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I need to know if my solution is correct and if i require more depth in the solution.

Let U=Span{v1,v2,v3,v4,v5} where

v1 := (1, 0, -2, 1); v2 := (0, 1, 2, 0); v3 := (2, 3, 0, 2); v4 := (0, 1, 1, 0); v5 := (1, -2, -2, 1);

What is the dimension of U?

My Answer: The dimension of U is 5 since the U spans 5 vectors.

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  • $\begingroup$ What if your vectors were $(1,0,0)$, $(2,0,0)$, and $(3,0,0)$? Would the dimension of the subspace spanned by these vectors be $3$? $\endgroup$
    – Clayton
    Dec 3 '14 at 2:20
  • $\begingroup$ Do i require to do a Reduced Row Echelon Form of the matrix of vectors and see which columns contain a pivot then ? $\endgroup$
    – KSM
    Dec 3 '14 at 2:22
  • $\begingroup$ @KSM: No, getting a non-reduced row echelon form should be enough. $\endgroup$ Dec 3 '14 at 2:52
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Note that the vectors $\vec{v}_i$ have only four coordinates. Thus, they span a vector space of dimension, at most, four.

One way to solve your problem is to get the rank of the matrix

$$\left(\begin{array}{rrrr} 1 & 0 & -2 & 1 \\ 0 & 1 & 2 & 0 \\ 2 & 3 & 0 & 2 \\0 & 1 & 1 & 0 \\ 1 & -2 & -2 & 1\end{array}\right).$$

It is

$$\mathrm{ran}\left(\begin{array}{rrrr} 1 & 0 & -2 & 1 \\ 0 & 1 & 2 & 0 \\ 2 & 3 & 0 & 2 \\0 & 1 & 1 & 0 \\ 1 & -2 & -2 & 1\end{array}\right)\underbrace{=}_{A_3-2A_1,A_5-A_1}\mathrm{ran}\left(\begin{array}{rrrr} 1 & 0 & -2 & 1 \\ 0 & 1 & 2 & 0 \\ 0 & 3 & 4 & 0 \\0 & 1 & 1 & 0 \\ 0 & -2 & 0 & 0\end{array}\right)\\\underbrace{=}_{A_3-3A_2,A_4-A_2,A_5+2A_2}\mathrm{ran}\left(\begin{array}{rrrr} 1 & 0 & -2 & 1 \\ 0 & 1 & 2 & 0 \\ 0 & 0 & -2 & 0 \\0 & 0 & -1 & 0 \\ 0 & 0 & 4 & 0\end{array}\right)=3,$$ since clearly the first three arrows are linearly independent and $A_5=-2A_3$ and $A_4=\frac12A_3.$ That is, the given vectors span a vector subspace of dimension three.

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