Exercise 1.19

Let $x_1,\dots ,x_n$ be the basis of ${\Re }^n$ in question, and let

$$y=a_1{x}_1+\dots +a_n{x}_n$$

(1)

be the representation of $y$ with respect to $x_1,\dots ,x_n$. Recall that ${\Re }^n$ also has standard basis $e_1,\dots ,e_n$ consisting of vectors $e_j$ that have $1$ as $j$th component and $0$ everywhere else. Each of the basis vectors $x_i$, $1\le i\le n$ can be represented as a linear combination of the standard basis vectors

$$\begin{array}{cc}{x}_1\hfill & =x_{11}{e}_1+\dots +x_{n1}{e}_n\hfill \\ \hfill & ⋮\hfill \\ x_n\hfill & =x_{1n}{e}_1+\dots +x_{\mathit{nn}}{e}_n\hfill \end{array}$$

(2)

Combining (1) and (2), we can find the representation of $y$ with respect to the standard basis $e_1,\dots ,e_n$:

$$\begin{array}{ccc}\hfill y& \hfill =\hfill & a_1{x}_1+\dots +a_n{x}_n\hfill \\ \\ \hfill & \hfill =\hfill & a_1\left(x_{11}{e}_1+\dots +x_{n1}{e}_n\right)\hfill \\ \hfill & \hfill \hfill & +\dots \hfill \\ \hfill & \hfill \hfill & a_n\left(x_{1n}{e}_1+\dots +x_{\mathit{nn}}{e}_n\right)\hfill \\ \\ \hfill & \hfill =\hfill & e_1\left(a_1{x}_{11}+\dots +a_n{x}_{1n}\right)\hfill \\ \hfill & \hfill \hfill & +\dots \hfill \\ \hfill & \hfill \text{=}\hfill & e_n\left(a_1{x}_{n1}+\dots +a_n{x}_{\mathit{nn}}\right)\hfill \end{array}$$

(3)

But $y$ has another convenient representation with respect to the standard basis:

$$y=y_1{e}_1+\dots +y_n{e}_n.$$

(4)

Setting both of them equal, we obtain the following system of linear equalities

$$\{\begin{array}{c}{a}_1{x}_{11}+\dots +a_n{x}_{1n}=y_1\hfill \\ ⋮\hfill \\ a_1{x}_{n1}+\dots +a_n{x}_{\mathit{nn}}=y_n\hfill \end{array}$$

which is equivalent to solving the following equation for $a$:

$$\left[\begin{array}{ccc}\hfill x_{11}\hfill & \hfill \text{…}\hfill & \hfill x_{1n}\hfill \\ \hfill ⋮\hfill & \hfill \text{⋱}\hfill & \hfill ⋮\hfill \\ \hfill x_{n1}\hfill & \hfill \text{…}\hfill & \hfill x_{\mathit{nn}}\hfill \end{array}\right]\left[\begin{array}{c}\hfill a_1\hfill \\ \hfill ⋮\hfill \\ \hfill a_n\hfill \end{array}\right]=\left[\begin{array}{c}\hfill y_1\hfill \\ \hfill ⋮\hfill \\ \hfill y_n\hfill \end{array}\right].$$

By Theorem 1.2 this linear system has a unique solution.