# A = bdb⁻¹

1. May 25, 2014

### Jhenrique

Given a matrix A, is possible to rewrite A like:

$A = B D B^{-1}$

$\begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \\ \end{bmatrix} = \begin{bmatrix} ?_{11} & ?_{12} \\ ?_{21} & ?_{22} \\ \end{bmatrix} \begin{bmatrix} \lambda_{1} & 0 \\ 0 & \lambda_{2} \\ \end{bmatrix} \begin{bmatrix} ?_{11} & ?_{12} \\ ?_{21} & ?_{22} \\ \end{bmatrix}^{-1}$

(if A is diagonalizable)

Being $\lambda_i$ the i-th root of the characterisc polynomial of A.

But, what is the definition of the matrix B in terms of A?

2. May 25, 2014

### HallsofIvy

Staff Emeritus
I believe you are asking if every matrix is "diagonalizable". The answer to that is "not every matrix"! A matrix is diagonalizable if and only if it has a "complete set" of eigenvectors. That is, an n by n matrix is diagonalizable if and only if it has a set of n independent eigenvectors. Since eigenvectors corresponding to distinct eigenvalues are always independent, if all eigenvalues of a matrix are distinct then it is diagonalizable. But a matrix with repeated eigenvalues may still be diagonalizable.

If A has n independent eigenvectors, then we can construct the matrix B having the eigenvectors of A as columns. Since the eigenvectors are independent, B is invertible and then we have $A= BDB^{-1}$.

The matrix $\begin{bmatrix}8 & -3 \\ 10 & -3\end{bmatrix}$ has eigenvalues 2 and 3. Eigenvectors corresponding to eigenvalue 2 are multiples of $\begin{bmatrix}1 \\ 2\end{bmatrix}$ and eigenvectors corresponding to eigenvalue 3 are multiples of $\begin{bmatrix}3 \\ 5 \end{bmatrix}$. So if we let $B= \begin{bmatrix}1 & 3 \\ 2 & 5\end{bmatrix}$, we have $B^{-1}= \begin{bmatrix}-5 & 3 \\ 2 & -1\end{bmatrix}$.

And then $$BDB^{-1}= \begin{bmatrix}1 & 3 \\ 2 & 5\end{bmatrix}\begin{bmatrix}2 & 0 \\ 0 & 3\end{bmatrix}\begin{bmatrix}-5 & 3 \\ 2 & -1\end{bmatrix}= \begin{bmatrix}8 & -3 \\ -10 & -3\end{bmatrix}$$

But, again, not every matrix is diagonalizable. The matrix $\begin{bmatrix}1 & 1 \\ 0 & 1\end{bmatrix}$ has 1 as a double eigenvalue but the only eigenvectors are the multiples of $\begin{bmatrix}1 \\ 0 \end{bmatrix}$.

Last edited by a moderator: Jun 1, 2014