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Let $ b \in \mathbb{R}^4, \space A\in M_{4\times4} (\mathbb{R}). $ Suppose $$ \begin{bmatrix} 0 \\ 1 \\ 1 \\ 1\end{bmatrix}, \begin{bmatrix} 1 \\ 0 \\ 1 \\ 1\end{bmatrix},\begin{bmatrix} 1 \\ 1 \\ 0 \\ 1\end{bmatrix},\begin{bmatrix} 1 \\ 1 \\ 1 \\ 0\end{bmatrix}, \begin{bmatrix} 1 \\ 1 \\ 1 \\ 1\end{bmatrix}$$ are all solutions of the equation $Ax=b$. Prove that $$b = \begin{bmatrix} 0 \\ 0 \\ 0 \\ 0\end{bmatrix}.$$

I had an intuiton that the given vectors are linear-dependent since there are five of them and the dim of the vector space is four. The professor solved that question after the exam using linear transformation which seemed much easier and clean solution however he didn't explain the intuition behind. I would like to get an explanation regarding that. Thanks

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5 Answers 5

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One thing that stands out is that if you sum the first four vectors together you get

$\begin{bmatrix} 0 \\ 1 \\ 1 \\ 1 \end{bmatrix} + \begin{bmatrix} 1 \\ 0 \\ 1 \\ 1 \end{bmatrix} + \begin{bmatrix} 1 \\ 1 \\ 0 \\ 1 \end{bmatrix} + \begin{bmatrix} 1 \\ 1 \\ 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 3 \\ 3 \\ 3 \\ 3 \end{bmatrix} = 3 \begin{bmatrix} 1 \\ 1 \\ 1 \\ 1 \end{bmatrix}$

If we write those vectors as $x_1, x_2, \ldots, x_5$ in the order they're given, we can just write $x_1 + x_2 + x_3 + x_4 = 3 x_5$.

So it's very tempting to see what happens when you multiply that equality by $A$:

$\begin{eqnarray} x_1 + x_2 + x_3 + x_4 & = & 3 x_5 \\ A(x_1 + x_2 + x_3 + x_4) & = & A(3 x_5) \\ A x_1 + A x_2 + A x_3 + A x_4 & = & 3 A x_5 \\ b + b + b + b & = & 3 b \\ 4 b & = & 3 b \\ \implies b & = & \mathbf{0} \end{eqnarray}$

You can do this more generally by taking your observation that there are 5 vectors in a 4-dimensional vector space, and hence there must be a linear dependency between them, except that you need to be able to span $\mathbb{R}^4$ with subsets of the vectors since you'll need to be in a situation where you have something like $(\lambda_1 + \lambda_2 + \ldots + \lambda_4 - 1) b = \mathbf{0}$ and you need to be able to escape the possibility that the scalar part is equal to zero.

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Call $e_i$ your vectors in the order you listed them, and note that $e_1+e_2+e_3+e_4 = 3e_5$. Then $A(e_1+e_2+e_3+e_4) = 4b = 3Ae_5 = 3b$. Hence $b=0$.

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  • $\begingroup$ I see, so it's mainly about paying attention to that summing pattern? $\endgroup$
    – X4J
    Mar 4, 2023 at 5:27
  • $\begingroup$ Surely this is one way to solve it. $\endgroup$
    – Gibbs
    Mar 4, 2023 at 5:28
  • $\begingroup$ Do you see any other way to solve it (with basic linear algebra)? $\endgroup$
    – X4J
    Mar 4, 2023 at 5:30
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    $\begingroup$ @X4J I think the most natural way to arrive at this would be to realise that if $Ax = b$ and $Ay = b$ then $A(y-x)=0$ so you know the difference between any of those vectors is annihilated. But the differences are all the standard basis vectors. So the whole space must be annihilated $\endgroup$
    – Zoe Allen
    Mar 4, 2023 at 5:31
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    $\begingroup$ @X4J "I see, so it's mainly about paying attention to that summing pattern?" <<< Not really. You said it yourself when you wrote the question: "I had an intuiton that the given vectors are linear-dependent since there are five of them and the dim of the vector space is four.". It just happens that the dependence relation is particularly easy to spot, but even if it wasn't, the reasoning would be the same. $\endgroup$
    – Stef
    Mar 4, 2023 at 15:49
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$$\begin{bmatrix} 0 \\ 1 \\ 1 \\ 1\end{bmatrix} + \begin{bmatrix} 1 \\ 0 \\ 1 \\ 1\end{bmatrix} + \begin{bmatrix} 1 \\ 1 \\ 0 \\ 1\end{bmatrix} + \begin{bmatrix} 1 \\ 1 \\ 1 \\ 0\end{bmatrix} - 3\begin{bmatrix} 1 \\ 1 \\ 1 \\ 1\end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \\ 0\end{bmatrix}$$ $$A\begin{bmatrix} 0 \\ 1 \\ 1 \\ 1\end{bmatrix} + A\begin{bmatrix} 1 \\ 0 \\ 1 \\ 1\end{bmatrix} + A\begin{bmatrix} 1 \\ 1 \\ 0 \\ 1\end{bmatrix} + A\begin{bmatrix} 1 \\ 1 \\ 1 \\ 0\end{bmatrix} - 3A\begin{bmatrix} 1 \\ 1 \\ 1 \\ 1\end{bmatrix} = A\begin{bmatrix} 0 \\ 0 \\ 0 \\ 0\end{bmatrix}$$ $$b + b + b + b -3b = 0$$ $$b=0$$

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I want to emphasize that it is not merely linear dependence, contrary to what your question seems to suggest you had thought. For example:

$\pmatrix{1&1&1&1\\1&1&1&1\\1&1&1&1\\1&1&1&1} \pmatrix{0&4&4&4&3\\4&0&4&4&3\\4&4&0&4&3\\4&4&4&0&3} = \pmatrix{12&12&12&12&12\\12&12&12&12&12\\12&12&12&12&12\\12&12&12&12&12}$

This is a concrete example of what another answer was referring to by "you need to be able to escape the possibility that the scalar part is equal to zero.".

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    $\begingroup$ I like the (counter)-example. Another way to phrase this: this matrix A has null space of dimension 3 not 4. $\endgroup$
    – user934527
    Mar 6, 2023 at 3:49
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There's an alternative phrasing to this problem that I think may be worthwhile to see which some comments have alluded to obliquely. It yields an algorithm which requires no tricks and I think gives you more insight into this question. It's also more in linear transformation flavor.

Algorithm: Call the given solution vectors $v_1,\dots,v_5$. Compute the rank of the matrix whose columns are $v_5 - v_1, v_4 - v_1, v_3 - v_1, v_2 - v_1$. If the rank of that matrix is 4, the constant term $b$ must be 0. In fact, the matrix $A$ must be the 0 matrix.

Reasoning:

For any $m\times n$ matrix $A$, the matrix equation $Ax = b$ is consistent if and only if $b\in\text{col}(A)$. A standard result yields $\dim(\text{col}(A)) = \text{rank}(A)$, and the rank-nullity theorem tells you that $\text{rank}(A) = n - \dim(N(A))$.

In this particular problem, we are given a $4 \times 4$ matrix and are told that for some specific $b\in\mathbb{R}^4$ that 5 specific vectors are solutions to $Ax = b$. Call them $v_1,v_2,\dots,v_5 \in \mathbb{R}^4$. We are asked to show $b=0$, which is equivalent to showing that $\text{col}(A) = \{0\}$, which is equivalent to $\dim(\text{col}(A)) = 0$, i.e., $\dim(N(A)) = 4$ by rank-nullity.

Note that for $1 < j \leq 5$, $$A(v_j - v_1) = Av_j - Av_1 = b - b = 0.$$

So the list of vectors $v_5 - v_1, v_4 - v_1, v_3 - v_1, v_2 - v_1$ are all elements of $N(A)$. Call the list $S$. In particular, the span of $S$ is a subspace of $N(A)$.

In fact, we can compute from the given vectors that $\dim(N(A)) = 4$. Since $N(A) \leq \mathbb{R}^4$, $\dim(N(A)) \leq 4$. Less trivially, we can compute that the vectors in $S$ are linearly independent by considering the rank of the matrix formed by placing the vectors in $S$ as columns. That matrix is:

$$\begin{pmatrix} 1 & 1 & 1 & 1 \\ -1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \end{pmatrix} $$

and you can compute easily that the RREF is the identity matrix. Since $S$ is a linearly independent set of vectors 4 and $S\subseteq N(A)$, $\dim(N(A)) \geq 4$. Thus $\dim(N(A)) = 4$.

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