= Solution
Since every <eigenvalue> has modulus less than one, $1$ is not an <eigenvalue> and $I-H$ is invertible. Hence the iteration has a unique <fixed point> $\widehat x=(I-H)^{-1}v$. The stated consistency implication gives $A\widehat x=b$, and nonsingularity of $A$ identifies $\widehat x=x^*$. This justifies using the fixed-point relation even though consistency was stated in only one direction.
Expand the initial error in the eigenvector <basis>, $x^0-x^*=\sum_i c_iw_i$. Subtract the fixed-point equation from the <stationary iterative method for a linear system> and iterate:
$$
x^k-x^*=H^k(x^0-x^*)=\sum_{i=1}^n c_i\lambda_i^k w_i.
$$
With $q=\max_i|\lambda_i|<1$, any <norm> obeys $\|x^k-x^*\|\le q^k\sum_i|c_i|\|w_i\|\to0$. Thus
$$
\boxed{x^k\longrightarrow x^*\quad\text{for every initial vector}}.
$$
The argument uses the given diagonalizable eigenbasis; the more general convergence criterion is <spectral radius> less than one.
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