Proving Norm of Matrix Inequality for Homework

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Homework Help Overview

The discussion revolves around proving an inequality related to the norm of a matrix, specifically showing that the maximum absolute value of its entries is less than or equal to the matrix norm, which is also bounded above by a square root of the sum of the squares of its entries.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to establish a lower bound for the matrix norm using a specific choice of vector. They express uncertainty regarding how to approach the upper bound. Another participant suggests a potential direction involving the Cauchy-Schwarz inequality.

Discussion Status

The discussion is ongoing, with participants exploring different aspects of the problem. Some guidance has been offered regarding the use of the Cauchy-Schwarz inequality, but there is no explicit consensus or resolution yet.

Contextual Notes

The original poster indicates a lack of clarity on how to proceed with the upper bound of the inequality, which may suggest that additional information or context is needed for a complete understanding of the problem.

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Homework Statement


Let A = [a_{ij}] be a mxn matrix. Show that max[tex]_{ij}[/tex]|a[tex]_{ij}[/tex]| ≤ ‖A‖ ≤ √(∑[tex]_{ij}[/tex]|a[tex]_{ij}[/tex])|

Homework Equations


The Attempt at a Solution



By the definition ‖A‖=max_{||x||≤1}‖A(x)‖ for all x ∈ Rⁿ.So, ‖A‖≥‖A∘(x₁,..,x_{n})[tex]^{T}[/tex]‖ for x = (0,...,1,...0) with 1 is in the i[tex]^{ij}[/tex] position and so ‖A‖ ≥ ‖A∘(x₁,..,x_{n})[tex]^{T}[/tex]‖ = ||(a[tex]_{i1}[/tex],a[tex]_{i2}[/tex],...,a[tex]_{ij}[/tex])|| = √(a[tex]_{i1}[/tex][tex]^{2}[/tex]+...+a[tex]_{in}[/tex]) ≥ max[tex]_{ij}[/tex]|a[tex]_{ij}[/tex]|.
I do not know what how to do the upper bound.
 
Last edited:
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Anyone?
 
That's sort of hard to read - do you want to prove that [itex]\| A \|^2 \leq \sum_{ij} |a_{ij}|^2[/itex]?

If so, the Cauchy-Schwarz inequality will be very useful.
 
Thank you for replying!
I will think about this.
 
Last edited:

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