Problem 1: Let T:R2→R2 be the transformation that rotates each point in R2 about the origin through an angle ϕ, with counterclockwise rotation for a positive angle. Assuming such a transformation is linear, find the standard matrix A of this transformation.
Solution: It is enough to find T(e1),T(e2) where e1 and e2 are the columns of the identity matrix. According to the description of T, we have
T(e1)=(cosϕsinϕ),T(e2)=(−sinϕcosϕ).
Therefore the standard matrix
A=(T(e1)T(e2))=(cosϕsinϕ−sinϕcosϕ).
Problem 2: Let T be the linear transformation whose standard matrix is
A=100−4208−10135
Does T maps R4 onto R3? Is T a one-to-one mapping?
Solution: Because the standard matrix is already an echelon form with each row containing a pivot position, its columns span R3. In other words, T maps R4 onto R3. On the other hand, since not every columns is a pivot column, Tx=0 has a non-trivial solution. In other words, T is not one-to-one.
Problem 3: Let T(x1,x2)=(3x1+x2,5x1+7x2,x1+3x2). Show that T is a one-to-one linear transformation. Does T map R2 onto R3?
Solution: First we shall find the standard matrix of T. Let x1=1 and x2=0. We obtain
T(e1)=T(10)=351.
Let x1=0 and x2=1. We obtain
T(e2)=T(01)=173.
Therefore the standard matrix of T is
A=351173
Since the echelon form of A contains two pivot columns, T is one-to-one. On the other hand, since A has three rows, not every row has a pivot position. Therefore T does not map R2 onto R3. In general, a linear transformation never maps Rm onto Rn if m<n.