00:01
So we want to ask the following question, is the sum of two eigenvectors also an eigenvector? so let's take some matrix a, and let's suppose that v1 and v2 are eigenvectors, and the corresponding eigenvalues are lambda 1 and lambda 2.
00:24
So lambda 1 and lambda 2 are the current -spiding eigenvalues.
00:30
Okay, so we know that a v1 is equal to lambda 1 v1 and a v2 is equal to lambda 2 v2 and the question is, does there exist? does there exist some, let's say, mu, such that a times v1 plus v2 is equal to mu times v1 plus v2 because this is what it would mean for v1 plus v2, i .e.
01:05
The sum of the two eigenvectors to be an eigenvector, it would mean there exists some constant such that a of v1 plus v2 is just some constant times v1 plus v2.
01:16
Okay, so let's look at a of v1 plus v2.
01:20
Well, we can just use linearity.
01:22
This is av1 plus av2, and we know what a of v1 and a of v2 are.
01:29
This is lambda 1 v1 plus lambda 2 v2 okay so we know this so from here we need to split into two cases case 1 lambda 1 and lambda 2 are equal so if this is the case then we have a of v1 plus v2 is equal to lambda 1 v1 plus lambda 2 v2 but now we can just use this fact that lambda 1 and lambda 2 are equal and we can write this is lambda 1 v1 plus lambda 1 v2.
02:01
We can put out a factor of lambda and we get lambda 1 v1 plus v2.
02:07
So in this case the sum of the two eigenvectors is indeed an eigenvector so in case that lambda 1 equals lambda 2 then v1 plus v2 is an eigenvector and then we need to look at case 2 lambda 1 and lambda 2 and not equal so then once again, we know that a of v1 plus v2 is equal to lambda 1 v1 plus lambda 2 v2.
02:42
And we want to ask, does there exist? does there exist, mu, such that a of v1 plus v2 equals mu of v1 plus v2? so let's suppose there does, and see that we arrive at the contradiction...