1144: For Commutative Ring, Transpose of Product of Matrices Is Product of Transposes of Constituents in Reverse Order|T.B.P.

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description/proof of that for commutative ring, transpose of product of matrices is product of transposes of constituents in reverse order

Topics

About: matrices space

The reader will have a description and a proof of the proposition that for any commutative ring, the transpose of the product of any matrices is the product of the transposes of the constituents in the reverse order.

R

\in {the commutative rings}

M_{1}

\in {the m \times n R matrices}

M_{2}

\in {the n \times o R matrices}

//

Statements:

(M_{1} M_{2})^{t} = {M_{2}}^{t} {M_{1}}^{t}

Whole Strategy: Step 1: let

M_{1} = (\begin{matrix} {M_{1}}_{l}^{j} \end{matrix})

and

M_{2} = (\begin{matrix} {M_{2}}_{m}^{l} \end{matrix})

; Step 2: see the components of

(M_{1} M_{2})^{t}

; Step 3: see the components of

{M_{2}}^{t} {M_{1}}^{t}

, and conclude the proposition.

Step 1:

Let

M_{1} = (\begin{matrix} {M_{1}}_{l}^{j} \end{matrix})

and

M_{2} = (\begin{matrix} {M_{2}}_{m}^{l} \end{matrix})

.

Step 2:

M_{1} M_{2} = (\begin{matrix} {M_{1}}_{l}^{j} {M_{2}}_{m}^{l} \end{matrix})

{(M_{1} M_{2})^{t}}_{m}^{j} = {M_{1}}_{l}^{m} {M_{2}}_{j}^{l}

.

Step 3:

({M_{2}}^{t} {M_{1}}^{t})_{m}^{j} = {M_{2}}^{t}_{l}^{j} {M_{1}}^{t}_{m}^{l} = {M_{2}}_{j}^{l} {M_{1}}_{l}^{m} = {M_{1}}_{l}^{m} {M_{2}}_{j}^{l}

, because

R

is a commutative ring,

= {(M_{1} M_{2})^{t}}_{m}^{j}

, by the result of Step 2.

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