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Let T1 : R3 −→ R2 be the linear transformation given by the matrix 1 3 2 2 0 −1 and let T2 : R2 −→ R1 be the linear transformation given by the matrix [5 3]. (a) The kernel of T1 is one dimensional. Find a vector u1 which spans Ker(T1) (i.e., a basis for Ker(T1)).
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that is, you want to find the solution to the system 1x + 3y + 2z = 0 2x + 0y +-1z = 0 because that's what it means for the a column vector (x y z)^t to be in the kernel of T1.
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