Need for Separate Basis for Kernel: Explained by Hello

[vish_maths]

hello :)

I was trying to prove the following result :
for a linear mapping L: V --> W
dimension of a domain V = dimension of I am (L) + dimension of kernel (L)

So, my doubt actually is that do we really need a separate basis for the kernel ?
Theoretically, the kernel is a subspace of the domain V . So, the basis for V can be used to express any vector in the kernel .

If we really need it, i think this might be the reason ( please check whether i am right or not )

The kernel directs the mapping to the zero vector. We are not interested in this trivial mapping. So, even though the basis vectors of V can clearly describe the vectors in kernel, we keep the vectors in kernel aside and are interested in only those vectors in V which do not produce the zero vector.

Then, since T(ki) = 0 where ki represents any vector in the kernel , ki too ought to be a part of the solution for any mapping. Hence, a separate basis which only can express any vector in the kernel is needed

Am i thinking on the right line ?

 

[Bacle2]

Hi,

In what sense do you need to find a basis for the kernel?

 

[SteveL27]

hello :)

I was trying to prove the following result :
for a linear mapping L: V --> W
dimension of a domain V = dimension of I am (L) + dimension of kernel (L)

So, my doubt actually is that do we really need a separate basis for the kernel ?
Theoretically, the kernel is a subspace of the domain V . So, the basis for V can be used to express any vector in the kernel .

It's true that any basis for V spans the kernel. But the representation would not be unique (unless the kernel is trivial). So a basis for V will be a spanning set, but not a basis, for the kernel.

 

[Bacle2]

Ah, I see , that's what he/you meant.

Then Stevel27 is right; unless L==0, the dimension of the space is larger than that

of the kernel, so that a basis for the space contains a linearly-dependent set.

 

[HallsofIvy]

For example, the linear transformation, from R² to R², defined by (x, y)-> (x- y, y- x) has {(x, y)| y= x} as kernel. The "standard basis" for R², {(1, 0), (0, 1)} spans the kernel but does not contain a basis for the kernel. On the other hand, a basis for the kernel can always be extended to a basis for the entire space: {(1, 1)} is a basis for the kernel and any set containing (1, 1) and a vector not a multiple of that, such as {(1, 1), (1, 0)} or {(1, 1), (0, 1)} or {(1, 1), (1, -1)}, is a basis for the entire space.

 

[vish_maths]

thank you all :)