Kade Reese
2022-07-25
Answered

If A is an n x n matrix , where are the entries on the main diagonal of A-A^T? Justify yoyr answer.

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Seromaniaru

Answered 2022-07-26
Author has **12** answers

A is nby n matrix

diagonal elements of matrix = diagonal elements of A^T

=> A - A^T=0 for elements in main diagonal of matrix A

there is no change in diagonal elements

diagonal elements of matrix = diagonal elements of A^T

=> A - A^T=0 for elements in main diagonal of matrix A

there is no change in diagonal elements

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Let $K=\mathbb{R}$ and $V:=\mathbb{R}[t{]}_{2}$ and let $F:\mathbb{R}[t{]}_{2}\to \mathbb{R}[t{]}_{2},f\mapsto f(0)+f(1)\cdot (t+{t}^{2})$

Want to calculate ${M}_{S}^{S}(F)$, the transformation Matrix of F with Basis S, being the standardbasis for the polynomialring $S=(1,t,{t}^{2})$

So we have to use the coordinate isomorphism:

${M}_{S}^{S}(F)=\left(\begin{array}{ccc}{I}_{S}(F(1))& {I}_{S}(F(t))& {I}_{S}(F({t}^{2}))\end{array}\right)$

Already know the solution, however I have no idea why the following matrix is the transformation Matrix of F.

${M}_{S}^{S}(F)=\left(\begin{array}{ccc}1& 0& 0\\ 1& 1& 1\\ 1& 1& 1\end{array}\right)$

Looking at the transformation matrix we know that $F(1)=1+t+{t}^{2}$?

Want to calculate ${M}_{S}^{S}(F)$, the transformation Matrix of F with Basis S, being the standardbasis for the polynomialring $S=(1,t,{t}^{2})$

So we have to use the coordinate isomorphism:

${M}_{S}^{S}(F)=\left(\begin{array}{ccc}{I}_{S}(F(1))& {I}_{S}(F(t))& {I}_{S}(F({t}^{2}))\end{array}\right)$

Already know the solution, however I have no idea why the following matrix is the transformation Matrix of F.

${M}_{S}^{S}(F)=\left(\begin{array}{ccc}1& 0& 0\\ 1& 1& 1\\ 1& 1& 1\end{array}\right)$

Looking at the transformation matrix we know that $F(1)=1+t+{t}^{2}$?

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