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Let T:V→V be a linear operator on a vector space V over C:

(a) Give an example of an operator T:C2→C2 such that R(T)∩N(T)={0} but T is not a projection

(b) Find a formula for a linear operator T:C^3→C^3 over C such that T is a projection with R(T)=span{(1,1,1)} and N(T)=span{(1,1,0);(0,1,1)}

I know that if T is a projection, then T2=T, but I don't know how to create one where R(T)∩N(T)={0}. As for (b), I try setting T(1,1,0)=(0,0,0) and T(0,1,1)=(0,0,0) but I'm not sure if this is correct or what to do afterwards.

R(T) is the range of T (I think it is also the image), N(T) is the null space or kernel of T

Thanks for any help

For a), an operator that has N(T) = {0} would work. This is saying that there are no vectors in C^2 that map to zero, i.e., the null space for T is empty. Thus, the vectors y in the transformation (with x ∈ C^2)

Tx = y ∈ C^2

span C^2 and so are not 0. Therefore R(T)∩N(T) = {0}.

For b), you started out right, stating that the transformation T maps vectors lying in the null space, in particular (1,1,0) and (0,1,1), to (0,0,0). You can also add the fact that T maps the vector (1,1,1) to (1,1,1); T(1,1,1) = (1,1,1). Note that a vector v ∈ N(T) can also be a linear combination of (1,1,0) and (0,1,1). Likewise, any vector along the line defined by (1,1,1), i.e., any multiple of (1,1,1), should also map to a vector along the line.

The transformation T can be represented by a matrix (since it is finite). Writing out the above relationships we have

T･(1,1,1)^t = (1,1,1)^t <-- I'm using the notation, ^t, for a transpose to turn (1,1,1) into a column vector just to be consistent

T･(1,1,0)^t = (0,0,0)^t

T･(0,1,1)^t = (0,0,0)^t

A more convenient way to write this is in matrix form

T･[ 1 1 0

1 1 1

1 0 1 ]

= [ 1 0 0

1 0 0

1 0 0 ].

T･X = Y.

If X has an inverse, then we could solve for T as

T = Y･X^-1.

First check det(X) = 1, which ≠ 0 so it does have an inverse. This is a good time for you to practice finding the inverse of a 3x3 matrix; it should be

X^-1 = [ 1 -1 1

0 1 -1

-1 1 0 ]

Multiplying X^-1Y =

T = [ 1 -1 1

1 -1 1

1 -1 1 ].

If you multiply T by any vector v ∈ c^3, you should get that all its components are equal and so lie along (1,1,1). Any linear combination of (1,1,0) and (0,1,1) should give (0,0,0).

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Comment | Thanks a lot for the help you provided. I have definitely gained a better understanding to the question. I just didn't realize how the range and kernel would relate to each other, but by rewriting the transformation as a matrix cleared it up. Thanks again for your help. |

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