## Related questions with answers

Find a basis of the space of solutions in R^n of the equation

$x_1 + 2x_2 + 3x_3 + \ldots + nx_n = 0.$

Solution

VerifiedLet $W$ be the space of solutions of this equation. Let

$X = \begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{bmatrix}$

be in $W$. Then

$x_1 + 2x_2 + \ldots + n x_n = 0 \Longrightarrow x_1 = -2x_2 - \ldots - nx_n$

Therefore,

$X = \begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{bmatrix} = x_2 \begin{bmatrix} -2 \\ 1 \\ \vdots \\ 0 \end{bmatrix} + \ldots + x_n \begin{bmatrix} -n \\ 0 \\ \vdots \\ 1 \end{bmatrix} \tag{1}$

Now let

$S = \left(\begin{bmatrix} -2 \\ 1 \\ \vdots \\ 0 \end{bmatrix}, \ldots, \begin{bmatrix} -n \\ 0 \\ \vdots \\ 1 \end{bmatrix} \right)$

Every element of $S$ solves the given equation, and by (1) it spans the entire space of solutions. To conclude that this is a basis of $W$, we need to check that it is linearly independent.

## Create a free account to view solutions

## Create a free account to view solutions

## Recommended textbook solutions

## More related questions

- algebra

1/4

- algebra

1/7