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I'm struggling to solve this proof:

Suppose that $\boldsymbol{u}_1, \boldsymbol{u}_2, . . . , \boldsymbol{u}_n$ are solutions of an homogeneous system of linear equations $\boldsymbol{Ax} = \boldsymbol{0}$, $\boldsymbol{x} \in \mathbb{R}^n$.

Suppose that $k_i \in \mathbb{R}$. Show that

$k_1\boldsymbol{u}_1 + k_2\boldsymbol{u}_2 + · · · + k_n\boldsymbol{u}_n$

is also a solution of the system $\boldsymbol{Ax}=\boldsymbol{0}$.

I know that I should use the proof for showing that if A is invertible then the only solution is $\boldsymbol{x}=\boldsymbol{0}$ when $\boldsymbol{Ax}=\boldsymbol{0}$, but am unsure about the most efficient way of doing this.

Any help is greatly appreciated.

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  • $\begingroup$ Use linearity of matrix multiplication. $\endgroup$ Commented Oct 20, 2017 at 18:06

1 Answer 1

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Hint: Let $\boldsymbol{x} = k_1\boldsymbol{u}_1+\ldots+k_n\boldsymbol{u}_n$ $$\boldsymbol{A}\boldsymbol{x}=\boldsymbol{A}(k_1\boldsymbol{u}_1+\ldots+k_n\boldsymbol{u}_n)=\sum_{i=1}^{n}\left[k_i\left(\boldsymbol{Au}_i\right)\right]$$

Remeber that all $\boldsymbol{u}_i$ are solutions to $\boldsymbol{A}\boldsymbol{x}=\boldsymbol{0}$. What can you conclude from this

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  • $\begingroup$ Surely that the sum of the series is 0? Is this enough to prove this statement though as it seems very short $\endgroup$ Commented Oct 20, 2017 at 18:09
  • $\begingroup$ I would say that this is enough. $\endgroup$ Commented Oct 20, 2017 at 19:09

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