I was doing some exercises to prepare on my exam when I ran into this problem and it got me confused in a way that made me lose more than 6 hours figuring it out.

Basically, the exercise is something like this (this is an example)

Find the basis and dimension of the following subspace in R3:

$ \begin{cases} (x,y,z) \in R3 | x+2y+3z =0 \end{cases} $

Now I go , put the coordinates into a "matrix" , then do rref, which is useless in this case, so I take free variables x2 and x3, as the only pivot is x1

$ x3 = k $

$ x2 = s $


$ x1 = -2s -3k $

$ x2 = s $

$ x3 = k $

$ x = s * [ -2, 1 0 ] + k * [ -3 0 1 ] $

and i get basis vectors: $ v1 = [ -2 , 1, 0 ] \\ v2 = [ -3, 0, 1 ] $ And dimension of 2.

Until here it's clear, but what confuses me is that i was using the exact same process to find the Null space of a matrix.

You take the matrix, do RREF, find pivot columns, those are the ones that correspond with the basis columns, from the original matrix, and are also the rank.

Then we have the free variables columns, with no pivot, and those are going to represent the dimension of the Kernel/Null space.

Once i have those columns, i go and put the equations into parametric form and solve exactly like i did now, and instead of the basis for the subspace, i get the null space of the matrix, this got me very confused, same process, two different outputs.

One is the Basis for the Null space (in case of matrix) , and this is the basis for the subspace, starting from an equation, what am i missing ?

In the case of the matrix, the basis is the basis of the column space, which are the columns indicated by the pivots, but in this case with one equation, the basis for the subspace is what would be the null space for a matrix.

How is it possible ?

Thank you

  • 4
    $\begingroup$ You have a typo in your equation: The second $x$ should be a $z$. And you should not use capital $X,Y,Z$ when you're using $x,y,z$. But, anyway, what you seem to be missing is that the description of the subspace is precisely giving it as the nullspace of the matrix $\begin{bmatrix} 1&2&3\end{bmatrix}$. $\endgroup$ Commented Jun 9, 2019 at 2:04
  • 1
    $\begingroup$ "One is the Basis for the Null space (in case of matrix) , and this is the basis for the subspace, starting from an equation, what am i missing" This subspace is in fact the null space of the matrix $[1,2,3]$. $\endgroup$
    – littleO
    Commented Jun 9, 2019 at 2:53
  • 1
    $\begingroup$ What is the definition of the nullspace of a matrix? Now, write down that definition for the matrix that littleO and I gave you. $\endgroup$ Commented Jun 9, 2019 at 3:21
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    $\begingroup$ If someone came up to you on the street one day and asked you what is the null space of the matrix $[1,2,3]$, what would you tell them? $\endgroup$
    – littleO
    Commented Jun 9, 2019 at 3:21
  • 1
    $\begingroup$ The null space of the 3x1 matrix $M=[1,2,3]$ is $\{(x,y,z): 1x+2y+3z=0\}$ when we treat $M$ as a function from $\Bbb R^3$ to $\Bbb R$. That is $M((x,y,z))=x+2y+3z.$ $\endgroup$ Commented Jun 9, 2019 at 3:33

2 Answers 2


The subspace in your question is defined as the solution set of a homogeneous linear equation. Recall that the solution set to a system of homogeneous linear equations is precisely the null space of the coefficient matrix. This is true no matter how many equations are in the system, including when there’s only a single equation. In that case, the coefficient matrix only has one row, but the general process of finding its null space is the same: compute its RREF (which might involve no work at all), identify the pivot column, and so on.

  • $\begingroup$ So basically because my linear equation is homogeneous, it's already set as Ax = 0, meanwhile when i get a matrix and they tell me to find the null space, i have to augment the matrix with the 0 vector myself, in order to find the null space. So when there is a homogeneous linear equation, by default its basis it's always the null space, so the elements that map that equation to the 0 vector. Right ? $\endgroup$
    – AndrewM
    Commented Jun 9, 2019 at 5:20
  • $\begingroup$ @AndrewM That’s right. Note that when compute the null space of a matrix $A$, there’s no particularly good reason to augment it with a column of zeros before row-reducing. Elementary row operations don’t change that extra column, and it gives you no additional information about the solution once you’ve computed the RREF. $\endgroup$
    – amd
    Commented Jun 9, 2019 at 7:01
  • $\begingroup$ Yes but in the end, i have to look for the variables whose values will form vectors that if then used as coefficients instead of the variables themselves, they will map my elements to 0, so will map my linear combination to the 0 vector, thus the null space/kernel. $\endgroup$
    – AndrewM
    Commented Jun 9, 2019 at 7:07

Here is another example of this type of exercise:

enter image description here

Here is how i solved it, and i looked at the teacher's results and are the same.

enter image description here My problem again is... i found the null space, and they say it's the basis for the subspace.. but why ? Isn't it the null space ?

And the basis for the subspace should be the rows that are not 0, or the pivot columns equivalent from the original matrix ? What am i missing ?

  • $\begingroup$ If we let $M$ denote the matrix you wrote down, then a vector $x$ is in $V$ if and only if $Mx = 0$, or in other words, if and only if $x$ is in the null space of $M$. That's the whole point of the way you chose the matrix $M$, to make this true. So the nullspace of $M$ is exactly $V$. And "find a basis for $V$" and "find a basis for the nullspace of $M$" are just two different ways to ask the same question. $\endgroup$ Commented Jun 9, 2019 at 4:28
  • $\begingroup$ Because V is a subspace of a vector space and it's basis is the null space of M , which is the null space of the linear transformation, no ? So when i see those exercises to find basis for the subspace given one or two equations, i shall always look for the null space of the matrix of the coefficients of those equations ? $\endgroup$
    – AndrewM
    Commented Jun 9, 2019 at 4:31
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    $\begingroup$ Please be more careful with your language - you are mixing up "$V$" with "basis for $V$". But yes, if you are given a subspace defined as the set of vectors satisfying some collection of homogeneous linear equations, then treating as the nullspace of the corresponding matrix is probably a good approach. $\endgroup$ Commented Jun 9, 2019 at 4:33
  • $\begingroup$ "then a vector x is in V if and only if Mx=0" this because the equation is equal to 0 right ? So we have to find a basis that spans the 0 vector or how should i say, make a linear combination with the existing coefficients, such that, we get the result of that equation, which in this case, it's = 0 $\endgroup$
    – AndrewM
    Commented Jun 9, 2019 at 4:33

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