Small Matrix Norm & Spectral Radius: Examples & Solutions

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In summary: In other words, for any two matrices $A$ and $B$, the following relation holds:||A-B|| = ||AB||This is known as the norm equivalence theorem.In summary, the norm equivalence theorem states that any two matrices have the same norm.
  • #1
sarrah1
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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
 
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  • #2
sarrah said:
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 $10$.
 
  • #3
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
 
  • #4
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.
 
  • #5
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
 
  • #6
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
 
  • #7
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.
 
  • #8
sarrah said:
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:
$$\frac 1{\sqrt n}\|A\|_\infty \le \|A\|_2 \le \sqrt m\|A\|_\infty \quad\implies\quad \frac{1}{\sqrt 3}\cdot 3\times 10^9 \le 2.2\times 10^9 \le \sqrt 3\cdot 3\times 10^9$$

sarrah said:
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.
 
  • #9
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
 

1. What is a small matrix norm?

A small matrix norm is a measure of the size or magnitude of a matrix. It is typically used to compare the size of different matrices or to determine the convergence of a matrix iteration method. A small matrix norm is usually defined as the maximum absolute value of the elements in the matrix.

2. How is the small matrix norm calculated?

The small matrix norm is calculated by taking the absolute value of each element in the matrix and then finding the maximum value among these absolute values. This can be represented mathematically as ||A||small = maxi,j|aij|, where A is the matrix and aij is the element in the i-th row and j-th column.

3. What is the significance of the small matrix norm?

The small matrix norm is significant because it provides a way to measure the size of a matrix and to compare the size of different matrices. It is also useful in determining the convergence of a matrix iteration method, as a smaller norm indicates faster convergence.

4. What is the spectral radius of a matrix?

The spectral radius of a matrix is the largest absolute value of its eigenvalues. It is a measure of the spread of the eigenvalues and is often used to determine the stability of a matrix iteration method.

5. How is the spectral radius related to the small matrix norm?

The spectral radius is related to the small matrix norm through the Gelfand formula, which states that the spectral radius is equal to the limit of the small matrix norm raised to the power of 1/n as n approaches infinity. This means that the spectral radius provides a tighter bound on the convergence of a matrix iteration method compared to the small matrix norm.

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