Let $v_1,v_2,v_3\in \mathbb {R}^3$ be basis vectors. Suppose there exists $3 \times 3$ matrix A and distinct numbers $\lambda_1,\lambda_2,\lambda_3$. Then $Av_i = \lambda_iv_i$ for $i = 1, 2, 3$. And $B=(v_1, v_2, v_3)$.
Let $X = (e_1, e_2, e_3)$ be the standard basis, $e_1= $$\begin{bmatrix}1\\0\\0\end{bmatrix}$ and so on. Prove that $C(B,X)AC(X,B)=$$\begin{bmatrix}\lambda_1 & 0 & 0 \\ 0 & \lambda_2 & 0 \\ 0 & 0 &\lambda_3\end{bmatrix}$
$C(X,B)$, $C(B,X)$ are change of basis matrices.
I understand that $C(X,B)=B$ because $X=I_3$ and $C(B,X)=B^{-1}$, which leaves me with $$B^{-1}AB= \begin{bmatrix}\lambda_1 & 0 & 0 \\ 0 & \lambda_2 & 0 \\ 0 & 0 &\lambda_3\end{bmatrix}$$
but I got stuck here.
Assuming $C(B2, B1)$ as the change of basis matrix:
$[v]_{B2} = C(B2, B1)[v]_{B1}$
where $[v]_{B2}$ is the coordinate with respect to the ordered basis $B2$, and similarly $[v]_{B1}$ for basis $B1$.
As $X$ contains the standard basis:
$C(X, B) = B$ and $C(B, X) = B^{-1}$
Now, for $Av_i = λ_i v_i$ for $i$ in $[1,3]$:
$$AB = A[v_1 \, v_2 \, v_3] = [Av_1 \, Av_2 \, Av_3]$$
Using matrix right multiplication:
$$= [λ_1 v_1 \, λ_2 v_2 \, λ_3 v_3]$$
Now, $B^{-1}(AB) = B^{-1}[λ_1 v_1 \, λ_2 v_2 \, λ_3 v_3]$:
$$= [λ_1 (B^{-1}v_1) \, λ_2 (B^{-1}v_2) \, λ_3 (B^{-1}v_3)]$$
Since $B^{-1}$ is the change matrix for the basis from $X$ to $B$:
$B^{-1}v_i = e_i$ where $e_i$ is the $i$-th standard vector.
Hence, the matrix will become:
$$[λ_1 e_1 \, λ_2 e_2 \, λ_3 e_3] = \begin{bmatrix} λ_1 & 0 & 0 \\ 0 & λ_2 & 0 \\ 0 & 0 & λ_3 \end{bmatrix}$$