# Picking a particular column to norm in matrix

2 vues (au cours des 30 derniers jours)
Jiapeng le 21 Nov 2022
Déplacé(e) : Steven Lord le 21 Nov 2022
A = [1.02 0.95 0.77 0.67 0.56 0.30 0.16 0.01];
b = [0.39 0.32 0.22 0.18 0.15 0.12 0.13 0.15];
da = [1.02+(1.2e-3) 0.95+(6.4e-3) 0.77+(-4.8e-3) 0.67+(2.8e-3) 0.56+(-1.4e-3) 0.30+(3.2e-3) 0.16+(3.6e-3) 0.01+(-9.0e-4)];
db = [0.39+(-4.8e-3) 0.32+(1.9e-3) 0.22+(3.2e-3) 0.18+(4.8e-3) 0.15+(-2.1e-3) 0.12+(2.4e-3) 0.13+(-5.0e-3) 0.15+(2.2e-3)];
for i = 1:4
num = 10.^(-i);
a2 = A+num;
b2 = b+num;
da2 = da+num;
db2 = db+num;
[U1,S1,V1] = svd(a2,0);
[U2,S2,V2] = svd(da2,0);
x1(:,:,i) = V1.*(S1.^(-1)).*(U1').*b2;
x2(:,:,i) = V2.*(S2.^(-1)).*(U2').*db2;
end
disp(x1);
disp(x2);
I want to calculate the norm of the first column of x1 and x2 and show them. But I cannot be able to do it by picking a particular column to norm.
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John D'Errico le 21 Nov 2022
A = [1.02 0.95 0.77 0.67 0.56 0.30 0.16 0.01];
b = [0.39 0.32 0.22 0.18 0.15 0.12 0.13 0.15];
da = [1.02+(1.2e-3) 0.95+(6.4e-3) 0.77(-4.8e-3) 0.67+(2.8e-3) 0.56+(-1.4e-3) 0.30+(3.2e-3) 0.16+(3.6e-3) 0.01+(-9.0e-4)];
Invalid expression. When calling a function or indexing a variable, use parentheses. Otherwise, check for mismatched delimiters.
Firtst, you don't need to use parens around numbers. 1.2e-3 is just a number. MATLAB knows it is that.
But you DO need to use an operator. MATLAB does not recognize implicit multiplication or addition.
da = [1.02+(1.2e-3) 0.95+(6.4e-3) 0.77(-4.8e-3) 0.67+(2.8e-3) 0.56+(-1.4e-3) 0.30+(3.2e-3) 0.16+(3.6e-3) 0.01+(-9.0e-4)];
?
Next,
db = [0.39+(-4.8e-3) 0.32+(1.9e-3) 0.22+(3.2e-3) 0.18+(4.8e-3) 0.15+(-2.1e-3) 0.12+(2.4e-3) 0.13+(-5.0e-3) 0.15+(2.2e-3)];
for i = 1:4
num = 10.^(-i);
a2 = A+num;
b2 = b+num;
da2 = da+num;
db2 = db+num;
[U1,S1,V1] = svd(a2,0);
[U2,S2,V2] = svd(da2,0);
x1(:,:,i) = V1.*(S1.^(-1)).*(U1').*b2;
x2(:,:,i) = V2.*(S2.^(-1)).*(U2').*db2;
end
So you are apparently creating 3-dimensional arrays. What is the first column of a 3-dimensional array? What is the second column?

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### Réponses (1)

Star Strider le 21 Nov 2022
Déplacé(e) : Steven Lord le 21 Nov 2022
Consider using the vecnorm function.
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