Answer to Question #329409 in Linear Algebra for peac_eboy

Question

QUESTION 2 2.1. Find the change of basis matrix P∁←ℬ for the bases

ℬ = {(9, 2), (4, −3)} and ∁= {(2, 1), (−3, 1)} of ℝ2 .

2.2.Verify [v]∁ = P∁←ℬ[v]ℬ for v = (−5, 3). 

Answer 1

2.1. Denote $a_1=(9,2),a_2=(4,-3)$ and $b_1=(2,1),b_2=(-3,1)$. Any vector $v$ can be presented as: $v=\alpha_1a_1+\alpha_2a_2$ or $v=\beta_1b_1+\beta_2b_2$ with real $\alpha_i,\beta_j$ . The aim is to find the matrix $T$ such that: $\left(\begin{array}{ll}\beta_1\\\beta_2\end{array}\right)=T\left(\begin{array}{ll}\alpha_1\\\alpha_2\end{array}\right)$. We may consider two vectors: $v_1=\alpha^1_1a_1+\alpha^1_2a_2,$$v_1=\beta^1_1b_1+\beta^1_2b_2$ and $v_2=\alpha^2_1a_1+\alpha^2_2a_2,v_2=\beta^2_1b_1+\beta^2_2b_2$. Then consider equation: $\left(\begin{array}{ll}\beta^1_1&\beta^2_1\\\beta^1_2&\beta^2_2\end{array}\right)=T\left(\begin{array}{ll}\alpha^1_1&\alpha^2_1\\\alpha^1_2&\alpha^2_2\end{array}\right)$ and receive: $\left(\begin{array}{ll}\beta^1_1&\beta^2_1\\\beta^1_2&\beta^2_2\end{array}\right)\left(\begin{array}{ll}\alpha^1_1&\alpha^2_1\\\alpha^1_2&\alpha^2_2\end{array}\right)^{-1}=T$ . As an example we take: $v_1=a_1$ and $v_2=a_2$. Then, $\left(\begin{array}{ll}\alpha^1_1&\alpha^2_1\\\alpha^1_2&\alpha^2_2\end{array}\right)=\left(\begin{array}{ll}1&0\\0&1\end{array}\right)$. $a_1=\beta^1_1b_1+\beta^1_2b_2, a_2=\beta^2_1b_1+\beta^2_2b_2$ provide two systems of equations: $9=2\beta_1^1-3\beta_2^1$, $2=\beta_1^1+\beta_2^1$. We receive: $\beta_1^1=3,\beta_2^1=-1$. Another system of equations is: $4=2\beta_1^2-3\beta_2^2$, $-3=\beta_1^2+\beta_2^2$ . We get: $\beta_1^2=-1,\beta_2^2=-2.$ Thus, matrix $T$ has the form: $T=\left(\begin{array}{ll}3&-1\\-1&-2\end{array}\right)$.

2.2. The aim is to find representations of vector $v$ in basis $B$ and basis $C$. We have: $v=\beta_1b_1+\beta_2b_2$. We receive equations: $2\beta_1-3\beta_2=-5,\beta_1+\beta_2=3.$ The solution is: $\beta_1=\frac45,\beta_2=\frac{11}{5}$. From representation: $v=\alpha_1a_1+\alpha_2a_2$ we get: $9\alpha_1+4\alpha_2=-5,2\alpha_1-3\alpha_2=3.$ We get: $\alpha_1=-\frac{3}{35},\alpha_2=-\frac{37}{35}$. In order to check the transformation, multiply: $T\left(\begin{array}{ll}\alpha_1\\\alpha_2\end{array}\right)=\left(\begin{array}{ll}3&-1\\-1&-2\end{array}\right)\left(\begin{array}{ll}-\frac{3}{35}\\-\frac{37}{35}\end{array}\right)=\left(\begin{array}{ll}\frac{4}{5}\\\frac{11}{5}\end{array}\right)$. Thus, the transformation is correct.