Diagonalisation and eigenvalues

Question

I have been wondering about the following:

If I have a matrix A, and I have some basis β, and I wish to find A in basis β, and I do that by applying the formula: β^-1 *A *β, and as a result I get a diagonal matrix C, does that mean that C is in fact also the diagonal matrix of A(with respect to standard basis) consisting of eigenvalues of A(with respect to standard basis) on its main diagonal, and that β has, in fact, eigenvectors of A(with respect to standard basis) in columns?

Thank you!

Answer 1

I don't know what is “the diagonal matrix of $A$”, but you are right about the rest: the entries of the main diagonal of $C$ are the eigenvalues of $A$ and the columns of $\beta$ are the eigenvectors of $A$.

Answer 2

Yes. Say you have $$ \beta^{-1}A\beta = D$$ where $D$ is diagonal with entries $\lambda_1,\ldots,\lambda_n.$. Let $v_i$ be the $i$-th column of $\beta.$ Then we have $$ Av_i = \beta D\beta^{-1}v_i.$$ Using the definition of inverse, we have $\beta^{-1}v_i = e_i$ where $e_i$ is the $i$-th standard basis element, so $$ Av_i = \beta De_i = \beta\lambda_ie_i.$$ And finally,we have $\beta e_i = v_i,$ so $$ A v_i = \lambda_i v_i.$$

Answer 3

With regard to your "eigenvalues of A(with respect to standard basis)", the eigenvalues of a linear transformation or matrix do NOT depend on the basis (the matrix corresponding to a given linear transformation depends on the basis but all equivalent matrices have the same eigenvalues).