Prove Existence of Basis with Zero Second Column in Linear Transformation Matrix

To prove this, we will use the Rank-Nullity theorem.

Since Null(T) is not equal to {0}, it means that there exists a non-zero vector v such that T(v) = 0. Let's call this vector v_0.

Now, let's extend v_0 to a basis of Null(T). Since Null(T) is a subspace of R^2, it must have dimension at most 2.

If dim(Null(T)) = 1, then v_0 is already a basis for Null(T). We can choose any other vector u in R^2 that is not a scalar multiple of v_0, and together with v_0, they form a basis for R^2. Let's call this basis a = {v_0, u}.

In this case, the matrix [T]_a will have the second column as the zero vector, since T(v_0) = T(u) = 0.

If dim(Null(T)) = 2, then Null(T) is the entire R^2 space, which means T is the zero transformation. In this case, we can choose any basis for R^2, and the second column of [T]_a will be the zero vector.

Therefore, in both cases, we have shown that there exists a basis a of R^2 such that the second column of [T]_a is the zero vector.