Question

If S and T are matrices, and tilde (~) denotes the transpose of the matrix. \[\widetilde{\textbf{S}\textbf{T}}=\tilde{\textbf{T}}\tilde{\textbf{S}} \]
How can I prove this statement.

Answer 1

similary
\[\left({\textbf{S}\textbf{T}}\right)^{-1}={\textbf{T}}^{-1}{\textbf{S}^{-1}} \]

Answer 2

T(ST)' = T T'S' = S'
ST(ST)' = SS' = 1
(ST)(ST)' = 1

Answer 3

So
\[\textbf S \textbf T \widetilde{\textbf S \textbf T}=1\]\[\textbf T \widetilde{\textbf S \textbf T}=\textbf S^{-1}\]\[\widetilde{\textbf S \textbf T}=\tilde{\textbf T} \tilde{\textbf S}\]
that is the whole proof

Answer 4

excusing the typographical error

Answer 5

hehe .. no probs man

Answer 6

How come we can assume the first line?

Answer 7

dosent it rely on the result?

Answer 8

can't seem to remember.
I think it got something to do with the property of multiplication of Matrix with it's transpose.
TT' = must equal to something
Since if we consider matrix to be system of n vectors with m components. It gives you dot product with itself.

Answer 9

the inner product of a normalized vector with it self which must be unitary

Answer 10

well this is the special case of unitary matrix.
It must be for general too.

Answer 11

hmm ok,

Answer 12

(a)ij 1<=i<=m, 1<=j<=n represent matrix A
(b)kj 1<=j<=n, 1<=k<=0 represent matrix B
AxB = summation j (a)ij(b)kj
Must be something like that ... and since transpose changes rows into colums, A' = (a)ji

Answer 13

A = (a)ij, B = (b)jk, AB = summation j (a)ij(b)jk = (ab)ik
(AB)T = (ab)ki = summation j (b)jk(a)ij = BTxAT
excellent proof here
https://www.google.com/search?rlz=1C1ASUT_enNP447NP447&sourceid=chrome&ie=UTF-8&q=how+to+prove+AB+transpose+is+B+transpose+A+transpose

Answer 14

thank you
experimentX

Answer 15

summation j (b)jk(a)ij <--- it cannot be a matrix multiplication unless it is transpose (because rows are not equal to columns)
You are welcome.

Answer 16

yeah because it works even if the matrices are not square .

Answer 17

That is a great article thank you again,
the author even alludes to the outer product,