ for every $n\times n$ matrix $A$. That will apply in particular to the operator norm, spectral norm, $1$-norm and $\infty$-norm. The value of $c$ is of the order of $\sqrt n$. See this Wikipedia article.
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,736
Again for any real matrix A, true the spectral norm is equal to its larger singular value, but consider this matrix

2 10^9 -2x10^9
-10^-9 5 -3
2x10^-9 -3 2

this matrix has as eigenvalues 1 , 2 , 6 and singular values 0 , 2.23 , 3.6 and norm infinity 3x10^9
No worries.

The singular values are actually $1.5\times 10^{-9}$, $3.6$, and $2.2\times 10^9$.

The largest singular value obeys the inequalities that Opalg has just referred to.
That is:
$$\frac 1{\sqrt n}\|A\|_\infty \le \|A\|_2 \le \sqrt m\|A\|_\infty \quad\implies\quad \frac{1}{\sqrt 3}\cdot 3\times 10^9 \le 2.2\times 10^9 \le \sqrt 3\cdot 3\times 10^9$$

another pathological but bad example for me is

-149 -50 -154
537 180 546
-27 -9 -25

its spectral radius is 3 , but its spectral norm is 817. ||A|| infinity is 1263 (spectral norm is big and norm infinity is comparable so bad example for me).
We see the same relationship that spectral norm and infinity norm are within a factor $\sqrt 3$ of each other.
 
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sarrah

Member
Jun 20, 2014
60
I don't know how to thank you for your extreme help and time spent to help me out. You are a kind person Also for Oplag I am grateful to him he helped me in another situation. God bless you both.
Sarrah