Question

Please correct me if I go wrong at any point.
Let A be a matrix.
A is symmetric if A = transpose(A)
A is orthogonal if transpose(A) = inverse(A)
A is orthogonal AND symmetric implies
A = transpose(A) = inverse(A)

Correct?

Answer 1

Then
A is normal if \[AA ^{T} = A ^{T}A\]

But how can it be implied that A is normal if A is either symmetric or orthogonal? (Looking for a short proof I feel like I just need a small leap in my logic to wrap my head around this)

Answer 2

hmm i think i prooved the orthogonal case:

considering a seperate property that is true for all invertible matrices
\[AA ^{-1} = A ^{-1}A = I\]

so if
\[A ^{T} = A ^{-1}\]
(definition of orthogonal matrix A)

this implies

\[AA ^{T} = A ^{T}A = I\]

proves that an orthogonal matrix is also normal



Still trying to work out why a symmetric matrix is also normal

Answer 3

If A is symmetric, then we have:\[AA^T=(A)A^T=(A^T)A^T=A^T(A^T)=A^T(A)=A^TA\]I know it looks weird, i added in more parenthesis to emphasis where I was using the fact that A^T = A since A is symmetric.