How do you show the following is diagonalizable? So if $A$ is diagonalizable, how would you show the following is diagonalizable?
a)$p(A),p(x)$ (any polynomial)
b)$kI+A$ for any scalar $k$
c)$U^{-1}AU$ for any invertible matrix $U$.
So for the first one, I have no idea how polynomials associate with diagonalization. And neither can think of anything for the third.
For the second one, I assume that $P^{-1}(kI+A)P=P^{-1}kIP+P^{-1}AP=kI+D$, since $kI$ and$D$ each is diagonalizable, $kI + D$ is diagonalizable. But I felt I have a wrong idea about this, someone help?
 A: If $A$ is diagonalizable, then there is $P$ invertible with $D = P^{-1}AP$ diagonal. If $p(x) = \sum_{k=0}^na_kx^k$, look at: $$P^{-1}p(A)P = P^{-1}\left(\sum_{k=0}^na_kA^k\right)P = \sum_{k=0}^n a_k D^k.$$Is that a diagonal matrix? How can you justify the steps I did above?
Your work in the second item is okay, because in the same way above: $$P^{-1}(kI+A)P = kI + D,$$which is a diagonal matrix.
A: It is good to recall that being diagonalisable is (or can be) defined in terms of the linear operator$~\phi$ represented by a matrix, rather than directly in terms of the matrix. Namely, an operator on a vector space $V$ is diagonalisable if the sum of its eigenspaces equals all of$~V$: every vector can be written as a sum of eigenvectors for$~\phi$.
Now on the eigenspace of$~\phi$ for eigenvalue$~\lambda$, any polynomial $P[\phi]$ of$~\phi$ acts as multiplication by the scalar$~P[\lambda]$. In particular any eigenvector for$~\phi$ is also an eigenvector for$~P[\phi]$. It follows that if $\phi$ is diagonalisable, then so is $P[\phi]$. (And on a basis where the matrix of $\phi$ is diagonal, so is that of $P[\phi]$).
Your question b) is a special case of a), for $P=X+k$.
Your question c) is immediate, since one can interpret this as a change of basis for one same linear operator.
