# Orthogonally diagonalize the 2 by 2 matrix. where a11= 1 a22= root2 b11=root2 and b22=0

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### 1 Answer

You need to orthogonally diagonalize the 2x2 matrix `A = ((1,sqrt2),(sqrt2,0))` , hence, you need to find first the eigenvalues lambda of the matrix A, such that:

`det(A - lambda*I_2) = 0`

`A - lambda*I_2 = ((1,sqrt2),(sqrt2,0)) - ((lambda,0),(0,lambda))`

`A - lambda*I_2 = ((1-lambda,sqrt2),(sqrt2,-lambda))`

`det(A - lambda*I_2) = [(1-lambda,sqrt2),(sqrt2,-lambda)]`

`det(A - lambda*I_2) = (-lambda)(1 - lambda) - sqrt2*sqrt2`

`det(A - lambda*I_2) = lambda^2 - lambda - 2`

You need to solve for lambda the equation `det(A - lambda*I_2) = 0` , such that:

`lambda^2 - lambda - 2 = 0`

`lambda_(1,2) = (1+-sqrt(1 + 8))/2 => lambda_(1,2) = (1+-3)/2`

`lambda_1 = 2; lambda_2 = -1`

You need to evaluate the associated eigenvectors `v_1, v_` 2, such that:

`lambda_1 = 2 => A*X = 2X => (A - 2I_2)*X = O_2`

`A - 2I_2 = ((1,sqrt2),(sqrt2,0)) - ((2,0),(0,2))`

`A - 2I_2 = ((-1,sqrt2),(sqrt2,-2))`

`(A - 2I_2)*X = O_2 =` > `((-1,sqrt2),(sqrt2,-2))*((x),(y)) = ((0),(0))`

`{(-x + sqrt2*y = 0),(sqrt2*x - 2y = 0):}`

`{(-sqrt2x + 2*y = 0),(sqrt2*x - 2y = 0):} => x = y = 0`

The eigenvector for `lambda_1 = 2 is v_1 = ((0),(0))`

`lambda_2 = -1=>A*X = -X => (A + I_2)*X = O_2`

`A + I_2= ((1,sqrt2),(sqrt2,0)) - ((1,0),(0,1))`

`A + I_2= ((0,sqrt2),(sqrt2,-1))`

`((0,sqrt2),(sqrt2,-1))*((x),(y)) = ((0),(0))`

`{(sqrt2*y = 0),(sqrt2x - y = 0):} => x = y = 0 => v_2 = ((0),(0))`

**The matrix P whose eigenvectors v_1 and v_2 are its columns,` P =((0,0),(0,0))` cannot be invertible since its determinant `Delta_P = 0,` hence, the given matrix cannot be diagonalizable.**