Question - Summarized

Given a linear system of homogeneous equations, we know that there exists a trivial solutions where all the variables are zero. Thus, we can only have a system with one solution (the trivial one), or infinitely many solutions.

Can the ranks of the system's coefficient matrix and augmented matrix immediately tell us which?

My thoughts

I have the understanding that yes, it can.

Let $A$ be the system's coefficient matrix, and $A'$ be the system's augmented matrix, and assume we have a linear system of homogeneous equations, with the augmented matrix:

$$ A' = \begin{bmatrix} a_{11} & a_{12} & a_{13} & 0 \\ a_{21} & a_{22} & a_{23} & 0 \\ a_{31} & a_{32} & a_{33} & 0 \\ \end{bmatrix} $$

and coefficient matrix:

$$ A= \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \\ \end{bmatrix} $$

If we reduce both these to row-echelon form, assume we end up with the following results, and let $*$ be any non-zero real number.

$$\text{ref}(A) = \begin{bmatrix} * & * & * \\ 0 & * & * \\ 0 & 0 & * \\ \end{bmatrix}$$

$$\text{ref}(A') = \begin{bmatrix} * & * & * & 0 \\ 0 & * & * & 0 \\ 0 & 0 & * & 0 \\ \end{bmatrix}$$

Let $n$ be the number of variables (so $n=3$ for this example).

We can see that $\text{rank}(A) = \text{rank}(A') = n$, which leads us to the conclusion that we have one distinct solution, which would be the trivial $(0, 0, 0)$.

However, if the third row ended up being the zero vector, we would have $\text{rank}(A) = \text{rank}(A') < n$, and we'd end up with a free variable, and thus infinitely many solutions.

My conclusion, which I'm hoping for verification to

We must conclude that if $\text{rank}(A) = \text{rank}(A') = n$, the system has a single solution. And if $\text{rank}(A) = \text{rank}(A') < n$, the system has infinitely many solutions in addition to the trivial one.

Addendum: The case where $\text{rank}(A) < \text{rank}(A') $ cannot occur for inhomogeneous systems, since it would require a pivot element in the right-most columns of $A'$. If that was the case, the system wouldn't be homogeneous. Since the assumption leads to a contradiction, the ranks must always be the same.

Would love to get some feedback on my thought process.

  • $\begingroup$ What if you have fewer equations than variables? $\endgroup$
    – amd
    Mar 17, 2020 at 18:52
  • $\begingroup$ @amd - Then $\text{rank}(A) = \text{rank}(A') < n$, right? Same as if I end up with a row of zeroes in the row-echelon form. $\endgroup$
    – Alec
    Mar 17, 2020 at 20:46
  • $\begingroup$ But that's what I said. They would have the same rank, but the ranks would be lower than the number of variables, thus leading to the case where we have infinite solutions? $\endgroup$
    – Alec
    Mar 17, 2020 at 21:27
  • $\begingroup$ Sorry, I misread what you wrote there. $\endgroup$
    – amd
    Mar 17, 2020 at 22:42
  • $\begingroup$ Anyway, what you’ve got here is a version of the Rank-Nullity theorem. $\endgroup$
    – amd
    Mar 17, 2020 at 22:42

1 Answer 1


Yes, you can discern the number of solutions for a homogeneous linear system of equations immediately from either the rank of the coefficient matrix or the rank of the augmented matrix. Let's consider the general case that $A = \begin{bmatrix} a_1 & a_2 & \cdots a_n \end{bmatrix}\in\mathbb{R}^{m\times n}$, where $a_i\in\mathbb{R}^m$ is the $i$th column of $A$.

First, as you correctly pointed out, the rank of the coefficient matrix equals that of the augmented matrix. To see why, note that the rank of a matrix is usually defined as the dimension of its range. Defining $A' = \begin{bmatrix} A & 0_{m\times 1} \end{bmatrix}$, we have that

\begin{align*} \text{rank}(A) ={}& \dim\{ Ax : x\in\mathbb{R}^n \} \\ ={}& \dim\left\{ \begin{bmatrix} A & 0_{m\times 1} \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} : (x,y)\in\mathbb{R}^n \times\mathbb{R} \right\} \\ ={}& \dim\{A'z : z\in\mathbb{R}^{n+1}\} \\ ={}& \text{rank}(A'). \end{align*}

Therefore, let's focus our attention on just the coefficient matrix $A$. There are two cases:

  • Case 1: $\text{rank}(A)=n$. In this case, the set $\{a_1,a_2,\dots,a_n\}$ is linearly independent, and therefore $x_1a_1 + x_2a_2 + \cdots + x_na_n = 0$ only if $x_1=x_2=\cdots=x_n=0$. Hence, $Ax=0$ implies that $x=0$, so there is one solution (the trivial solution).
  • Case 2: $\text{rank}(A)<n$. In this case, the set $\{a_1,a_2,\dots,a_n\}$ is linearly dependent, and therefore there exists a nonzero linear combination of the columns of $A$ that sum to zero. That is, $Ax = x_1a_1 + x_2a_2 + \cdots + x_na_n = 0$ for some $x\in\mathbb{R}^n\setminus\{0\}$. Now, let $\alpha\in\mathbb{R}$ and define $\tilde{x} = \alpha x$. We have that $A\tilde{x} = \alpha Ax = 0$. Since $\alpha$ was chosen arbitrarily, we see that there is an infinite number of solutions to the homogeneous system of equations.

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .