Why are these entries 0's? Look at the formula given for T. Do a21,a22,a23 appear anywhere in the transformed 2x2 matrix?_F_ said:N(T) = {[a/2, a, -2*a; 0, 0, 0] | a in F}
Basis(N(T)) = {[1/2 1 -2; 0, 0, 0]}
nullity(T) = 6
But nullity(T) should be 4...
Defennder said:Why are these entries 0's? Look at the formula given for T. Do a21,a22,a23 appear anywhere in the transformed 2x2 matrix?
The null space of a linear transformation T is the set of all vectors in the domain of T that map to the zero vector in the codomain. In other words, it is the set of all vectors that are "ignored" or "mapped to nothing" by the transformation.
The nullity of a linear transformation is the dimension of its null space. In other words, it is the number of vectors in the null space that are linearly independent. This is an important concept in linear algebra as it helps us understand the structure of a linear transformation and its effects on vectors.
The nullity of a linear transformation can be calculated by finding the dimension of the null space. This can be done by finding the basis for the null space and counting the number of vectors in the basis. Alternatively, the nullity can also be determined by subtracting the rank of the transformation from the dimension of the domain.
A nullity of 4 for a linear transformation T:M2x3(F) → M2x2(F) indicates that the dimension of the null space is 4, meaning there are 4 linearly independent vectors in the domain of T that are mapped to the zero vector in the codomain. This could suggest that the transformation is not one-to-one (since there are vectors in the domain that are mapped to the same vector in the codomain) and may have a non-trivial null space.
No, a linear transformation with a nullity of 4 cannot be invertible. In order for a linear transformation to be invertible, it must be both one-to-one (have a nullity of 0) and onto (have a rank equal to the dimension of the codomain). A nullity of 4 indicates that the transformation is not one-to-one, therefore it cannot be invertible.