On page $5$ of Linear Algebra from Hoffman and Kunze's book they prove the following Theorem:

Theorem 1. Equivalent systems of linear equations have exactly the same solutions

I will add the definition of equivalent system of linear equations.

Two systems of linear equations are equivalent if each equation in each system is a linear combination of the equations in the other system.

I was wondering if the converse is true. The converse appears to be true, but maybe I am missing something. All examples that I found (systems having exactly the same solutions), the systems were equivalent.

I think I am missing something trivial!

  • $\begingroup$ What do you mean by the converse? Theorem 1 is not of the form of an 'if-then' statement. IF you have a statement of the form $P \Rightarrow Q$, then the converse is that $Q \Rightarrow P$. In this case, there is no converse. $\endgroup$ – Samuel Reid Feb 21 '12 at 19:19
  • $\begingroup$ @SamuelReid: $P$ is "The systems are Equivalent". $Q$ is "They have exactly the same solutions". $\endgroup$ – Aryabhata Feb 21 '12 at 19:52
  • $\begingroup$ @SamuelReid: Sorry, I am not a Mathematician, but I thought that every theorem was of the form "if $P$ then $Q$". What is wrong with the following? If two systems of linear equations are equivalent, then they have exactly the same solutions. Please, help me to improve my undestanding about it because I was undertanding it in a wrong way. Thanks! $\endgroup$ – spohreis Feb 21 '12 at 19:52
  • $\begingroup$ You understand it fine, I was being overly picky about Hoffman's exact wording. Watch out later in the book, he says "if" when he should say "if and only if" quite a lot. I'm sure that this is the case here as well. $\endgroup$ – Samuel Reid Feb 21 '12 at 20:12
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    $\begingroup$ There is one particular problem: two inconsistent systems always have the same set of solutions, but they need not be equivalent. For example, the system $x+y=0$, $x+y=1$; and the system $x+2y=0$, $x+2y=1$; are both inconsistent, but we cannot obtain $x+y=0$ as a linear combination of the second system. $\endgroup$ – Arturo Magidin Feb 22 '12 at 3:28

There is one fly in the ointment, which is inconsistent systems. Two inconsistent systems have the same set of solutions, but they need not be equivalent in the sense you give. They may not even be systems in the same number of variables! But even if you require that they be systems in the same number of variables, you run into trouble. Here are two systems that have the exact same solutions (to wit, none): $$\begin{array}{rcccl} x & + & y & = & 0;\\ x & + & y & = & 1; \end{array}\qquad\text{and}\qquad \begin{array}{rcccl} x & + & 2y & = & 0;\\ x & + & 2y & = & 1. \end{array}$$ But $x+y=0$ cannot be obtained from the second system, since any combination of the equations in the second system will give you an equation in which the coefficient of $y$ is twice the coefficient of $x$.

But if you remove this bad case, then the result is true: two consistent systems that have the same (nonempty) set of solutions are equivalent in the sense you give.

Consider first the case of homogeneous systems (which are always consistent). We can write the system as $A\mathbf{x}=\mathbf{0}$, where $A$ is the $n\times m$ coefficient matrix of your system, with $n$ equations and $m$ unknowns.

A vector $\mathbf{x}_0$ is a solution if and only if it lies in the orthogonal complement of the subspace of $\mathbb{R}^m$ spanned by the rows of $A$ (which corresponds to the equations). If $A\mathbf{x}=\mathbf{0}$ and $B\mathbf{x}=\mathbf{b}$ have the same solution set, then that means that $\mathbf{b}=\mathbf{0}$ (since the solution that assigns every variable to $0$ is a solution to the first system, hence to the second).

But that means that the row space of $A$ has the same orthogonal complement as the rowspace of $B$. In finite dimensional vector spaces, $(\mathbf{W})^{\perp\perp}=\mathbf{W}$. Thus, the row space of $A$ and the rowspace of $B$ have to be equal.

That means that every row of $B$ (every equation in the second system) is a linear combination of the rows of $A$ (the equations of the first system), and every row of $A$ (equations in the first system) is a linear combination of the rows of $B$ (equations of the second system). Thus, the two systems are equivalent.

Now, to consider the more general case of systems of the form $A\mathbf{x}=\mathbf{a}$, note that the solutions to $A\mathbf{x}=\mathbf{a}$ are of the form $$\mathbf{s}_0 + \mathbf{n}$$ where $\mathbf{n}$ is a solution to $A\mathbf{x}=\mathbf{0}$ and $\mathbf{s}_0$ is a specific solution to $A\mathbf{x}=\mathbf{a}$. This follows from the fact all those are solutions, since $$A(\mathbf{s}_0+\mathbf{n}) = A\mathbf{s}_0 + A\mathbf{n} = \mathbf{a} + \mathbf{0} = \mathbf{a}.$$ And, if $\mathbf{s}_1$ is any solution, then $\mathbf{s}_1 = \mathbf{s}_0 + (\mathbf{s}_1-\mathbf{s}_0)$, and $\mathbf{n}=\mathbf{s}_1 -\mathbf{s}_0$ is a solution to $A\mathbf{x}=\mathbf{0}$: $$A(\mathbf{s}_1-\mathbf{s}_0) = A\mathbf{s}_1 - A\mathbf{s}_0 = \mathbf{a}-\mathbf{a}=\mathbf{0}.$$

Suppose that $A\mathbf{x}=\mathbf{a}$ has the same solution set as $B\mathbf{x}=\mathbf{b}$, and that both have at least one solution. Let $S_A$ be the solutions to $A\mathbf{x}=\mathbf{0}$ and let $S_B$ be the solutions to $B\mathbf{x}=\mathbf{0}$. I claim that $S_A = S_B$.

Indeed, let $\mathbf{s}_0$ be a particular solution to $A\mathbf{x}=\mathbf{a}$; then it is also a solution to $B\mathbf{x}=\mathbf{b}$ by assumption; so for every $\mathbf{n}\in S_B$, $\mathbf{s}_0+\mathbf{n}$ is a solution to $B\mathbf{x}=\mathbf{b}$, hence to $A\mathbf{x}=\mathbf{a}$, hence $\mathbf{n}\in S_A$. Thus, $S_B\subseteq S_A$, and a symmetric argument shows that $S_A\subseteq S_B$, so the two are equal.

But we know that if the solution set to $A\mathbf{x}=\mathbf{0}$ is the same as the solution set to $B\mathbf{x}=\mathbf{0}$, then the two systems are equivalent. If we take a row of $B$ and ignore the right hand side, then we can express it as a linear combination of the rows of $A$. If, taking into account the right hand side, we were to get an equation different from the equation we have in $B$, then this would tell us that the solutions to $B\mathbf{x}=\mathbf{b}$ satisfy two equations with identical left hand sides but different right hand sides; this is impossible, since we are assuming the system is consistent. Thus, the linear combination of equations of $A$ that yields the equation of $B$ will also give the same right hand side; so every equation in the second system is a linear combination of the equations in the first system; the converse argument also holds, so the two systems are equivalent.


Assuming the equations are of the form $Ax = 0$.

The row space and the null space are orthogonal complements to each other.

So over $\mathbb{R}$, what you say is true, I think.

In case the equations are not homogenous, the solutions spaces are translates of the null space. We can pick the same vector to translate by, and so the null spaces are the same and the above arguments holds.


If they have the same solutions, the canonical echelon form of the matrix of the coefficients of the two systems must be the same. Then the respective matrices of coefficients of the two systems are row equivalent, and from this you can conclude that each row of one matrix is a linear combination of the rows of the second matrix.

  • $\begingroup$ (I thought about this argument for the case when the set of solutions is not empty). $\endgroup$ – alpha.Debi Feb 22 '12 at 6:10

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