QuestionConsider the weighted Euclidean inner product defined by $<u, v>=3 u_{1} v_{1}+7 u_{2} v_{2}+4 u_{3} v_{3}$ . The generating matrix for this inner product is given by $A=$ $\left[\begin{array}{ccc} \sqrt{3} & 0 & 0 \\ 0 & \sqrt{7} & 0 \\ 0 & 0 & 2 \end{array}\right]$ True False

Studdy Solution

STEP 1

1. We are given a weighted Euclidean inner product.
2. We need to determine if the provided matrix $A$ is the generating matrix for this inner product.
3. The generating matrix $A$ should satisfy the condition $<u, v> = u^T A^T A v$ .

STEP 2

1. Understand the properties of the generating matrix for a weighted Euclidean inner product.
2. Verify if the given matrix $A$ satisfies the condition for the inner product.

STEP 3

Understand that for a weighted Euclidean inner product defined as $<u, v> = \sum w_i u_i v_i$ , the generating matrix $A$ should satisfy $A^T A = \text{diag}(w_1, w_2, w_3)$ .

STEP 4

Calculate $A^T A$ for the given matrix $A$ :
$A = \begin{bmatrix} \sqrt{3} & 0 & 0 \\ 0 & \sqrt{7} & 0 \\ 0 & 0 & 2 \end{bmatrix}$
Calculate $A^T$ :
$A^T = \begin{bmatrix} \sqrt{3} & 0 & 0 \\ 0 & \sqrt{7} & 0 \\ 0 & 0 & 2 \end{bmatrix}$
Calculate $A^T A$ :
$A^T A = \begin{bmatrix} \sqrt{3} & 0 & 0 \\ 0 & \sqrt{7} & 0 \\ 0 & 0 & 2 \end{bmatrix} \begin{bmatrix} \sqrt{3} & 0 & 0 \\ 0 & \sqrt{7} & 0 \\ 0 & 0 & 2 \end{bmatrix}$
$= \begin{bmatrix} (\sqrt{3})^2 & 0 & 0 \\ 0 & (\sqrt{7})^2 & 0 \\ 0 & 0 & 2^2 \end{bmatrix}$
$= \begin{bmatrix} 3 & 0 & 0 \\ 0 & 7 & 0 \\ 0 & 0 & 4 \end{bmatrix}$

STEP 5

Compare $A^T A$ with the weights in the inner product definition $<u, v> = 3 u_1 v_1 + 7 u_2 v_2 + 4 u_3 v_3$ .
The matrix $A^T A = \begin{bmatrix} 3 & 0 & 0 \\ 0 & 7 & 0 \\ 0 & 0 & 4 \end{bmatrix}$ matches the weights $3, 7, 4$ .
Therefore, the given matrix $A$ is indeed the generating matrix for the inner product.
The statement is:
$\boxed{\text{True}}$

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