Hi,
I wanna understand if the eigenbasis for this task can be computed directly from the given info or should I assume some potential basis and check if it worked out?
If I’m not mistaken I can only find two eigenvalues from this matrix: λ1 = 2 and λ2 = 1. Meaning I can only find two eigenvectors and therefore determine they’re eigenbasis.
Should I just assume and try some other bases to see if they’re actual eigenbasis or I was supposed to somehow get the third eigenvector from the calculation?
Solving the system for λ = 2 I can get multiple variations of the eigenvectors - how am I supposed to define which one works best?
Guys, I still have a problem understanding the choice of eigenvectors. We had only 2 eigenvalues, how did we obtain 3 eigenvectors?
I’m not sure if I can give the best explanation of this, but I’ll give it a shot. I assume that you’ve already figured out the eigenvector corresponding to the eigenvalue of 1.
For the eigenvalue of 2, I get the following corresponding system of equations (I’m using the variables x_1, x_2, x_3 for the vector x in the equation Px = 2x):
2x_1 = 2x_1
x_1 + 2x_2 + x_3 = 2x_2
-x_1 + x_3 = 2x_3
The first equation tells us nothing useful while the third equation tells us that x_1 = -x_3. Finally, if you plug x_1 = -x_3 into the second equation, you end up with 2x_2 = 2x_2, which tells us that x_2 can be anything.
Therefore, you can break down all possible solutions to Px = 2x as a linear combination of the two vectors [1, 0, -1] and [0, 1, 0]. The first vector ensures that x_1 = -x_3, while the second vector allows x_2 to be any real number.
For example, the vectors [6, 2, -6], [-1, 5, 1], [0.5, -15, -0.5] are all valid solutions to Px = 2x, but they cannot all be described as a scalar multiple of a single basis vector. In other words, there is no single vector that all possible solutions will be a multiple of. The proper basis for this solution set requires two basis vectors, in this case [1, 0, -1] and [0, 1, 0].
A more conceptual way of thinking about this is that the solution space of Px = x (eigenvalue of 1) are all points in 3-D space (since the vectors we are working with are three-dimensional) that happen to lie on the same line (1-D space), and thus only requires a single basis vector to describe the entire solution space. However, the solution space of Px=2x (eigenvalue of 2) are points in 3-D space that do not lie on the same line, but instead lie on the same plane (2-D space), and thus will require two basis vectors to describe the entire solution space. All linear combinations of a single vector will form a line, while all linear combinations of two independent vectors will form a plane.
Hope this helps.
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Thanks, it helped me understand the solution better, especially about vector [0, 1, 0] that allows x_2 to be any real number, but I still have some questions:
Why do we choose [-1, 0, 1] as a basis vector instead of [1, 0, -1]? According to the system of equations, we have that x_1 = x_1 and x_1 = -x_3, thus both vectors I mentioned above seem to be solutions.
Similarly couldn’t we have the eigenvector corresponding to eigenvalue 1 to be [0, 1, -1] instead of [0, -1, 1]?
Yup, you’re right. There is no unique solution, so any multiple of [-1, 0, 1] will do. [1, 0, -1] is perfectly valid, as are an infinite number of alternate basis vectors as well. As long as some multiple of each of the three different basis vectors (one basis vector coming from eigenvalue 1 and two basis vectors coming from eigenvalue 2), your solution is valid.
Think of it as three separate “families” of vectors: one consisting of all multiples of [0, 1, -1], one consisting of all multiples of [-1, 0, 1], and one consisting of all multiples of [0, 1, 0]. As long as one member of each “family” is represented in the eigenbasis, any valid solution will be a linear combination of the three vectors in your eigenbasis.
You could even specify an eigenbasis of [0, -10, 10], [pi, 0. -pi], and [0, sqrt(2), 0] if you wanted to. But for the sake of keeping things neat, it’s probably good practice to scale your eigenvectors so that they contain as many ones and zeros as much as possible.
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