Orthogonal vectors & subspaces
Two vectors $v_1$ and $v_2$ are orthogonal if $v_1^T v_2=\left\lVert v_1 \right\rVert \left\lVert v_2 \right\rVert \cos\theta=0$.
Two subspaces $M_1$ and $M_2$ are orthogonal if $v_1 \bot v_2$ for all $v_1 \in M_1$ and $v_2 \in M_2$.
Orthogonal complement
The orthogonal complement to a subspace $M \subset R^n$ is defined by $M^\bot=\{w \in R^n: w^T v=0 \: \forall v \in M \}$.
$M^\bot$ is the largest (unique) subspace orthogonal to $M$.
The following holds:
$M^\bot$ is a subspace.
$R^n=M \oplus M^\bot$
If $\text{dim}\:M=k$, then $\text{dim}\:M^\bot=n-k$.
$(M^\bot)^\bot=M$
The four fundamental subspaces
Definitions
$A=\begin{pmatrix} a_{11} & \cdots & a_{1n} \\ \vdots & \cdots & \vdots \\ a_{m1} & \cdots & a_{mn} \end{pmatrix}=\begin{pmatrix} \vec{a_1} & \cdots & \vec{a_n} \end{pmatrix}=\begin{pmatrix} \overline{a}_1^T \\ \overline{a}_2^T \\ \vdots \\ \overline{a}_m^T \end{pmatrix}$
Column space
$C(A)=\{Ax\:|\:x \in R^n \}=\{ \sum_{j=1}^n \vec{a_j}x_j\:|\:x_j \in R, j=1,2,\cdots,n\}$
Nullspace
$N(A)=\{x \in R^n\:|\:Ax=0\}=\{x \in R^n\:|\:\overline{a}_i^T x=0, i=1,2,\cdots,m\}$
Row space
$C(A^T)=\{A^Tu\:|\:u \in R^m \}=\{ \sum_{i=1}^m \overline{a}_i u_i\:|\:u_i \in R, i=1,2,\cdots,m\}$
Left nullspace
$N(A^T)=\{u \in R^m\:|\:A^T u=0\}=\{u \in R^m\:|\:\vec{a_j}^T u=0, j=1,2,\cdots,n \}=\{u \in R^m\:|\:u^T A=0\}$
The left nullspace is the space of all vectors y such that $A^Ty=0$, or equivalently $y^TA=0$. Thus the term left nullspace.
Geometric picture [later]
Relation to RREF [later]
Summary
Let $A$ be an $m\times n$ matrix, and $r$ be its rank.
| subspace | notation | subspace of | dimension |
| row space | $C(A^T)$ | $R^n$ | $r$ |
| column space | $C(A)$ | $R^m$ | $r$ |
| nullspace | $N(A)$ | $R^n$ | $n-r$ |
| left nullspace | $N(A^T)$ | $R^m$ | $m-r$ |
$\text{dim}\:C(A^T)+\text{dim}\:N(A)=\text{dim}\:R^n\\
\text{dim}\:C(A)+\text{dim}\:N(A^T)=\text{dim}\:R^m$
Fundamental Theorem of Linear Algebra
Let A be an $m \times n$ matrix.
1. $N(A)$ is the orthogonal complement of $C(A^T)$ in $R^n$.
2. $N(A^T)$ is the orthogonal complement of $C(A)$ in $R^m$.
3. $N(A) \oplus C(A^T)=R^n$
4. $N(A^T) \oplus C(A)=R^m$
The following orthogonality relations hold:
$C(A)^\bot=N(A^T)\\
N(A^T)^\bot=C(A)\\
C(A^T)^\bot=N(A)\\
N(A)^\bot=C(A^T)$
Explanations:
Orthogonal subspaces
We will examine an $m \times n$ matrix $A$. Then both the row space and the nullspace are subspaces of $R^n$. [$A^T$ is $n \times m$ and $u$ is $m \times 1$. By matrix multiplication, the row space is $n \times 1$.] Suppose $x$ is a vector in the nullspace of $A$. Then $Ax=0$. From the definition of matrix multiplication, we have $$\begin{pmatrix} \text{row 1} \\ \text{row 2} \\ \vdots \\ \text{row m} \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ \vdots \\ x_m \end{pmatrix}=0.$$ As the row space is the linear combination of the rows, the dot product of $x$ with any vector in the row space must be $0$. So if $v\in C(A^T)$ and $w\in N(A)$, we must have $v^T w=0$ Therefore, the row space and the nullspace of $A$ are orthogonal.
Similarly, every vector in the left nullspace of $A$, $N(A^T)$, is perpendicular to every vector in the column space of $A$, $C(A)$. Hence, the column space of A and the left nullspace of A are orthogonal.
Orthogonal complements
We have established that the row space and the nullspace are orthogonal. It turns out the nullspace is in fact the orthogonal complement of the row space. The dimension of the nullspace is $n-r$, where $r$ is the rank (dimension of row space) of $A$. If there were a vector orthogonal to the row space but not to the nullspace, then the dimension of $C(A^T)^\bot$ would be at least $n-r+1$. But this would be too large for both $C(A^T)$ and $C(A^T)^\bot$ to fit in $R^n$. Therefore, the nullspace is the largest subspace perpendicular to the row space, and we have $C(A^T)^\bot=N(A)$. Using similar arguments, we have $N(A)^\bot=C(A^T)$.
What is so useful about complements is that we can use them to break down vectors into components. Namely, for an $m \times n$ matrix $A$, any vector $x \in R^n$ can be written as the sum of a component $x_r$ in the row space and a component $x_n$ in the nullspace: $x=x_r+x_n$.
Now, let's consider $Ax=b$. We claim that every output vector $b$ comes from one and only one vector in the row space. Write $x=x_r+x_n$. Then $Ax=Ax_r+Ax_n=Ax_r+0=Ax_r$. Thus, we know there is a vector $x_r$ in the row space such that $Ax_r=b$. Suppose $x_r'$ is another vector in the row space such that $Ax_r'=b$. Note that $x_r-x_r'$ is in row space since it is the difference of two vectors in the row space. Now $A(x_r-x_r')=Ax_r-Ax_r'=b-b=0$. Then $x_r-x_r'$ is in the nullspace. As the row space and nullspace are orthogonal complements, the only vector in both of them is $0$. So, $x_r-x_r'=0$ and $x_r=x_r'.\:\Box$
Relation to domain and codomain
Consider an $m \times n$ matrix. The matrix $A$ maps from $R^n$ to $R^m$; we say that its domain is $R^n$ and its codomain is $R^m$. Since a vector in the domain either maps to zero or elsewhere, we can say that the domain includes the nullspace of $A$ and the preimage of the range of $A$. Also, the codomain of $A$ includes the range of $A$ and the part we cannot reach with $A$. Now, the matrix $A^T$ maps from $R^m$ to $R^n$. We can say that the range of $A^T$ is the preimage of the range of $A$ and that the nullspace of $A^T$ is the subspace we cannot reach with $A$.
To sum up, we have the four fundamental subspaces as
• an orthogonal decomposition of the domain of $A$:
nullspace of $A$ $\oplus$ range of $A^T$
• an orthogonal decomposition of the codomain of $A$:
range of $A$ $\oplus$ nullspace of $A^T$
Useful:
http://xoax.net/math/crs/linear_algebra_mit/lessons/Lecture10/
Reference:
https://rip94550.wordpress.com/2009/09/07/linear-algebra-the-four-fundamental-subspaces/
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