Question

Question #23912

Answer 1

See below

Explanation:

`A = ((1,0),(9,4))`

To diagonalise, we need the matrix's eigenvectors. And .... the matrix needs to offer up 2 distinct eigenvectors in order to be diagonalisable.

Handy trick: For a triangular matrix, the eigenvalues are its diagonal entries.

But we can show that here by going the long way round: they are also the solutions to the characteristic equation, which for a `2 times 2` is:

`lambda^2 - Tr(A) lambda + det A = 0`

`implies lambda^2 - (1+4) lambda + (1*4 - 9*0) = 0`

`implies lambda = 1, 4`, the diaginal entries!

The eigenvectors (`mathbf alpha_(1,2)`) follow from the basic geometric eigenvalue idea that `A mathbf alpha_(1,2) = lambda_(1,2) \ mathbf alpha_(1,2)`:

• `lambda = 1`

`x_1 + 0 x_2 = x_1`
`9 x_1 + 4 x_2 = x_2`

From the first equation, `x_1` can be anything: we choose `x_1 = 1`. This means from the 2nd equation that `x_2 = -3`. Thus:

`mathbf alpha_1 = ((1),(-3))`

• `lambda = 4`

`x_1 + 0 x_2 = 4x_1`
`9 x_1 + 4 x_2 = 4 x_2`

From the first equation, `x_1 = 0` is the only solution. This means from the 2nd equation that `x_2 = 1`. Thus:

`mathbf alpha_2 = ((0),(1))`

The diagonalisation matrix P is then simply:

`P = (mathbf alpha_1, mathbf alpha_2) = ((1,0),(-3,1))`

The matrix D is a diagonal matrix that has the eigenvalues as it only non-zero entries:

`D = ((1,0),(0,4))`

We are asserting that:

`A = PDP^(-1)`

Now, we have everything apart from `P^(-1)`, which is:

`P^(-1) = ((1,0),(3,1))`

So our assertion here is that:

`((1,0),(9,4)) = ((1,0),(-3,1)) ((1,0),(0,4)) ((1,0),(3,1))`

Try it and see.

You can then do some seriously cool stuff with matrix. There is a point to diagonalisation :)