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geometrically describe both.

the kernell would be just the zero vector, correct? and would also merely live in the 1st dimension, but geometrically speaking be non existent?

and for the range of the matrix, it would just be $\{(1,0,0),(0,1,0),(0,0,1)\}$. geometrically this would span the whole 3 dimensional space, correct?

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When to consider $Id$ as a linear transformation, and know its kernel is trivial, then the range should be clear once you remember the dimension formula, which is an alternative way to argue from the one you use. – gnometorule Mar 1 '13 at 16:21
up vote 1 down vote accepted

The kernel of the identity is indeed the zero vector. This can be thought of as the origin in three dimensions. This is a zero-dimensional point.

The image of the identity is the whole space itself, i.e. all of the three dimensional space.

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The kernel is a subspace of the domain, regarding the matrix as a transformation. So the kernel is written as $\{ (0, 0, 0) \}$. Now, you can certainly regard this kernel as a space as well, but it is properly called 0-dimension, not non-existant.

Your range is correct. All of $\mathbb{R}^3$.

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