Prove that any $n$ orthonormal vectors in $\mathbb{R}^{n}$ form a basis for $\mathbb{R}^{n}$.

Let $A$ be a real symmetric $n \times n$ matrix with $n$ orthonormal eigenvectors $\mathbf{e}_{i}$ and corresponding eigenvalues $\lambda_{i}$. Obtain coefficients $a_{i}$ such that

$$\mathbf{x}=\sum_{i} a_{i} \mathbf{e}_{i}$$

is a solution to the equation

$$A \mathbf{x}-\mu \mathbf{x}=\mathbf{f},$$

where $\mathbf{f}$ is a given vector and $\mu$ is a given scalar that is not an eigenvalue of $A$.

How would your answer differ if $\mu=\lambda_{1}$ ?

Find $a_{i}$ and hence $\mathbf{x}$ when

$$A=\left(\begin{array}{ccc} 2 & 1 & 0 \\ 1 & 2 & 0 \\ 0 & 0 & 3 \end{array}\right) \quad \text { and } \quad \mathbf{f}=\left(\begin{array}{l} 1 \\ 2 \\ 3 \end{array}\right)$$

in the cases (i) $\mu=2$ and (ii) $\mu=1$.