Statement 1: if `D=`diag`[d_1, d_2, ,d_n]`,then `D^(-1)=`diag`[d_1^(-1),d_2^(-1),...,d_n^(-1)]` Statement 2: if `D=`diag`[d_1, d_2, ,d_n]`,then `D^n=`diag`[d_1^n,d_2^n,...,d_n^n]`

Explanation

Let
`f(x)=a_(0)+a_(1)x+a_(2)x^(2)+...+a_(n)x^(n)`
`implies f(D)=a_(0)I+a_(1)D+a_(2)D^(2)+...+a_(n)D^(n)`
`=a_(0)xx"diag. "(1, 1, ... , 1)+a_(1)xx"diag. "(d_(1), d_(2), ..., d_(n))+a_(2)xx"diag. "(d_(1)^(2), d_(2)^(2), ..., d_(n)^(2))`
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`="diag."(a_(0)+a_(1)d_(1)+a_(2)d_(1)^(2)+...+a_(n)d_(1)^(n), a_(0)+a_(1)d_(2)+a_(2)d_(2)^(2)+...+a_(n)d_(2)^(n), a_(0)+a_(1)d_(3)+a_(2)d_(3)^(2)+...+ a_(n)d_(3)^(n)`,
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`a_(0)+a_(1)d_(n)+a_(2)d_(n)^(2)+...+a_(n)d_(n)^(2))`
`="diag."(f(d_(1)), f(d_(2)), ..., f(d_(n)))`