Can I assign a variable to a number in a matrix from another matrix?

Using the matrix that comes from eig(), I would like to assign individual values in it to 3 separate variables, and then plug those into a separate matrix. However, I get an error. How do I do this?

E = eig(A)

E(1,1) = w1;
E(2,1) = w2;
E(3,1) = w3;

ans = [m1*w1^2 0 0;
       0    m2*w2^2 0;
       0 0 m3*w3^2 0]


Unrecognized function or variable 'w1'.

Error in eigen (line 23)
E(1,1) = w1;

Answer 1

[Walter Roberson]

A = magic(7)

A = 7×7

30 39 48 1 10 19 28 38 47 7 9 18 27 29 46 6 8 17 26 35 37 5 14 16 25 34 36 45 13 15 24 33 42 44 4 21 23 32 41 43 3 12 22 31 40 49 2 11 20

E = eig(A)

E = 7×1

175.0000 -56.4848 -31.0882 -25.3967 56.4848 25.3967 31.0882

w1 = E(1,1)

w1 = 175

w2 = E(2,1)

w2 = -56.4848

w3 = E(3,1)

w3 = -31.0882

m1 = 3; m2 = 11; m3 = -7;

output = [
            m1*w1^2     0       0;
            0           m2*w2^2 0;
            0           0       m3*w3^2]

output = 3×3

9.1875 0 0 0 3.5096 0 0 0 -0.6765

[Anastasia Zistatsis]

After that change, I have a new error:

Error using vertcat
Dimensions of arrays being concatenated are not consistent.

Error in eigen (line 30)
output = [m1*(w1)^2 0 0;

[Anastasia Zistatsis]

Nevermind, I had an extra zero!

[Anastasia Zistatsis]

@Walter Roberson, what happens with output2? I can tell it isn't a simple w1^2 * m1. Is there an inverse happening?.

These are my numbers: m2 = 1, w2 = 6. The answer should be 2.449, but I am getting 36

[Walter Roberson]

I would need your A matrix to test with, and all your m* values.

[Anastasia Zistatsis]

I apologize for the weight, here you go @Walter Roberson

m1 = 1;
m2 = 1;
m3 = 1;
k = 2;

A = [2*k -k -k;
    -k 2*k -k;
    -k -k 2*k];

E = eig(A)

w1 = E(1,1)
w2 = E(2,1)
w3 = E(3,1)

output = [m1*(w1)^2    0          0;
            0       m2*(w2)^2     0;
            0          0      m3*(w3)^2]

[Walter Roberson]

syms k
A = [2*k -k -k;
    -k 2*k -k;
    -k -k 2*k];
E = eig(A)

E = 

If you are expecting 2.449 then your k would have to be roughly 0.816

Answer 2

[Steven Lord]

There's a way to do this without creating 2*n individual variables.

A = magic(7);
E = eig(A);

w = E(1:3);
m = [3; 11; -7];
output1 = diag(m.*(w.^2))

output1 = 3×3

9.1875 0 0 0 3.5096 0 0 0 -0.6765

w1 = E(1,1);
w2 = E(2,1);
w3 = E(3,1);
m1 = 3; m2 = 11; m3 = -7;
output2 = [ m1*w1^2     0       0;
            0           m2*w2^2 0;
            0           0       m3*w3^2]

output2 = 3×3

9.1875 0 0 0 3.5096 0 0 0 -0.6765

You can compare output1 and output2. The first block of code scales to larger matrices with more eigenvalues.