Finding the 2-norm of a 2x2 matrix

  • Thread starter TheFerruccio
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In summary, the conversation is about finding the matrix norm and using Lagrange multipliers. The norm of the matrix is found by adding the squares of the indices and taking the square root. However, this method does not work for all matrices and can lead to incorrect solutions.
  • #1
TheFerruccio
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Homework Statement


Find the matrix norm.

Homework Equations



[itex]\left(
\begin{array}{cc}
4 & 2 \\
2 & 1 \\
\end{array}
\right)[/itex]

The Attempt at a Solution



There are definitely different ways to solve this. I will use Lagrange multipliers.

[itex]\textbf{A}x =
\left(
\begin{array}{cc}
4x_1+2x_2 \\
2x_1+x_2 \\
\end{array}
\right)[/itex]

[itex](\left|\textbf{A}\textbf{x}\right|)^2=(4x_1+2x_2)(4x_1+2x_2)+(2x_1+x_2)(2x_1+x_2)=18x_1^2+20x_1 x_2+5x_2^2[/itex]

Use Lagrange multipliers at this step, with the condition that the norm of the vector we are using is x. The goal is to find the unit vector such that A maximizes its scaling factor.

[itex]f = 18x^2+20x_1x_2+5x_2^2+\mu(x_1^2+x_2^2)[/itex]

Find the derivatives in the ::x_1:: and ::x_2:: directions and set each to 0.

[itex]\frac{\partial f}{\partial x_1}=36x_1+20x_2+2\mu x_1=0[/itex]
[itex]\frac{\partial f}{\partial x_2}=20x_1+10x_2+2\mu x_2=0[/itex]

I am not sure where to go from here. I know that the norm of the matrix is 5, and I tried comparing coefficients to get different values for ::\mu:: and pick the right ::\mu:: for the right unit vector. If I were to directly compare coefficients, I get two different values for ::\mu::.
First value:
[itex]\mu = -18[/itex]
Second value:
[itex]\mu = -5[/itex]

These are associated with two unit vectors which would make the the two partial derivative equations hold true:
(1,0)
and
(0,1)

These are definitely wrong, because if I use the first unit vector (1,0), the associated magnitude of Ax would be root 18. If I were to use the second unit vector (0,1), the associated magnitude would be root 5.

What am I doing wrong?
 
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  • #2
I know this has to be the right method, but the last step that I am taking is wrong. Does anyone have any insight into what is going on? I'm wasting hours on this thing.
 
  • #3
I guess that's a no.

I take it I can just add up the squares of the indices, take the square root of the whole thing, and get the answer for the norm, which is 5.

But, why doesn't this work for the following matrix?

[itex]\left(
\begin{array}{ccc}
-1 & 0 & 0 \\
0 & 2 & 0 \\
0 & 0 & 4 \\
\end{array}
\right)[/itex]

The norm of this matrix is 4, but, when I add the squares, I get [itex]\sqrt{21}[/itex].
 
  • #4
I guess my question's not that interesting. Oh well. The test is in a few minutes. Too late now.
 

FAQ: Finding the 2-norm of a 2x2 matrix

1. What is the 2-norm of a 2x2 matrix?

The 2-norm of a 2x2 matrix is a way to measure the size of the matrix. It takes into account both the magnitude and direction of the matrix's elements.

2. How do you find the 2-norm of a 2x2 matrix?

To find the 2-norm of a 2x2 matrix, you first need to square each element of the matrix. Then, add the squared values together. Finally, take the square root of the sum to get the 2-norm.

3. Can the 2-norm of a 2x2 matrix be negative?

No, the 2-norm of a 2x2 matrix cannot be negative. It is always a positive value since it involves taking the square root of a sum of squared values.

4. What is the significance of the 2-norm in mathematics?

The 2-norm is commonly used in mathematics to measure the size of a matrix. It is also used in various applications, such as optimization problems and data analysis.

5. How does the 2-norm differ from other matrix norms?

The 2-norm is a specific type of matrix norm that is commonly used due to its relationship with the Euclidean distance. Other matrix norms may have different properties and uses in various mathematical applications.

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