MATH319 Slides 27 Polynomials of a diagonable matrix 29 Functions of the matrix

28 Diagonalizing the matrix

Proof. Let $X_{j}\neq 0$ be a $n\times 1$ column such that $AX_{j}=\lambda_{j}X_{j}$ , and let $S=[X_{1}\,X_{2}\,\dots X_{n}]$ ; then the $X_{j}$ are linearly independent by Chapter 6 of MATH220 and hence $S$ has column rank $n$ . Hence $S$ is invertible. Now let

D=\begin{bmatrix}\lambda_{1}&0&0&\dots\\ 0&\lambda_{2}&0&\ddots\\ \vdots&\vdots&\ddots\\ 0&\dots&0&\lambda_{n}\end{bmatrix}

and note the chain of identities

AS=A\begin{bmatrix}X_{1}&\dots&X_{n}\end{bmatrix}=\begin{bmatrix}AX_{1}&\dots AX% _{n}\end{bmatrix}

=\begin{bmatrix}\lambda_{1}X_{1}&\dots\lambda_{n}X_{n}\end{bmatrix}=\begin{% bmatrix}X_{1}&\dots X_{n}\end{bmatrix}D=SD

where $S$ is invertible, so $A=SDS^{-1}$ .