How to tell if a set of vectors spans R4
First year linear algebra.
Question is Exercise 5.1.3.b from "Linear Algebra w Applications, K. Nicholson"
Determine if the given vectors span $R^4$: {$(1,3,-5,0), (-2,1,0,0), (0,2,1,-1), (1,-4,5,0)$}
From class I only understand that the vectors (call them a, b, c, d) will span $R^4$ if $t_1a+t_2b+t_3c+t_4d=some vector$ but I'm not aware of any tests that I can do to answer this. The book only gives the answer "No"...
Thanks
$\endgroup$ 13 Answers
$\begingroup$The easiest test is to show that the determinant $$\begin{vmatrix} 1 & -2 & 0 & 1 \\ 3 & 1 & 2 & -4 \\ -5 & 0 & 1 & 5 \\ 0 & 0 & -1 & 0 \end{vmatrix} \neq 0 $$ This works since the determinant is the ($n$-dimensional) volume, and if the subspace they span isn't of full dimension then that value will be 0, and it won't be otherwise.
$\endgroup$ $\begingroup$In a matrix the vectors form:$$M=\begin{bmatrix} 1 & -2& 0& 1\\ 3 & 1& 2& -4\\ -5& 0& 1& 5\\ 0 & 0& -1& 0 \end{bmatrix}$$then, using row operations, convert M into RREF.
$$M\sim A=\begin{bmatrix} 1 & 0& 0& -1\\ 0 & 1& 0& -1\\ 0& 0& 1& 0\\ 0 & 0& 0& 0 \end{bmatrix}_{RREF}$$
As $A$'s columns are not linearly independent ($R_{4}=-R_{1}-R_{2}$), neither are the vectors in your questions. $4$ linear dependant vectors cannot span $\mathbb{R}^{4}$.
This comes from the fact that columns remain linearly dependent (or independent), after any row operations.
This method is not as quick as the determinant method mentioned, however, if asked to show the relationship between any linearly dependent vectors, this is the way to go.
$\endgroup$ 1 $\begingroup$$S\subseteq \mathbb R^4$, with
$$S=\{(1,3,−5,0),(−2,1,0,0),(0,2,1,−1),(1,−4,5,0)\}.$$ We say $S$ span $\mathbb R^4$ if for all $v\in \mathbb{R}^4$, $v$ can be expressed as linear combination of $S$, i.e.
$$v=c_1(1,3,−5,0)+c_2(−2,1,0,0)+c_3(0,2,1,−1)+c_4(1,−4,5,0).$$
Let $v=(v_1,v_2,v_3,v_4)$.
We have system of linear equation
$$ \begin{bmatrix} 1&-2 & 0 & 1\\ 3&1&2&-4\\ -5&0&1&5\\ 0&0&-1&0 \end{bmatrix} \begin{bmatrix} c_1\\ c_2\\ c_3\\ c_4 \end{bmatrix} = \begin{bmatrix} v_1\\ v_2\\ v_3\\ v_4 \end{bmatrix}. $$Now we must check system of linear have solutions $c_1,c_2,c_3,c_4$ or not. If the system of linear equation not have solution, the $S$ is not span $\mathbb R^4$.
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