Question

let A be a matrix of 3×2 order with real entites. H=A(A^TA)^-1A^T . where A^T is the transpose of the matrix. let I be the identity matrix of the order 3×3. H^2=?

Answer 1

Answer to Question #90371 – Math – Linear Algebra

Question

Let $A$ be a matrix of $3 \times 2$ order with real entities. $H = A(A^{\wedge}T A)^{\wedge} - 1A^{\wedge}T$, where $A^{\wedge}T$ is the transpose of the matrix. Let $I$ be the identity matrix of the order $3 \times 3$. $H^{\wedge}2 = ?$

Solution

Given that $H = A_{3\times 2}(A^T_{3\times 3}A_{3\times 2})^{-1}A^T_{2\times 3}$

$$= A _ {3 \times 2} (B) ^ {- 1} _ {2 \times 3} A ^ {T} _ {2 \times 3} (\text {Since the order of } B = A ^ {T} _ {2 \times 3} A _ {3 \times 2} \text {is} 2 \times 2)$$

Consider, $H^{2} = \left(A_{3\times 2}(B)^{-1}_{2\times 3}A^{T}_{2\times 3}\right)^{2}$

$$= \left(A _ {3 \times 2} (B) ^ {- 1} _ {2 \times 3} A ^ {T} _ {2 \times 3}\right) \left(A _ {3 \times 2} (B) ^ {- 1} _ {2 \times 3} A ^ {T} _ {2 \times 3}\right)$$ $$= A _ {3 \times 2} (B) ^ {- 1} _ {2 \times 3} \left(A ^ {T} _ {2 \times 3} A _ {3 \times 2}\right) (B) ^ {- 1} _ {2 \times 2} A ^ {T} _ {2 \times 3}$$

(Since Matrix multiplication is associative)

$$\begin{array}{l} = A _ {3 \times 2} (B) ^ {- 1} _ {2 \times 3} (B _ {3 \times 2}) (B) ^ {- 1} _ {2 \times 3} A ^ {T} _ {2 \times 3} (\text {Since}, B _ {2 \times 2} = A ^ {T} _ {2 \times 3} A _ {3 \times 2}) \\ = A _ {3 \times 2} (B) ^ {- 1} _ {2 \times 3} \left(I _ {2 \times 2}\right) A ^ {T} _ {2 \times 3} (\text {Since}, \left(B _ {2 \times 2}\right) (B) ^ {- 1} _ {2 \times 3} = I _ {2 \times 2}) \\ = A _ {3 \times 2} (B) ^ {- 1} _ {2 \times 3} A ^ {T} _ {2 \times 3} (\text {Since}, I _ {2 \times 2} A ^ {T} _ {2 \times 3} = A ^ {T} _ {2 \times 3}) \\ = A _ {3 \times 2} C _ {2 \times 3} (\text {Since}, (B) ^ {- 1} _ {2 \times 2} A ^ {T} _ {2 \times 3} = C _ {2 \times 3}) \\ = D _ {2 \times 3} \\ \end{array}$$

Therefore, $H^{2} = \left(A_{3\times 2}(B)^{-1}_{2\times 3}A^{T}_{2\times 3}\right)^{2} = D_{3\times 3}$.

In this problem we cannot apply $(AB)^{-1} = (B)^{-1}(A)^{-1}$ (Since, A and Transpose A are not invertible in the given problem as they are rectangular matrices)

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