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Question

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

x1+2x2+3x3++nxn=0.x_1 + 2x_2 + 3x_3 + \ldots + nx_n = 0.

Solution

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Step 1
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Let WW be the space of solutions of this equation. Let

X=[x1x2xn]X = \begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{bmatrix}

be in WW. Then

x1+2x2++nxn=0x1=2x2nxnx_1 + 2x_2 + \ldots + n x_n = 0 \Longrightarrow x_1 = -2x_2 - \ldots - nx_n

Therefore,

X=[x1x2xn]=x2[210]++xn[n01](1)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=([210],,[n01])S = \left(\begin{bmatrix} -2 \\ 1 \\ \vdots \\ 0 \end{bmatrix}, \ldots, \begin{bmatrix} -n \\ 0 \\ \vdots \\ 1 \end{bmatrix} \right)

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

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