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A matrix A belonging to $\displaystyle M_n$ is a square root of B belonging to $\displaystyle M_n$ if $\displaystyle A^2$=B. Show that every diagonalizable matrix in $\displaystyle M_n$ has a square root.
I know that a diagonal matrix has all zeros outside of the diagonal and the diagonal may or maynot have zeros. I want to say that these matrices must be similar as well.
I am not sure how to prove this. Any help would be great. Thanks!
Here is the jist of it. If a matrix is diagonalizable then there exists a matrix P such thatQuote:
Originally Posted by page929
$\displaystyle \mathbf{A}=\mathbf{PDP^{-1}}$ where
$\displaystyle \mathbf{D}= \begin{bmatrix}\lambda_1 & 0 & \cdots & 0 & 0 \\ 0 & \lambda_2 & \cdots & 0 & 0 \\ \vdots & & & 0& 0 \\ \vdots & & & & 0 \\ 0 & 0 \cdots & 0 & 0 & \lambda_n \end{bmatrix}$
Now define the matrix B as
$\displaystyle \mathbf{D}_B= \begin{bmatrix}\sqrt{\lambda_1} & 0 & \cdots & 0 & 0 \\ 0 & \sqrt{\lambda_2} & \cdots & 0 & 0 \\ \vdots & & & 0& 0 \\ \vdots & & & & 0 \\ 0 & 0 \cdots & 0 & 0 & \sqrt{\lambda_n} \end{bmatrix}$
Where $\displaystyle \mathbf{B}=\mathbf{PD}_B\mathbf{P^{-1}} $
Now we have that
$\displaystyle \mathbf{B}^2=(\mathbf{PD}_B\mathbf{P^{-1}})(\mathbf{PD}_B\mathbf{P^{-1}})=\mathbf{PD}_B^2\mathbf{P^{-1}}=\mathbf{PD}\mathbf{P^{-1}}=\mathbf{A}$
