# Check whether the matrices A and B are diagonalisable?Diagonalise those matrices which are diagonalisable. (i) A={[-2,-5,-1],[3,6,1],[-2,-3,1]} (ii) B={[-1,-3,0],[2,4,0],[-1,-1,2]}.

Feb 15, 2018

$A$ is not diagonalizable.
$B$ is diagonalizable and we have $B = S D {S}^{- 1}$ where one diagonalizing matrix is $S = \left(\begin{matrix}1 & 0 & 3 \\ - 1 & 0 & - 2 \\ 0 & 1 & 1\end{matrix}\right)$ and the diagonal matrix is $D = \left(\begin{matrix}2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 1\end{matrix}\right)$

#### Explanation:

For the matrix $A$, the characteristic equation is
$| \left[- 2 - \lambda , - 5 , - 1\right] , \left[3 , 6 - \lambda , 1\right] , \left[- 2 , - 3 , 1 - \lambda\right] | = 0$
After some algebra this can be recast in the form
$4 - 8 \lambda + 5 {\lambda}^{2} - {\lambda}^{3} = 0$
or $\left(1 - \lambda\right) {\left(2 - \lambda\right)}^{2} = 0$
Thus the eigenvalues are 1,2 and 2.

Since the eigenvalue 2 is repeated (it is algebraically degenerate) there is a possibility that there may not be enough independent eigenvectors to form a diagonalizing matrix. To check that, let us try to find the independent eigenvector(s) corresponding to $\lambda = 2$ :

$\left(\begin{matrix}- 4 & - 5 & - 1 \\ 3 & 4 & 1 \\ - 2 & - 3 & - 1\end{matrix}\right) \left(\begin{matrix}x \\ y \\ z\end{matrix}\right) = \left(\begin{matrix}0 \\ 0 \\ 0\end{matrix}\right)$

By elementary row operations (or just by appropriately adding or subtracting the equations) we get $x = - y = z$ so that the only independent eigenvector corresponding to $\lambda = 2$ is ${\left(1 , - 1 , 1\right)}^{T}$.
Since $A$ has only two independent eigenvectors (one for $\lambda = 1$ and one for $\lambda = 2$) it is not diagonalizable.

It is easy to see that $B$ has the same characteristic equation as $A$. Once again, let us examine the case for $\lambda = 2$ closely.
Here the eigenvector(s) satisfy

$\left(\begin{matrix}- 3 & - 3 & 0 \\ 2 & 2 & 0 \\ 1 & 1 & 0\end{matrix}\right) \left(\begin{matrix}x \\ y \\ z\end{matrix}\right) = \left(\begin{matrix}0 \\ 0 \\ 0\end{matrix}\right)$

It is obvious that in this case the only constraint on the eigenvectors is $x + y = 0$. This means that we can write down two independent eigenvectors, say ${\left(1 , - 1 , 0\right)}^{T}$ and ${\left(0 , 0 , 1\right)}^{T}$

To complete the diagonalization, we need to find the eigenvector corresponding to $\lambda = 1$. It is easy to check that one such eigenvector is ${\left(3 , - 2 , 1\right)}^{T}$

So, a diagonalizing matrix is
$S = \left(\begin{matrix}1 & 0 & 3 \\ - 1 & 0 & - 2 \\ 0 & 1 & 1\end{matrix}\right)$