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Math  /  Algebra

QuestionConsider the weighted Euclidean inner product defined by ⟨u,v⟩=3u1v1+7u2v2+4u3v3\langle\mathbf{u}, \mathbf{v}\rangle=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=A= [300070002].\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 given matrix A A is the generating matrix for this inner product.
3. The generating matrix A A should satisfy the condition ⟨u,v⟩=(Au)T(Av)\langle \mathbf{u}, \mathbf{v} \rangle = (\mathbf{Au})^T (\mathbf{Av}).

STEP 2

1. Understand the relationship between the generating matrix and the inner product.
2. Verify if the given matrix A A satisfies the condition for the weighted Euclidean inner product.

STEP 3

Recall that the generating matrix A A for a weighted Euclidean inner product is such that:
⟨u,v⟩=(Au)T(Av)\langle \mathbf{u}, \mathbf{v} \rangle = (\mathbf{Au})^T (\mathbf{Av})
This implies that:
(Au)T(Av)=(Au)TAv=uTATAv(\mathbf{Au})^T (\mathbf{Av}) = (A\mathbf{u})^T A\mathbf{v} = \mathbf{u}^T A^T A \mathbf{v}
The matrix ATA A^T A should equal the diagonal matrix with weights 3,7,4 3, 7, 4 .

STEP 4

Calculate ATA A^T A for the given matrix A A :
A=[300070002]A = \begin{bmatrix} \sqrt{3} & 0 & 0 \\ 0 & \sqrt{7} & 0 \\ 0 & 0 & 2 \end{bmatrix}
Calculate AT A^T :
AT=[300070002]A^T = \begin{bmatrix} \sqrt{3} & 0 & 0 \\ 0 & \sqrt{7} & 0 \\ 0 & 0 & 2 \end{bmatrix}
Now calculate ATA A^T A :
ATA=[300070002][300070002]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}
=[(3)2000(7)200022]=[300070004]= \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 the resulting matrix ATA A^T A with the diagonal matrix of weights 3,7,4 3, 7, 4 .
Since they are equal, the given matrix A A is indeed the generating matrix for the weighted Euclidean inner product.
The statement is:
True\boxed{\text{True}}

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