# Thread: Prove that if A is diagonalizable, so is its inverse

1. ## Prove that if A is diagonalizable, so is its inverse

Let $\displaystyle A$ be an invertible matrix. Prove that if $\displaystyle A$ is diagonalizable, so is $\displaystyle A^{-1}$.

I'm afraid I don't quite know how this could be proven. This is the closest thing I have: proving that if $\displaystyle A$ is diagonalizable, so is $\displaystyle A^T$.
$\displaystyle P^{-1}AP=D\Rightarrow P^TA^T(P^T)^{-1}=D$ Since $\displaystyle D=D^T=(P^{-1}AP)^T=P^TA^T(P^T)^{-1}$
I doubt that proof is even relevant to the question. Does anyone know how to prove it?

2. A matrix is diagonizable iff it acts on a basis by scaling each vector by a certain amount. Then its inverse acts on that basis by scaling each vector by the inverse amount. Equivalently, if a_{kk} are the diagonal entries of a diagonal matrix, its inverse is the diagonal matrix with diagonal entries 1/x_{kk}.

3. [ Different wording from maddas but essentially the same thing: A matrix, A, is diagonalizable if and only if there is a basis for the vector space consisting of eigenvectors of A. That is, for every basis vector $\displaystyle v_i$, $\displaystyle Av_i= \lambda_i v_i$ (maddas' "scaling" each vector).

Applying $\displaystyle A^{-1}$ to both sides of that, $\displaystyle A^{-1}(Av_i)= A^{-1}(\lambda_iv_i)= \lambda_i A^{-1}v_i$ so that $\displaystyle A^{-1}v_i= \frac{1}{\lambda_i} v_i$.

Of course, in order to have an inverse, A must not have any 0 eigenvalues. Then every eigenvector of A is an eigenvector of $\displaystyle A^{-1}$.

4. That's a good one, but not in terms of what we were given. I found another source which put in this way.

$\displaystyle A=Q^{-1}DQ$, where $\displaystyle D$ is diagonal.
Since $\displaystyle A$ is an invertible matrix, $\displaystyle Q^{-1}DQ$ has an inverse as well, which means $\displaystyle D$ is invertible.
$\displaystyle A^{-1}=(Q^{-1}DQ)^{-1}=Q^{-1}D^{-1}(Q^{-1})^{-1}=Q^{-1}D^{-1}Q$
Since $\displaystyle A$ is written in such a form, we can say that it is diagonalizable.

Is this proof solid enough?

5. You must observe that $\displaystyle D^{-1}$ is diagonal (which is trivial) but yea, looks like a fine proof.

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# proof for A being diagonalizable if inverse of A is diagonalizable

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