Apologies in advance for any bad formatting etc. as this is my first post. Last night I was thinking about how one could prove that four vectors cannot span $\mathbb R^5$, and the solution below is what I came up with. I have tried searching related posts here on stackexchange and elsewhere to find errors in my solution, however I did not find too much information, except here: Can two vectors of 3-Tuples span $\mathbb R^3$? . I would be very grateful for any feedback regarding the correctness of the attempted proof. Also I believe that this proof is easily generalized to the question of why $n$ vectors cannot span $\mathbb R^{n+1}$, given that it is correct in the special case below. Any input on this is appreciated as well. Thanks for considering the question.
Let $\mathbb e_1$ denote the unit vector $[1,0,0,0,0]^T$, $\mathbb e_2$ denote the unit vector $[0,1,0,0,0]^T$ and so on. Then the matrix $[\mathbb e_1, \mathbb e_2, \mathbb e_3, \mathbb e_4, \mathbb e_5]$ is the $5\times5$ identity matrix, denoted $\mathbb E$. The $5$ vectors $\mathbb e_1…e_5$ span $\mathbb R^5$ since for $\mathbb x = [x_1, x_2, x_3, x_4, x_5]^T$ we get that $\mathbb Ex=b$ has a solution for every vector $\mathbb b \in \mathbb R^5$ given by $\mathbb x = [b_1, b_2, b_3, b_4, b_5]$. Any set of vectors that spans $\mathbb R^5$ must have the same solution set as $\mathbb Ex=b$.
Choose the four column vectors $\mathbb v_1,\mathbb v_2,\mathbb v_3,\mathbb v_4$ where each has dimension $5$, and form the matrix $\mathbb A = [v_1, v_2, v_3, v_4]$ as we did with $\mathbb E$. Now consider the solution set of $\mathbb Ax=b$ where $\mathbb b \in \mathbb R^5$. Firstly note that $\mathbb Ax$ is not defined, so consider instead $\mathbb A^tx$. This is now a $4\times5$ matrix, so it can have a maximum of $4$ pivot columns in reduced row echelon form, meaning at least one column in $A^t$ is linearly dependent upon the others. Without loss of generality, assume that $x_4$ is parametrized by $x_5$ and that we have found a particular solution $[x_1, x_2, x_3, x_4, x_5]^T = [b^*_1, b^*_2, b^*_3, b^*_4, b^*_5]$. Now consider the point $b^{'}= [b^*_1, b^*_2, b^*_3, b^*_4, b^*_5 +1]$. This point is not in our solution set since $x_4$ changes value at least one time when $x_5$ changes value (if it did not then $x_5$ would not parametrise $x_4$, which we have assumed). We have therefore found a point in $\mathbb R^5$ that is not in our solution set, and consequently $\mathbb A = [v_1, v_2, v_3, v_4]^T$ does not span $\mathbb R^5$.
Best Answer
When you are considering $Ax = b$ you can say that $A$ is not invertible by the Invertible Matrix Theorem so already you are going to have trouble finding some unique solution to your equation, which probably means that your set does not span $\mathbb{R}^{5}$.