# C1W4 Regarding co-ordinates of the new point using eigen-basis

In the 4th lecture of Eigenvalues and Eigenvectors at time 1:15 the instructor says that the vectors in the blue represent the basis vectors we are considering for given plane.

And then as shown in image he says that the vector (1, 0) moves to (2,0)
and vector (1,1) moves to (3,3).
But I think this is wrong and here is my explanation:
→ Basis Vectors for our plane: phi1 = (1, 0) and phi2 = (1, 1)
→ For (1, 0) if we apply the linear transformation we get (2, 0) this means
that we’re at point represented by 2 * phi1 + 0 * phi2 = 2 * phi1 (this is
because we chose phi1 and phi2 as our basis vectors) and since phi1 =
(1,0) so 2*phi1 = (2,0) which is correct.

→ Now here is the problem, for (1,1) if we apply the linear transformation
we get (3, 3) = 3phi1 + 3phi2 = 3(1,0) + 3(1,1) = (6,3) this also makes
sense as if you would walk 3 units along both the blue vectors you
would end-up at point (6,3).
But as you can see in image the result they show is that (1,1) ends-up at
(3,3).
I guess this would be true if we considered (1,0) and (0,1) as basis
vectors for our plane but since we considered (1,0) and (1,1) as basis, i
guess this is wrong.

What i think is happening here (but I am not familiar with this course), the blue matrix is the transformation matrix and as you mention it is composed of the blue vectors [2, 0] and [1, 3]. But when you tranform a point using these basis you have to multiply row of blue matrix with column of green point as 2×1+ 0×1 for the first coordinate and the second row with again the point l, to obtain the second coordinate of the transformed point!

Yeah that is correct and that’s what i showed above, but if you do this you get the point as a linear combination of the basis vectors you chose.
And since we chose the blue vectors as basis we would get (1,1) = (3,3) but (3,3) here would mean 3 * phi1 + 3 * phi2 as those were the basis we chose

So what he has shown is that (1, 0) and (1, 1) are eigenvectors of the transformation. They are also linearly independent, so they must be a basis for the transformation. But how does that get you to the equation you show?

Maybe I’m just missing your point, but if the question is how do you get (3,3) as a linear combination of the basis vectors, you have to solve this equation:

a * (1,0) + b * (1,1) = (3,3)

If you solve that, you get a = 0 and b = 3, right?

So basically my point was whenever we express a point in space we express it as a sum of basis vectors right.
Generally for cartesian co-ordinates our basis vectors are i and j so when we write (1,1) we mean 1i + 1j, i.e one unit along the first basis i and one unit along the second basis j.

But since in the lecture it was said that the eigen-vectors are our new basis vectors so every point in space would be expressed as their sum
thus (1,1) here would mean = 1phi1 + 1phi2 where phi1, phi2 correspond to the chosen basis vectors which in above case is (1,0) and (1,1).
So (1,1) here would mean 1 unit along basis1 (1,0) and 1 unit along basis2 (1,1).

Thus (3,3) would mean = 3phi1 + 3phi2 and if we change it to our usual i and j representation it would be 3*(1,0) + 3*(1,1) = (6,3).

When we describe a point, it is always with Cartesian coordinates. So (3,3) is with basis (1,0) and (0,1).

As I pointed out above, if you were going to rewrite in the way you described then it would be (0,3) in the new coordinate system. So that’s clearly not what is going on.