Faster way to calculate the inverse of matrice $C $(when diagonisable.. $C-1AC = D$)

Question

We found the orthonormal basis for the eigen spaces.

We got $C$ to be the matrix

[ 1/squareroot(2)        1/squareroot(6)         1/squareroot(3)
 -1/squareroot(2)       1/squareroot(6)         1/squareroot(3)
 0                      -2/squareroot(6)        1/squareroot(3) ]

And the original matrix $A $ is

[4 2 2
 2 4 2
 2 2 4]

After finding $C$, my notes jump to:

therefore $C^-1 A C = $
[2 0 0
 0 2 0
 0 0 8]

They do not show any steps on how to calculate the inverse of $C$. Is there an easy way of calculating it? How would I start off reducing it to RREF? How would I get rid of the square roots? (normally, I'm used to just dealing with regular integers).

Thanks in advance!

Answer 1

The matrix $C$ is orthogonal. Hence $C^{-1}=C^T$.

Answer 2

Notice that $ A = \begin{bmatrix}4 & 2 & 2 \\ 2 & 4 & 2 \\ 2 & 2 & 4 \end{bmatrix}$ has eigenvalues $2$ and $8$. Since there exists a basis of eigenvectors $v_1,v_2,v_3$ define the change of basis matrix $C = \begin{bmatrix} \vec{v_1} & \vec{v}_2 &\vec{v}_3\\\end{bmatrix}$. Then $$ C^{-1} A C$$ is the expression of $A$ in the basis $ \{v_1, v_2,v_3\}$. Since this is a basis of eigenvectors $C^{-1} AC$ is a diagonal matrix with the eigenvalues on the diagonal

$$ \begin{bmatrix} \lambda_1 & 0& 0 \\ 0 & \lambda_2 & 0 \\ 0 & 0 & \lambda_3 \end{bmatrix}$$

Where $\lambda_i$ is the eigenvalue corresponding to the eigenvector $v_i$. It is not necessary to explicitly calculate the inverse of $C$ and multiply the three matrices together. If you still wish to do it check out nicomezi's answer.