Norm of a Matrix Homework: Show ||A|| $\leq$ $\lambda \sqrt{mn}$

In summary, the norm of a matrix ||\textbf{A}|| is the smallest number C such that |\textbf{Ax}| \leq C |\textbf{x}| for all vectors \textbf{x} in ℝn. In this problem, we are asked to show that for an m x n matrix \textbf{A} and \lambda = \max\{ |a_{ij}| : 1 \leq i \leq m, 1 \leq j \leq n \}, the norm of the matrix ||\textbf{A}|| is less than or equal to \lambda \sqrt{mn}. By letting \textbf{y} = \textbf{Ax
  • #1
Yagoda
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Homework Statement


Let [itex]\textbf{A}[/itex] be an m x n matrix and [itex]\lambda = \max\{ |a_{ij}| : 1 \leq i \leq m, 1 \leq j \leq n \}[/itex].

Show that the norm of the matrix [itex]||\textbf{A}|| \leq \lambda \sqrt{mn}[/itex].

Homework Equations


The definition I have of the norm is that [itex]||\textbf{A}||[/itex] is the smallest number such that [itex]|\textbf{Ax}| \leq ||\textbf{A}|| \, |\textbf{x}|[/itex] for all [itex]\textbf{x}[/itex] in ℝn.


The Attempt at a Solution


I let [itex]\textbf{y} = \textbf{Ax} [/itex] and so [itex]|\textbf{y}| = \sqrt{\sum_{i=1}^{m}y_i^2}[/itex].
I started by looking at a matrix [itex]\textbf{A}[/itex] with all the same entries so that any entry could be thought of as [itex]\lambda[/itex]. So when you calculate [itex]|\textbf{y}|[/itex] you will get [itex]m \lambda^2[/itex]'s under the radical along with sum of the [itex]x_1, \dots, x_n[/itex] squared. So in the more general case that not all the entires of [itex]\textbf{A}[/itex] are equal I can see how [itex]\lambda[/itex] would provide an upper bound.
But I'm having trouble seeing how the [itex]\sqrt{n}[/itex] comes into it.
 
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  • #2
So let A = ##\begin{pmatrix}
a& a & a \\
a & a & a\\
\end{pmatrix}##

(It's easier to type a than ##\lambda##)

and multiply it by X = (x,y,z). You get AX = ##\begin{pmatrix}
ax +& ay + & az \\
ax+ & ay + & az\\
\end{pmatrix}##

You started correctly computing the ||AX|| and got to where |AX| = a##\sqrt 2 \sqrt{x^2 + y^2 + z^2}##.

To see where the n comes in take X = (1,1,1,). Now what is ||AX|| less than?

In finishing up, keep in mind that the < must work for every possible X. So it is perfectly possible there is an X which forces ||A|| lower. This is a bound, but for a given A it may not be the least upper bound.
 
  • #3
If X = (1,1,1) then [itex]|\textbf{Ax}| = a \sqrt{2}\sqrt{3}[/itex], but I guess I am missing what happens if X is a different vector, say (1,2,3). Then [itex]|\textbf{Ax}| = a \sqrt{2}\sqrt{14}[/itex], which isn't less than [itex] a \sqrt{2}\sqrt{3}[/itex]. I think in this case [itex]\textbf{||A||}= \frac{a\sqrt{72}}{\sqrt{14}}[/itex] so in this case [itex]\textbf{||A||} \, \textbf{|x|}< \lambda \sqrt{mn} [/itex]. But my trouble is generalizing it.
 
  • #4
After fiddling a little bit with the inequalities, it seems like in the general case it would be helpful to show that [itex]\frac{x_1 + x_2 + \dots + x_n}{\sqrt{x_1^2 + \dots + x_n^2}} < \sqrt{n}[/itex].
Would this be a proper approach? And if so, is there some glaringly obvious fact I'm missing that would help me show this?
 
  • #5
Let X = ##(x_1,x_2,x_3)##. Let |x| be the largest of the 3 components. From our previous work we have ||AX|| ## \le a \sqrt 2 \sqrt {x_1^2+x_2^2 + x_3^2} \le a \sqrt 2 \sqrt 3 |x|##.

Go back to your definition of ||AX||. This is the smallest number C such that ##||Ax|| \le C ||x||## for every vector X. We have shown (generalizing from the 3x2 case) that there is one vector x = (1,1,1...) such that ||Ax|| ## \le a \sqrt m \sqrt n |x| \le a \sqrt m \sqrt n||X||##. So there is no way that C can be greater than ##a \sqrt m \sqrt n##. It could be smaller than that, which is why we have the inequality.

All this computation depended on A consisting of all a's. To finish up, you need to redo this deduction when A has varying entries ##a_{ij}##. In this case, choose ##a = max|a_{ij}|## and proceed more or less as we did before.
 
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FAQ: Norm of a Matrix Homework: Show ||A|| $\leq$ $\lambda \sqrt{mn}$

What is the norm of a matrix?

The norm of a matrix is a measure of its size or magnitude. It is defined as the maximum value of its singular values.

What is the significance of the norm of a matrix?

The norm of a matrix is important in many areas of mathematics and science, including linear algebra, functional analysis, and signal processing. It is used to quantify the distance between matrices and to measure the stability of numerical algorithms.

What does the equation ||A|| $\leq$ $\lambda \sqrt{mn}$ mean?

This inequality means that the norm of a matrix A is less than or equal to the product of the largest singular value (represented by the Greek letter lambda) and the square root of the number of rows (m) and columns (n) in the matrix.

Why is this equation important in linear algebra?

This equation is important because it provides a bound on the norm of a matrix, which can be used to analyze the properties and behavior of linear systems. It also helps to determine the accuracy and stability of numerical algorithms used to solve linear equations.

How can this equation be applied in real-world situations?

This equation can be applied in many real-world situations, such as image and signal processing, data analysis, and control systems. It can be used to analyze the stability and performance of systems represented by matrices and to design efficient algorithms for solving linear equations.

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