Question

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]}`.

Answer 1

`A` is not diagonalizable.
`B` is diagonalizable and we have `B = SDS^{-1}` where one diagonalizing matrix is `S = ((1,0,3),(-1,0,-2),(0,1,1))` and the diagonal matrix is `D = ((2,0,0),(0,2,0),(0,0,1))`

Explanation:

For the matrix `A`, the characteristic equation is
`|[-2-lambda,-5,-1],[3,6-lambda,1],[-2,-3,1-lambda]| = 0`
After some algebra this can be recast in the form
`4-8 lambda + 5 lambda^2-lambda^3=0`
or `(1-lambda)(2-lambda)^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` :

`([-4,-5,-1],[3,4,1],[-2,-3,-1])([x],[y],[z]) = ([0],[0],[0])`

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 `(1,-1,1)^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

`([-3,-3,0],[2,2,0],[1,1,0])([x],[y],[z]) = ([0],[0],[0])`

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 `(1,-1,0)^T` and `(0,0,1)^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 `(3,-2,1)^T`

So, a diagonalizing matrix is
`S = ((1,0,3),(-1,0,-2),(0,1,1))`