Show that if $R^n$ has an eigenvector basis for a matrix A, then there exists an eigenvector basis with respect to the transpose of A

Having a basis of eigenvectors is the same as a matrix being diagonalizable.

i.e. there exist an invertible matrix P such that $P^{-1}AP=D$ where $D=diag(\lambda_{1},\lambda_{2},...,\lambda_{n})$ the diagonal matrix whose diagonal entries are the eigen values of $A$. The colums of this matrix $P$ are precisely the vectors in the eigen basis.

So $D=D^{t}=(P^{-1}AP)^{t}=P^{t}A^{t}(P^{t})^{-1}$ . So I assume you might know that $A$ is invertible $\iff$ $A^{t}$ is invertible. So you have your invertible matrix $(P^{t})^{-1}$ which is doing the job.

Hence $A^{t}$ has an eigen basis. In otherwords, if $P$ is the matrix whose columns are the eigenvectors of $A$. Then the eigen basis wrt which $A^{t}$ is diagonal is given by the columns of $(P^{t})^{-1}$.