Let’s take a close look at these red and blue vectors. In fact, let’s try looking at them in the original i hat j hat basis. In the original basis, the first vector can be written (1,0.5). That is to say it can be created with one i hat and a half a j hat. The second vector is (-1,1) so it can be created with minus one i hat and one j hat. Here, we’ve learned how to write the vectors of the new basis in the language of the old basis. So, we effectively know how to translate between them. Now, let’s see how we can use this information to take a vector expressed in one basis and find out how to express it in another basis. We know that we need two copies of the blue vector and one of the red vector to represent our original vector, let’s write that down. When we multiply it out, we see that we recover our vector as written in the i hat, j hat basis, which makes sense since we know that these vectors are also written in the i hat j hat basis. So, now we can see how translation between the two languages works. We need to write the new basis vectors in the language of the old basis, then the formula for building the vector in the language of the new basis gives us our vector in the language of the old basis. Conversely, if we were starting with (1,2) in our original language of i hat, j hat and we wanted to translate to this new basis, we’d have to find out how to write i hat and j hat in the new basis. Then, we’d multiply it by that matrix and outward pop our original vector in the new basis. It turns out that the matrix that tells you how to translate from the language of the old basis to the language of the new basis is just the inverse of the matrix that tells you how to translate from the language of the new basis to the language of the old basis. This makes sense because translating from the old basis to the new basis and back should have the effect of doing nothing and give you back what you started with. I’ll let you verify this for yourself.