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Let S = $\{u_1,u_2, u_3, u_4\}$ be a set in $\mathbb{R}^4$. After performing Gram-Schmidt process on S, $v_4 = 0$ but $ v_3 \ne 0$.

Which of the following statement is true?

(1) $\{v_1,v_2, v_3, v_4\}$ can be normalized to an orthonormal set.

(2) Set S is linearly independent.

(3) span(S) is a 3-dimensional subspace.

Is it correct that only (3) is true, since $u_4$ is a redundant vector? Is my reasoning correct? Thank you.

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1 Answer 1

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Yes, that is correct: $v_4=0$ implies that $u_4$ is a linear combination of $u_1$, $u_2$, and $u_3$. But $u_2$ is not a multiple of $u_1$ (otherwise, $v_2=0$) and $u_3$ is not a linear combination of $u_1$ and $u_2$ (otherwise, $v_3=0$). So, $\dim\operatorname{span}\{u_1,u_2,u_3,u_4\}=3$.

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