Question

A is n by n real matrix. Prove: If (A+I)(A+2I)=0, Then A must be diagonalizable.

Answer 1

here, (A+I)(A+2I)=0 is the minimal polynomial having distinct root of order one. therefore A is diagonolisable

Answer 2

\[(A + I)(A + 2I) = 0\]
\[({A^2} + 3A + 2)x = 0\]
\[({\lambda ^2} + 3\lambda + 2)x = 0\]
\[\lambda = - 1or\lambda = - 2\]
But A is n by n matrix, I want to prove it has n independent eigenvector.

Answer 3

The minimal polynomial has the form

\[m_A(t) = (t - \lambda_1)^{s _{1} } (t - \lambda_2)^{s _{2} } \dots (t - \lambda_r)^{s _{r} } \]
for some \[s_i \]. Morever, \[A\] is digonalisable iff all \[s_i = 1\]

Answer 4

...Sorry, I haven't learn that. Could you give me a link about that?

Answer 5

Maybe I have learn those knowledge, but in different form of it. From what I know, If A can be diagonalized, It must have n independent eigenvector.
\[{A^2} = A\]
Like that, I know it has eigenvalue 1 and 0.
If r(A)=r
then A the eigenvalue 1 repeated r times and 0 repeated n-r times.

Answer 6

Start with an arbitrary vector X. First show that (A+I)X = Y satisfies (A+2I)Y = 0. Then show (A+2I)X = Z satisfies (A+I)Z = 0. Then find a way to show that X is a linear combination of Y and Z. Then any vector in R^n is a combination of eigenvectors--i.e., there is a basis consisting of eigenvectors, so A must be diagonalizable.