Question

Let A be an n x n real diagonalizable matrix. Show that A + (alpha)I(n) is also real diagonalizable.

Answer 1

Show that this is also real diagonalizable.
\[A+\alpha I _{n}\]

Answer 2

@zzr0ck3r

Answer 3

To show that \[A+\alpha I_n\]is diagonalizable, we just need to show that it is similar to a diagonal matrix. That is, we need to show there exists an invertible matrix P such that:\[A+\alpha I_n = PDP^{-1}\]

Answer 4

where D is diagonal.

Answer 5

Since A is already diagonalizable, we can write A as:\[A=PD_1P^{-1}\]

Answer 6

So looking at\[A+\alpha I_n\]we see that:\[A+\alpha I_n=PD_1P^{-1}+\alpha PP^{-1}=P(D+\alpha I_n)P^{-1}\]

Answer 7

Note that \[D_1+\alpha I_n\] is a diagonal matrix. so you are done.

Answer 8

You make this seem so easy. I tried to do something similiar to that except I was putting all the entries to the matrix, and putting all the eigenvectors to v(n) in the equation, it was a mess... but when you put it like that it seems so easy... thank you very much... i hope you have a few moments, I have another I would like to ask.

Answer 9

and also in your equation, instead of the Diagonal, the book has A, for the original matrix... where D=P(-1)AP

Answer 10

Ah, thats fine. Its just a matter of perspective. \[A=PDP^{-1}\iff P^{-1}AP=D\]either one is fine. If your class writes it the other way, just change the proof around.

Answer 11

I don't think it matters to him either way, as long as I show what needs to be proven. I thank you again. Could you take a look at this next one I am stuck on.

Answer 12

Sure.