Expressing a vector as a linear combination in $\mathbb{R}^4$

Question

Is it possible to express any vector in $\mathbb{R}^4$ as a linear combination of the vectors $[−1,4,1,0]^T$, $[3,2,0,1]^T$, $[1,−1,0,2]^T$ and $[0,1,3,−2]^T$?

This is a problem on a practice test that I was working on but the test has no answers to check my work. I was curious how to prove the above statement.

Thanks!

Answer 1

Hint: See if the vectors are linearly independent or not!

Four vectors $v_1, v_2,v_3, v_4$ are linearly independent if $c_1v_1+ c_2 v_2 +c_3 v_3 +c_4 v_4 = 0$ implies $c_i = 0$ for each $i$.

Answer 2

Check to see if the vectors are linearly independent. That is,

$$av_1 + bv_2 + cv_3 + dv_4 =0$$ $$\left[\begin{matrix} -1 & 3 & 1 & 0 \\ 4 & 2 & -1 & 1 \\ 1 & 0 & 0 & 3 \\ 0 & 1 & 2 & -2 \end{matrix}\right] \left [\begin{matrix} a \\ b \\ c \\ d \end{matrix}\right ] = \left [\begin{matrix} 0 \\ 0 \\ 0 \\ 0 \end{matrix}\right ]$$ where not all of $a, b, c, d =0$. Which is equivalent to checking the determinant: $$\left|\begin{matrix} -1 & 3 & 1 & 0 \\ 4 & 2 & -1 & 1 \\ 1 & 0 & 0 & 3 \\ 0 & 1 & 2 & -2 \end{matrix}\right| = 0 .$$ Or you can do row operations to see if you can get the matrix into the identity matrix.

Answer 3

Since $$\begin{vmatrix} -1&4&1&0\\ 3&2&0&1\\ 1&-1&0&2\\ 0&1&3&-2 \end{vmatrix} \neq0, $$ $S=\{(−1,4,1,0), (3,2,0,1),(1,−1,0,2),(0,1,3,−2)\}$ forms a linealy independent set of vectors. Since $\dim\mathbb{R}^4=4$ and $S$ is a L.I. subset of $\mathbb{R}^4$ with $4$ vectors, so $S$ spans $\mathbb{R}^4$. Thus your statement is possible.