Upper bound to 2-norm of a matrix

[Simfish]

Homework Statement

Using the fact that [tex]||A||_2 = \sqrt { \rho ( A^* A )}[/tex], prove that
[tex]||A||_2 \leq \sqrt { ||A||_1 ||A||_\infty }[/tex]. This is an easy estimate to
find in practice for an upper bound on [tex]||A||_2[/tex].

Homework Equations

The Attempt at a Solution

Or, in other words, the magnitude of the largest eigenvalue of A*A <= the product of the maximum row sum and the maximum column sum.

So I've been trying to proceed and I'm not totally sure what to do. One thing I've figured: equality exists when the matrix is a diagonal matrix (so the entire matrix's values occur where they contribute to the eigenvalues).

There is an inequality [tex]||A||_2 < \sqrt{m} * ||A||_\infty[/tex] but I doubt this would help since the inequality would contain an extra factor of [tex]\sqrt{m}[/tex] (and an extra factor of sqrt(n) too, once you accounted for the 1-norm)

 

[mighty2000]

it is important to understand the mathematical principles behind the equations and not just rely on practical estimates. Therefore, let us start by understanding what the 2-norm, 1-norm, and infinity-norm represent.

The 2-norm of a matrix A, denoted by ||A||_2, is the maximum singular value of A. This is equivalent to the square root of the largest eigenvalue of A^*A, where A^* is the conjugate transpose of A. In other words, it represents the maximum stretch factor of A in any direction.

The 1-norm of a matrix A, denoted by ||A||_1, is the maximum absolute column sum of A. This represents the maximum absolute value of the sum of the elements in any column of A.

The infinity-norm of a matrix A, denoted by ||A||_\infty, is the maximum absolute row sum of A. This represents the maximum absolute value of the sum of the elements in any row of A.

Now, let us prove the given inequality using the fact that ||A||_2 = \sqrt { \rho ( A^* A )}. We can rewrite the inequality as:

\sqrt { \rho ( A^* A )} \leq \sqrt { ||A||_1 ||A||_\infty }

Squaring both sides, we get:

\rho ( A^* A ) \leq ||A||_1 ||A||_\infty

Now, let us consider the spectral decomposition of A^*A, which states that A^*A = U \Lambda U^*, where U is a unitary matrix and \Lambda is a diagonal matrix with the eigenvalues of A^*A on the diagonal. Therefore, we have:

\rho ( A^* A ) = \rho ( U \Lambda U^* ) = \rho ( \Lambda )

Using the fact that the maximum eigenvalue of a diagonal matrix is the maximum value on its diagonal, we can rewrite this as:

\rho ( A^* A ) \leq \max_i \lambda_i

Where \lambda_i represents the i-th diagonal element of \Lambda. Now, let us consider the maximum row sum and the maximum column sum of A. Using the definition of the 1-norm and infinity-norm, we have:

||A||_1 = \max