Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm fairly new to linear algebra and I'm trying to make sense of what I'm being taught in class. I'm a little confused as to what happens when all the values of a matrix's row are equal to 0: does this mean that the matrix has infinite solutions, no solutions, or something completely different?

share|cite|improve this question
Matrices don't have solutions. Matrices may represent systems of equations; systems of equations may have solutions. If all the entries in a row are zero, that row represents the equation $0=0$, which can be ignored in deciding how many, if any, solutions a system has. – Gerry Myerson Feb 26 '13 at 4:46
@Peter: $A$ isn't necessarily the zero matrix. It just has (at least) one row of all $0$s. Given that, don't we only need for the appropriate component(s) of $b$ to be $0$, rather than all of $b$? – Cameron Buie Feb 26 '13 at 4:51
@CameronBuie I misread a part of the question, you're right. – Pedro Tamaroff Feb 26 '13 at 5:06
up vote 2 down vote accepted

If you are interested in $Ax=b$, with $A$ of dimension $n \times n$ and $x,b$ of dimension $n \times 1$ with the first row of $A$ all $0$'s, when we do $Ax=b$ the first component of $x$ doesn't matter at all to $b$.

share|cite|improve this answer

Geometrically, if you have an all zero vector in a matrix (row or column; doesn't matter), it means that it represents a dimension-crushing transformation. For instance, a three-dimensional matrix with an all zero row will take three-dimensional vectors and project them into a some two-dimensional plane in three-dimensional space.

The converse isn't true: not all dimension-crushing matrices announce themselves by exhibiting such obvious zeros.

Such transformations are trap door functions: they have no inverse. Therefore, the corresponding matrices are not invertible. Matrices which have no inverse have a determinant of zero (and it's obvious from the way the determinant is calculated that this is so for matrices that are cross-cut by an all-zero row or column). Non-invertible matrices are called singular, or degenerate.

So, if you understand this dimension-crushing aspect, it's easy to see that $Ax = b$ might have no solutions, or many solutions, if $A$ is degenerate. Suppose that $A$ is a 3D matrix that crushes vectors into a 2D plane. Now suppose that vector $b$ lies outside of the plane. Well, then there are no solutions! There is no vector $x$ such that the matrix $A$ will project it onto $b$, because $A$ takes everything into some particular plane, and $b$ is not in that plane. Now suppose that $b$ is inside that plane. Then there are many solutions, because many vectors from 3D space project to any given point on that plane, including $b$.

So all of this gives you a whole new geometric view on systems of linear equations.

share|cite|improve this answer
Thanks for explaining it like that. It's easier to understand everything by picturing the matrices lying in a 3-d plane. – mathisnotmyforte Feb 26 '13 at 8:14

I love Linear Algebra!

I realize that the question has already been answered in the most basic way for the beginning of your course, but this might help you later, when you're given a matrix A, where one row is all 0's, then you are rank deficient. Which means that the dimension of the Column Space is not equal to the dimension of your matrix, say n (which also happens if your matrix is not square). It also means that there's a pivot value missing. A pivot value is, essentially, the values on the main diagonal of your matrix. With a pivot value missing, you can do Gaussian Elimination to solve for the x-values, and there's a theorem, when I took Linear Algebra, that is very useful when solving for x. However, my professor did not give the theorems he discussed in class names, so this theorem is true, but I'm not entirely sure what it's called.

Theorem: If $x_{part}$ is a solution to $Ax = b$,then every solution is of the form $x = x_{part}+x_{null}$ for some $x_{null}\in N(A)$. If you would like to see the proof of this, I will be happy to type it up later.

There are 4 fundamental subspaces: Column Space, Null Space, Left Null Space, and Row Space, denoted $C(A)$, $N(A)$, $N(A^{T})$, and $C(A^{T})$, respectively. You will learn more about this later, when determining which part of the matrix will form a basis (which is a subspace which is linearly independent vectors within it, and the span of the basis is the entire space). The Null Space takes your b-vector to be 0, and it sends x to 0. It is trivial if the only vector that is sent to 0 is the zero vector. However, if there is a row of all zeros in your matrix, then that means that $N(A)$ is nontrivial and there's a free variable.

Ex: Let A = \begin{array}{cccc} 1 & 0 & -1 & 0\\ 0 & 1 & 2 & 0\\ 0 & 0 & 0 & 1\\ 0 & 0 & 0 & 0\end{array}

b = \begin{array}{c} 5\\ -7\\ 0\\ 0\end{array}

This example gives you that $x_{3}$ is the only free variable, and you know that:

$x_{1} - x_{3} = 5$

$x_{2}+2x_{3} = -7$

$x_{4} = 0$

$0= 0$

x = \begin{array}{c} x_{1}\\ x_{2}\\ x_{3}\\ x_{4}\end{array}

= \begin{array}{c} 5+x_{3}\\ -7-2x_{3}\\ x_{3}\\ 0\end{array}

= \begin{array}{c} 5\\ -7\\ 0\\ 0\end{array} + $x_{3}$*\begin{array}{c} 1\\ -2\\ 1\\ 0\end{array}

So this means that each $x$ in your x-vector, whether they are all x's or other different variables, can be written such that $x_{part}$ are the constants left over after Gaussian Elimination was done on A concatonated with b. Sorry I couldn't figure out how to type the matrix and vectors for a more appealing viewing.

A row of all zeros also implies that A is not invertible, which is important when you start learning about how to take the inverse of a matrix, which must always be square to take the inverse, but the inverse is irrelevant when solving for x in the system of equations, unless you're using Matlab to solve for x.

Side note: Did you do Gaussian Elimination to get the row to become all 0's? If so, then that means that the rows of your matrix weren't linearly independent, which also implies that the dimension of your column space is also not the dimension of your matrix, so you are not full rank.

Side Note2: If you have any questions later on, feel free to ask me.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.