# Example of a linear map TϵL(R4→R4)T\epsilon \mathcal{L}(\mathbb{R}^4 \to \mathbb{R}^4) such that dim[null(T)]=dim[range(T)]\text{dim}[\text{null}(T)] = \text{dim}[\text{range}(T)]

Give an example of a linear map TϵL(R4→R4)T\epsilon \mathcal{L}(\mathbb{R}^4 \to \mathbb{R}^4) such that dim[null(T)]=dim[range(T)]\text{dim}[\text{null}(T)] = \text{dim}[\text{range}(T)].

I came up with Let (e1,e2,e3,e4) be a basis of R4 and let TϵL(R4→R4) such that\text{Let}\ (e_1, e_2, e_3, e_4)\ \text{be a basis of}\ \mathbb{R}^4\ \text{and let}\ T \epsilon \mathcal{L}(\mathbb{R}^4 \to \mathbb{R}^4)\ \text{such that} T(e1)=T(e2)=0T(e_1)=T(e_2)=0T(e3)=e3T(e_3)=e_3T(e4)=e4T(e_4)=e_4.

Is this example so trivial as to be invalid, and if so, what would have been a better example?

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Seems wrong. null(T)=span(e1,e2)null(T) = span(e_1,e_2), range(T)=span(e3,e4)range(T) = span(e_3,e_4).
– Simon
2 days ago

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This is wrong. You should put T(e3)=e2T(e_3)=e_2 and similarly for e4e_4 (for example) to make it valid. Besides that, no, it’s not that trivial, as in normal form they very much look like the one you wanted to produce
– b00n heT
2 days ago

Many apologies: I forgot to mention that I’m interested in dimension of the kernel and the image.
– HandsomeGorilla
2 days ago

By the rank-nullity theorem, any linear map with dim(null(T))=2\dim(\operatorname{null}(T)) = 2 will suffice.
– arkeet
2 days ago

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You are very close. You can fix it with:

T(e3)=T(e4)=0T(e_3) = T(e_4) = 0
T(e1)=e3T(e_1) = e_3
T(e2)=e4T(e_2) = e_4

Now you have the null(T)=span(e3,e4)=range(T).\text{null}(T) = \text{span}(e_3,e_4) = \text{range}(T).

Any 4×44 \times 4 matrix with two columns equal to zero and the other two linearly independent will give you a transformation with this property.

Apologies: I forgot to mention that I was comparing the dimension of the kernel to the dimension of the image
– HandsomeGorilla
2 days ago

@HandsomeGorilla No worries, I just didn’t want someone to come in and berate me for saying your example was wrong.
– Ken Duna
2 days ago