Is vector in the span?

Question

So, from my understanding, a span is simply the space that can be made up of specified vectors. So if I wanted to find if vector $(1,2,3,4,5)$ was in the span of some vectors, I assume I should use gaussian elimination to sort it in row echelon form and check whether the equations are consistent.

So my matrix looks like this at the moment:

$$\left[\begin{array}{rrrrr|r} 1 & 2 & -1 & -1 & 3 & 1\\ 0 & 3 & 1 & 1 & -2 & 2\\ -3 & 4 & 1 & -1 & 1 & 3\\ 1 & -4 & 2 & 2 & 2 & 4\\ 1 & 2 & 3 & -2 & 1 & 5 \end{array}\right]$$ My workings out got very messy with the rows I chose to add and subtract, so I was wondering whether there's an trick to make the math easier when choosing to minus or add certain rows.

Answer 1

In face of such questions, I usually just use a calculator to compute the determinant, which in this case is $-790$ and hence non-zero. An $n\times n$ matrix with nonzero determinant does not have zero as an eigenvalue, and hence multiplication by that $n\times n$ matrix has to be both injective and surjective, and we conclude that any vector $(a,b,c,d,e)$ lies in the image of this matrix, or equivalently the span of the column vectors of this matrix.

Answer 2

There is not trick, just to make the calculations in order to apply the Gaussian elimination in the following order $$\begin{bmatrix} 1\color{red}{\to \boxed{1}}& 2 & -1 & -1 & 3 & | & 1\\ 0 & 3\color{red}{\to \boxed{2}} & 1 & 1 & -2 & | & 2\\ -3 & 4 & 1\color{red}{\to \boxed{3}} & -1 & 1 & | & 3\\ 1 & -4 & 2 & 2\color{red}{\to \boxed{4}} & 2 & | & 4\\ 1 & 2 & 3 & -2 & 1\color{red}{\to\boxed{5}} & | & 5 \end{bmatrix}$$ where the $\color{red}{\to \boxed{N}}$ denote the order of the pivoting to $\color{blue}{\to \boxed{1}}$ and all the entries under the principal diagonal are $\color{green}{\to \boxed{0}}$. Hence your approach is correct. Also, using a bit a linear algebra theory notice that you have $5$ vectors in $\mathbb{R}^{5}$ and if you show that the determinant is different to zero. Then they will independent linearly and also since you have $5$ vectors in $\Bbb{R}^{5}$ so you have a basis and then of course the vector $\begin{bmatrix} 1 & 2 & 3 & 4 & 5\end{bmatrix}^{T}$ is in the span.