Hey @Sri_Vardhamanan, @bs80, @titocaput, @stesoye and @farmerinatechstack,
Thanks for raising the question, since after watching the video, this question arises trivially. But before beginning with this reply, I would like to mention that Week 4’s programming assignment highlights and describes this issue as well, along with, how to deal with it while programming. So, feel free to take a look at the assignment once before proceeding with the lecture videos further.
Now, let’s try to delve it a bit deeper into this (using the in-lecture quiz example). In order to find the eigenvalues, we try to do something like follows:
The next step is as follows:
Now, for finding the eigenvalues, we use the fact that this matrix has to have infinite solutions, and hence, this matrix has to be singular, i.e., the determinant of the matrix must be 0. This gives us the eigen-values.
Now, if we take a look at the above equation, \lambda here represents the eigen-values. That means when we substitute the computed eigen-values in place of \lambda, once again, this has to have infinite solutions. In simple words, there are infinite values of x and y, which will satisfy this equation, for both the eigen-values; the very observation which all of us have come across.
In other words, for a given value, we can find infinite eigen-vectors. But the observation to make here is that all of these eigen-vectors are scalar multiples of each other.
Let’s try to see if this is true. This matrix has 2 eigen-values, 1 and 11. Let’s take 1 as an example. In this case, we will substitute \lambda = 1 in the previous equation. We will get the follows:
which will lead to the following equations:
Now, both of these equations are pretty much the same, and will give us the single equation 2x + y = 0 or y = -2x. Now, some of the different vectors satisfying this are as follows:
(1, -2); (2, -4); (-1, 2); (-2, 4) …
But note that all these are scalar multiples of each other.
Now, coming to the last question, “So how do they have decided only one eigen-vector for each eigen-value, out of infinite many?”. Well, I haven’t found it anywhere written specifically, but I could make a teeny tiny observation of my own.
Let’s once again consider the same example. For \lambda = 1, we got the equation of x and y as 2x + y = 0. So, if you follow the below steps in order, you will find the eigen-vector listed in the quiz:
- Isolate one of the variables such that it has a leading coefficient of +1, and the other variable has an integer coefficient (no fraction). For our example, x = -0.5y and y = -2x both have leading coefficients of isolated variables as +1, but the former has a fractional coefficient for the other variable, so we rule the former one out.
- Now, for the variable with leading coefficient not equal to 1, substitute +1, and get the value for the other variable.
- If needed (to match the options), you can multiply the values by -1. Voila, we get the one eigen-vector.
Let’s test this approach out:
- For \lambda = 1, we get y = -2x, we put x = 1, y = -2. To match the options, multiplying by -1, voila, (-1, 2)
- For \lambda = 11, we get x = 2y, we put y = 1, x = 2, voila, (2, 1).
Please note that this approach is just for this in-quiz example, and is not some mathematically proven approach 
For any subjective question, you can write any of the infinite eigen-vectors, and you should be good to go, unless they have given you some constraint, for instance, write the eigen-vectors, such that their norm is 1. In that case too, you will have 2 eigen-vectors, since multiplying by -1 won’t change the norm 
For any objective question, choose all the answers will scalar multiples of what you have found as the eigen-vectors.
Resources
Cheers,
Elemento