User:Egm4313.s12.team8.lucas/R5.3

=R5.3=

Problem Statement
Verify that $$\ b_1$$ and $$\ b_2$$ in (1)-(2) p.7-34 are linearly independent using the Gramian.

$$\ b_1=2e_1+7e_2$$

$$\ b_2=1.5e_1+3e_2$$

Solution
For vectors, the Gramian is defined as:

$$\ \boldsymbol \Gamma(b_1,b_2)=\begin{bmatrix} \langle b_1,b_1 \rangle & \langle b_1,b_2 \rangle\\ \langle b_2,b_1 \rangle & \langle b_2,b_2 \rangle \end{bmatrix}$$

where:

$$\ \langle b_i,b_j \rangle=b_i \cdot b_j$$

For the given vectors, the dot products are:

$$\ \langle b_1,b_1 \rangle=(2e_1+7e_2) \cdot (2e_1+7e_2)=(2)(2)+(7)(7)=4+49=53$$

$$\ \langle b_1,b_2 \rangle=(2e_1+7e_2) \cdot (1.5e_1+3e_2)=(2)(1.5)+(7)(3)=3+21=24$$

$$\ \langle b_2,b_1 \rangle=(1.5e_1+3e_2) \cdot (2e_1+7e_2)=(1.5)(2)+(3)(7)=3+21=24$$

$$\ \langle b_2,b_2 \rangle=(1.5e_1+3e_2) \cdot (1.5e_1+3e_2)=(1.5)(1.5)+(3)(3)=2.25+9=11.25$$

So the Gramian matrix becomes:

$$\ \boldsymbol \Gamma(b_1,b_2)=\begin{bmatrix} 53 & 24\\ 24 & 11.25 \end{bmatrix}$$

Finding the determinant of the Gramian matrix gives the Gramian:

$$\ \Gamma=(53)(11.25)-(24)(24)=596.25-576=20.25 \ne 0$$

The Gramian does not equal 0, therefore the vectors $$\ b_1$$ and $$\ b_2$$ are linearly independent.