Math Is Fun Forum

GemmaJ1988

Member

Registered: 2008-10-08

Posts: 39

linear algebra

can anyone give an example of a linear map such that Im(f)=Ker(f)
im(f)=image of f
ker(f)=kernal of f

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: linear algebra

What is the kernel of a linear transformation?  Once you answer this, you'll see that there is really only one such linear map, and it should be rather straightforward as to what it is.

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: linear algebra

Are you sure? I'd have thought there are lots, unless I'm misremembering my definitions.
For example, f: R² --> R²; f(x,y) = (0,x) would be one, but there are plenty of others.

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Why did the vector cross the road?
It wanted to be normal.

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: linear algebra

The image of your mapping is 0 x R, the kernel is (0,0).

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: linear algebra

I'm not sure where you think I'm wrong, so here's my reasoning:

For a map f: X --> Y, I remember kernel to be defined as:

Using the f in my previous post, f(0,y) --> (0,0) for any y.
Therefore Im(f) = ker(f)

???

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Why did the vector cross the road?
It wanted to be normal.

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: linear algebra

Ah, you're right.  I was not considering mapping between spaces of different dimensions.

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."