Let A be a square matrix such that AA^T = I. | vedclass

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(D) Given that $A$ is a square matrix such that $AA^T = I$. Since $A$ is a square matrix,$AA^T = I$ implies $A^TA = I$ as well.Now,consider the expres

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$A^3+A^T$

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(D) Given that $A$ is a square matrix such that $AA^T = I$. Since $A$ is a square matrix,$AA^T = I$ implies $A^TA = I$ as well.Now,consider the expression $\frac{1}{2} A[(A+A^T)^2 + (A-A^T)^2]$.Expanding the squares inside the brackets:$(A+A^T)^2 = A^2 + AA^T + A^TA + (A^T)^2 = A^2 + I + I + (A^T)^2 = A^2 + (A^T)^2 + 2I$.$(A-A^T)^2 = A^2 - AA^T - A^TA + (A^T)^2 = A^2 - I - I + (A^T)^2 = A^2 + (A^T)^2 - 2I$.Adding these two expressions:$(A+A^T)^2 + (A-A^T)^2 = (A^2 + (A^T)^2 + 2I) + (A^2 + (A^T)^2 - 2I) = 2A^2 + 2(A^T)^2$.Substituting this back into the original expression:$\frac{1}{2} A[2A^2 + 2(A^T)^2] = A[A^2 + (A^T)^2] = A^3 + A(A^T)^2$.Since $A(A^T) = I$,we have $A(A^T)^2 = (AA^T)A^T = I A^T = A^T$.Therefore,the expression simplifies to $A^3 + A^T$.

Let $A$ be a square matrix such that $AA^T = I$. Then $\frac{1}{2} A[(A+A^T)^2 + (A-A^T)^2]$ is equal to

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