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(CAB)^T = B^T A^T

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(AB)T=BTAT(AB)^T = B^T A^T

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Okay, Christove, let's verify the matrix transpose property (AB)T=BTAT(AB)^T = B^T A^T. This property states that the transpose of the product of two matrices is equal to the product of their transposes in reverse order.

  1. Define two matrices AA and BB.

    Let's use 2×22 \times 2 matrices for this example:

    A=[1203],B=[4125]A = \begin{bmatrix} 1 & 2 \\ 0 & 3 \end{bmatrix}, \quad B = \begin{bmatrix} 4 & 1 \\ 2 & 5 \end{bmatrix}

  2. Calculate the product ABAB.

    AB=[1203][4125]=[(1)(4)+(2)(2)(1)(1)+(2)(5)(0)(4)+(3)(2)(0)(1)+(3)(5)]AB = \begin{bmatrix} 1 & 2 \\ 0 & 3 \end{bmatrix} \begin{bmatrix} 4 & 1 \\ 2 & 5 \end{bmatrix} = \begin{bmatrix} (1)(4) + (2)(2) & (1)(1) + (2)(5) \\ (0)(4) + (3)(2) & (0)(1) + (3)(5) \end{bmatrix} AB=[4+41+100+60+15]=[811615]AB = \begin{bmatrix} 4 + 4 & 1 + 10 \\ 0 + 6 & 0 + 15 \end{bmatrix} = \begin{bmatrix} 8 & 11 \\ 6 & 15 \end{bmatrix}

  3. Calculate the transpose of (AB)(AB), which is (AB)T(AB)^T.

    To find the transpose, we swap the rows and columns of ABAB.

    (AB)T=[861115](AB)^T = \begin{bmatrix} 8 & 6 \\ 11 & 15 \end{bmatrix}

  4. Calculate the transpose of AA, which is ATA^T.

    AT=[1023]A^T = \begin{bmatrix} 1 & 0 \\ 2 & 3 \end{bmatrix}

  5. Calculate the transpose of BB, which is BTB^T.

    BT=[4215]B^T = \begin{bmatrix} 4 & 2 \\ 1 & 5 \end{bmatrix}

  6. Calculate the product BTATB^T A^T.

    BTAT=[4215][1023]=[(4)(1)+(2)(2)(4)(0)+(2)(3)(1)(1)+(5)(2)(1)(0)+(5)(3)]B^T A^T = \begin{bmatrix} 4 & 2 \\ 1 & 5 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 2 & 3 \end{bmatrix} = \begin{bmatrix} (4)(1) + (2)(2) & (4)(0) + (2)(3) \\ (1)(1) + (5)(2) & (1)(0) + (5)(3) \end{bmatrix} BTAT=[4+40+61+100+15]=[861115]B^T A^T = \begin{bmatrix} 4 + 4 & 0 + 6 \\ 1 + 10 & 0 + 15 \end{bmatrix} = \begin{bmatrix} 8 & 6 \\ 11 & 15 \end{bmatrix}

  7. Compare the results from Step 3 and Step 6.

    We can see that (AB)T=[861115](AB)^T = \begin{bmatrix} 8 & 6 \\ 11 & 15 \end{bmatrix} and BTAT=[861115]B^T A^T = \begin{bmatrix} 8 & 6 \\ 11 & 15 \end{bmatrix}.
    Since both results are identical, the property is verified.

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