# Finding the column space from nullspace and particular solution

Let $A$ be a $3 \times 4$ matrix and it has a particular solution of $\begin{bmatrix} x_1\\ x_2\\ x_3\\ x_4 \end{bmatrix}= \begin{bmatrix} 1\\ 0\\ -1\\ 0 \end{bmatrix}$

The nullspace of $A$ is $\begin{bmatrix} x_1\\ x_2\\ x_3\\ x_4 \end{bmatrix}= t\begin{bmatrix} 1\\ 1\\ 0\\ 1 \end{bmatrix}+ s\begin{bmatrix} -2\\ 1\\ 1\\ 0 \end{bmatrix}$, where $t,s \in \mathbb{R}$.

From this, do I have enough information to find the column space of $A$? I am thinking that I could because I can find out the reduced row echelon form of $A$ by working backwards from the nullspace. And I could get the $rref(A)=\begin{bmatrix} 1 & 0 & 2 & -1\\ 0 & 1& -1 & -1\\ 0 & 0 & 0 & 0 \end{bmatrix}$. From here, isn't the column space of $A$ just the $span(\begin{bmatrix} 1\\ 0\\ 0 \end{bmatrix}, \begin{bmatrix} 0\\ 1\\ 0 \end{bmatrix})$?

We could also determine that the left-nullspace is going to be of rank 1.

But from the book that I am reading, it says that I should not have enough information to find the column space of $A$. Why is it so? Isn't the span I wrote above the column space of $A$ already?

1. It makes no sense to talk about "particular solution of a $3\times 4$ matrix." Matrices don't have "particular solutions", systems of equations have particular solutions. If you are interpreting $A$ as the augmented matrix of a $3\times 3$ system of linear equations, then you need to say that this is what you mean. If you are interpreting $A$ as the coefficient matrix of a $3\times 4$ system of homogeneous linear equations, then you need to say so. If you are interpreting $A$ as the coefficient matrix of a $3\times 4$ system of linear equations with unspecified "solution vector" $\mathbf{b}$, then you need to say so as well.

I am guessing that what you have is that $A$ is the coefficient matrix of a system $A\mathbf{x}=\mathbf{b}$, and that the vector you give happens to be a particular solution to that system.

2. Elementary row operations do not preserve the columnspace. That is, the columnspace of the reduced row echelon form of $A$ is not the same as the columnspace of $A$. For example, if $A$ is a matrix that consists only of $1$s in every entry, then the reduced row-echelon form of $A$ is a row of $1$s and all the other rows $0$. The column space of $REF(A)$ is then the span of $(1,0,\ldots,0)^t$; but the column space of $A$ is the span of $(1,1,\ldots,1)^t$, which is different.

3. However, say that $C$ is a matrix obtained from $A$ by the application of elementary row operations (for example, $C$ is the reduced row echelon form of $A$, or the row echelon form of $A$). If the $j_1,j_2,\ldots,j_k$ columns of $C$ are a basis for the columnspace of $C$, then the corresponding columns of $A$ are a basis for the columnspace of $A$. In your example, the first and second columns of the reduced row echelon form of $A$ are a basis for its columnspace. That means that the first and second columns of $A$ are a basis for the columnspace of $A$.

• Thanks. I am still a little confused between the second and third points. In the second point, elementary row operations will not preserve the column space. I suppose by that, it is referring to the row operations that do row exchanges and row subtractions will not preserve column space whereas if a row is multiplied by a factor, the column space is still preserved? In other words, not just the space that the basis vectors are spanning, but the direction of the basis also matter for a column space of a matrix? – xenon Oct 23 '11 at 1:16
• @xEnOn: Elementary row operations do not preserve the column space. Even simply multiplying a row by a nonzero, non-$1$ constant will not preserve the column space. Again, consider a matrix with all $1$s. the columns space is the span of $(1,1,\ldots,1)^t$. Now multiply the first row by $2$. The column space of the resulting matrix is spanned by $(2,1,\ldots,1)^t$, so it is not the same as the column space of the original matrix. I cannot parse your "In other words"; bases don't have "direction". – Arturo Magidin Oct 23 '11 at 3:58

What you wrote is the column space of $rref(A)$, not the column space of $A$. The column space changes when you perform row operations on the matrix. You have enough information to find the dimension of the column space (since that doesn't change when you perform row operations), but not the column space itself.

• But wouldn't the column space of $rref(A)$ spans the same space as the column space of $A$? If so, how has the row operations changed the column space? – xenon Oct 22 '11 at 18:04
• For example, if I interchange first 2 rows in the matrix, then the column space certainly can change. If the first 2 entries in the column were (1,2) before, now it's become the column (2,1). – Ted Oct 22 '11 at 18:06
• The direction of the basis of $A$ has changed if you swap the rows. But the space it spans is still the same as the column space wouldn't it? Unless the direction of the basis must be preserved to be considered column space despite having the same spanning space? – xenon Oct 22 '11 at 18:12
• Let $A$ be the 2 by 1 matrix which is just a single column (1,2). The column space of $A$ is the set of all multiples of the column vector (1,2). Now interchange the 2 rows. You get the 2 by 1 matrix $B$ with a column (2,1). The column space of $B$ is the set of all multiples of column vector (2,1). The column spaces of $A$ and $B$ are completely different, because (1,2) and (2,1) are not multiples of each other. The column vector (1,2) is in the column space of $A$ but not $B$, and the column vector (2,1) is in the column space of $B$ but not $A$. – Ted Oct 22 '11 at 18:19