Row And Column Spaces

Inlinear algebra, when studying a particular matrix, one is often interested in determiningvector spacesassociated with the matrix, so as to better understand how the correspondinglinear transformationoperates. Two important examples of associated subspaces are therow spaceandcolumn spaceof a matrix.

Suppose $A$is an $m$-by- $n$matrix, with rows $r_1, \cdots, r_m \in \mathbb{R}^n$and columns $c_1, \cdots, c_n \in \mathbb{R}^m$. Therow space $R(A)$is the subspace of $\mathbb{R}^n$spanned by the vectors $\{r_i\}_{1\le i \le m}$; similarly, thecolumn space $C(A)$is the subspace of $\mathbb{R}^m$spanned by the vectors $\{c_i\}_{1\le i \le n}$.

Consider the matrix

$A = \begin{pmatrix} 2 & 1 & 0 \\ 3& -1 & 2 \end{pmatrix}.$

Compute the row space $R(A)$and the column space $C(A)$.

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The rows of this matrix are $(2,1,0)$and $(3,-1,2)$, so the row space $R(A)$is the span of these two vectors in $\mathbb{R}^3$. In particular, since these rows are linearly independent, $R(A)$is a two-dimensional subspace of $\mathbb{R}^3$. To determine this subspace explicitly, note that both rows are orthogonal to $(-2,4,5)$, so they span the plane $-2x + 4y + 5z = 0$.

To determine the column space of $A$, first note the columns of the matrix are $(2,3)$, $(1,-1)$, and $(0,2)$. Since the first two of these vectors are linearly independent, it follows that their span $C(A)$is a two-dimensional subspace of $\mathbb{R}^2$, and hence $\mathbb{R}^2$itself. $_\square$

Note that, in the above computation, we have $\dim\big(R(A)\big) = \dim\big(C(A)\big)$. This is an instance of the linear algebra fact that "row rank equals column rank," and is discussed in the article onrank.

Let $A$be an $m$-by- $n$matrix, which corresponds to a linear transformation $T: \mathbb{R}^n \to \mathbb{R}^m$. One can interpret the row and column spaces of $A$in terms of this transformation.

Suppose the columns of $A$are $c_1, \ldots, c_n \in \mathbb{R}^m$. For any vector $(a_1, a_2, \ldots, a_n)\in \mathbb{R}^n$, one may compute $T(a_1, \ldots, a_n) = a_1 c_1 + a_2 c_2 + \cdots + a_n c_n \in \mathbb{R}^m.$This computation implies theimageof $T$is precisely the column space $C(A)$. One should think of this as a "coordinate-free" interpretation of the column space; no matter which matrix one chooses to represent the linear transformation $T$, the column space of that matrix will always equal the image of $T$.

If $A$is a matrix representing a linear transformation $T$, then the column space of $A$is the image of $T$. In symbols, $C (A) = \文本{Im} (T)。$

Similarly, one can interpret the row space of $A$as follows: If the rows of $A$are $r_1, \ldots, r_m \in \mathbb{R}^n$, then for any $a\in \mathbb{R}^n$, one computes $T(a) = (r_1 \cdot a, r_2 \cdot a, \ldots, r_m \cdot a),$where $\cdot$denotes thedot product. Recall that thekernelof $T$is the subspace of $\mathbb{R}^n$consisting of all the vectors $v$such that $Tv = 0 \in \mathbb{R}^m$. Then, the computation above shows that $a$is in the kernel of $T$if and only if it is orthogonal to each row $r_i$. In other words, the kernel of $T$is precisely the space of vectors $v$such that $v\cdot w = 0$forall $w\in R(A)$.

Let $V$be a vector space and $W \subset V$a subspace; assume that $V$has a dot (inner) product $\cdot$defined on it. Theorthogonal complementof $W$is the subspace $W^{\perp}$consisting of the vectors $a\in V$such that $a\cdot b = 0$forall $b \in W$.

If $A$is a matrix representing a linear transformation $T$, then the kernel of $T$is the orthogonal complement of $R(A)$. In symbols, $R (A) ^{\补}= {Ker} \文本(T)。$