Prove/disprove that $A = kI$

Any $n$ independent vectors form a basis, and any other vector can be expressed as a linear combination of them. Then applying the transform $A$, you get that vector times the unique Eigenvalue and $A=\lambda I$.

$$Ax=A(x_0 e_0+x_1 e_1+x_2e_2)=x_0 \lambda e_0+x_1 \lambda e_1+x_2\lambda e_2=\lambda x.$$

Since the matrix $A$ has $n$ linearly independent eigenvectors, then the matrix $P$ with columns being each eigenvector of $A$ is a full rank matrix. For example if the eigenvectors are $\begin{bmatrix}1\\2\end{bmatrix}$ and $\begin{bmatrix}3\\4\end{bmatrix}$ then the matrix $P$ would be:$$\begin{bmatrix}1&3\\2&4\end{bmatrix}$$which is full rank and invertible. Also according to definition if $v$ is an eigenvector of $A$ corresponding to eigenvalue $\lambda$ therefore$$Av=\lambda v$$since the columns of $P$ are all eigenvectors we have$$AP=A\begin{bmatrix}v_1&v_2&...&v_n\end{bmatrix}=\begin{bmatrix}Av_1&Av_2&...&Av_n\end{bmatrix}=\begin{bmatrix}\lambda_1v_1&\lambda_2v_2&...&\lambda_nv_n\end{bmatrix}$$where $v_i$'s and $\lambda_i$'s are eigenvectors and eigenvalues respectively. Also $\lambda_1=\lambda_2=...=\lambda_n=\lambda$ therefore we have$$AP=\begin{bmatrix}\lambda_1v_1&\lambda_2v_2&...&\lambda_nv_n\end{bmatrix}=\begin{bmatrix}\lambda v_1&\lambda v_2&...&\lambda v_n\end{bmatrix}=\lambda\begin{bmatrix}v_1&v_2&...&v_n\end{bmatrix}=\lambda P$$since $P$ is full rank then is invertible and we have$$AP=\lambda P\to APP^{-1}=\lambda PP^{-1}\to A=\lambda I$$ and our proof is complete