# Can anyone please check if i calculated it correctly? b=diag{1,0,3,5}, p(t)=t^2-1 is p(b)=diag{0,-1,8,24}?

## can anyone please check if i calculated it correctly? $b = \mathrm{di} a g \left\{1 , 0 , 3 , 5\right\} , p \left(t\right) = {t}^{2} - 1$ is $p \left(b\right) = \mathrm{di} a g \left\{0 , - 1 , 8 , 24\right\}$?

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I want someone to double check my answer

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SCooke Share
Feb 14, 2018

That is correct.

#### Explanation:

The transformation is of the b{diag} matrix through the p(t) function. Thus, applying the p(t) function to each matrix value derives the p(b) matrix.

$p \left(t\right) = {t}^{2} - 1$ where we substitute b{diag} for t.

$p \left(1\right) = {1}^{2} - 1 = 1 - 1 = \text{ } 0$
$p \left(0\right) = {0}^{2} - 1 = 0 - 1 = \text{ } - 1$
$p \left(3\right) = {3}^{2} - 1 = 9 - 1 = \text{ } 8$
$p \left(5\right) = {5}^{2} - 1 = 25 - 1 = \text{ } 24$

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