Proof that four ($n$) vectors cannot span $\mathbb R^5$ ($\mathbb R^{n+1}$)

Question

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 $\mathbb R^5$.

Answer 1

I would advise using the following theorems (if you're not comfortable with vector spaces, replace $V$ with $\Bbb{R}^n$):

Suppose $V$ is a vector space and $S$ is a spanning set for $V$. Then there exists some $S' \subseteq S$ that is a basis for $V$ (i.e. spanning sets can be reduced to bases).

Suppose $V$ is a vector space with bases $B_1, B_2$. Then $|B_1| = |B_2|$.

Then, a set of $4$ vectors can be reduced to a basis of at most $4$ vectors, but the standard basis consists of $5$ vectors. This contradicts the second theorem.

It's not that doing it with matrices is wrong, it's just using a hammer to kill a fly. A lot of the theorems about matrices use the above theorems (and the one about linearly independent sets being extended to bases) in their proofs. The above theorems are just more fundamental than the matrix theorems.

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Since you did use the proof-verification tag, let's critique your proof. You say,

Any set of vectors that spans $\mathbb R^5$ must have the same solution set as $\mathbb Ex=b$.

That statement is misleading, for a few reasons. Firstly, spanning sets don't have solution sets; only equations (and inequalities) do.

Secondly, the "solution set" of the equation is the set of $x$ satisfying the above equation (in this case, $\{b\}$). It does not refer to the set of $b$ such that solution set is non-empty, which is $\Bbb{R}^5$ if and only if the columns of the matrix are spanning. If you change to a different spanning set, the actual solution set, for a given $b$, will change dramatically!

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 $\mathbb A^tx$.

It's not clear what value you get from doing. By taking the transpose, it means you are now looking at the rows of $\Bbb A$, rather than the columns. It also means that $\Bbb{A}^\top x$ will be a column vector in $\Bbb{R}^4$, instead of $\Bbb{R}^5$. It's highly confusing that you continue to use $b$, and seem to put $5$ components in it.

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).

It's not clear why you can assume this without loss of generality. What if you get a column of $0$s in $\Bbb{A}^\top$, and four other linearly indpendent columns? Then, one variable will be free without anything else depending on it, and adding $1$ to $b^*_5$ may not generate a $b$ for which there is no solution.

Also, writing $[x_1, x_2, x_3, x_4, x_5]^T = [b^*_1, b^*_2, b^*_3, b^*_4, b^*_5]$ is a very confusing way of denoting that $\Bbb{A}$ times the left side is equal to the right side (doubly so if you remember the right side should be in $\Bbb{R}^4$).

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So, in summary, there are some very serious issues with the proof. If you are going to use matrices, definitely do not take the transpose. Instead, just take your vector of dummy variables $x$ from $\Bbb{R}^4$ instead of $\Bbb{R}^5$. The important thing is which $b$ is achievable via matrix multiplication with a column vector, not the size of the column vector itself.

Answer 2

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 $\mathbb{R}^{5}$.