Subjects Linear Algebra

Transpose Properties

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Transpose Properties


1. **Problem statement:** We are given matrices $$A = \begin{bmatrix} 2 & 3 \\ 1 & 4 \end{bmatrix}$$ and $$B = \begin{bmatrix} 1 & 2 \\ -1 & 1 \end{bmatrix}$$ We need to show two properties about transposes: (i) $$(A^T)^T = A$$ (ii) $$(A + B)^T = A^T + B^T$$ --- 2. **Recall the transpose of a matrix**: The transpose of a matrix is obtained by swapping its rows and columns. For any matrix $M$, its transpose $M^T$ satisfies $(M^T)_{ij} = M_{ji}$. --- 3. **Calculate $A^T$**: $$A^T = \begin{bmatrix} 2 & 1 \\ 3 & 4 \end{bmatrix}$$ 4. **Calculate $(A^T)^T$:** Transpose $A^T$ again, swapping rows and columns: $$(A^T)^T = \begin{bmatrix} 2 & 3 \\ 1 & 4 \end{bmatrix}$$ This matches the original matrix $A$, so $$(A^T)^T = A$$ --- 5. **Calculate $A + B$**: Add corresponding elements of $A$ and $B$: $$A + B = \begin{bmatrix} 2+1 & 3+2 \\ 1+(-1) & 4+1 \end{bmatrix} = \begin{bmatrix} 3 & 5 \\ 0 & 5 \end{bmatrix}$$ 6. **Calculate $(A + B)^T$:** Transpose the matrix $A + B$: $$(A + B)^T = \begin{bmatrix} 3 & 0 \\ 5 & 5 \end{bmatrix}$$ 7. **Calculate $A^T + B^T$:** Calculate $A^T$ and $B^T$ separately: $$A^T = \begin{bmatrix} 2 & 1 \\ 3 & 4 \end{bmatrix}, \quad B^T = \begin{bmatrix} 1 & -1 \\ 2 & 1 \end{bmatrix}$$ Sum: $$A^T + B^T = \begin{bmatrix} 2+1 & 1+(-1) \\ 3+2 & 4+1 \end{bmatrix} = \begin{bmatrix} 3 & 0 \\ 5 & 5 \end{bmatrix}$$ 8. **Compare $(A + B)^T$ and $A^T + B^T$:** They are equal: $$(A + B)^T = A^T + B^T$$ --- **Final answers:** (i) $$(A^T)^T = A$$ (ii) $$(A + B)^T = A^T + B^T$$