Can I find a standard matrix when only two linear transformations are given?

Question

Hi!

I study linear algebra in Swedish so sorry if I get some of the words wrong.

My question is, how do you find A when A*(1,1,1)=(1,0,1) and A*(1,0-1)=(1,1,2) where A is a 3x3 matrix.

I know when three of these linear transformations are given you can easily work out L(e1), L(e2) and L(e3) and use those to build a matrix.

The question more specifically asks "Is there a linear transformation L so L(1,1,1)=(1,0,1) and L(1,0-1)=(1,1,2)?

If yes, how many L are there? Name such a matrix. If no, explain"

Thanks!

Answer 1

Yes, there is such a linear transformation. You could probably figure out the matrix for one by inspection but here's a more systematic way to do it.

Let {e1, e2, e3} be the standard ordered basis for R^3. Then the matrix A for the linear transformation L has columns L(e1), L(e2), and L(e3) and is given by A = [ L(e1) L(e2) L(e3)].

By expressing the product of A with (1, 1, 1) as a linear combination of the columns of A we get that L(e1) + L(e2) + L(e3) = (1, 0, 1).

Similarly, expressing A*(1, 0, -1) as a linear combination of the columns of A gives us the equation L(e1) - L(e3) = (1, 1, 2).

We now have a system of two equations in the three unknown column vectors.

Solving the second system equation for L(e1) gives us that L(e1) = (1, 1, 2) + L(e3). Substituting that into the first system equation and then solving for L(e2) gives us that L(e2) = (0, -1, -1) - 2L(e3).

We now have L(e1) and L(e2) solved for in terms of L(e3). Like a free variable you can let L(e3) be any column vector in R^3 you want. Hence, there are infinitely many such matrices for this linear transformation.

For example, let L(e3) = (0, 0, -1). Then L(e1) = (1, 1, 1) and L(e2) = (0, -1, 1). Using these column vectors as the column vectors for A gives us a matrix for L such that A*(1, 1, 1) = (1, 0, 1) and A*(1, 0, -1) = (1, 1, 2), as desired.