Ngô Quốc Anh

April 23, 2010

Comparison of some subspaces associated to A and A^TA

Filed under: Các Bài Tập Nhỏ — Ngô Quốc Anh @ 19:10

Let us start with a given $n\times m$ matrix $A$ $\displaystyle {A_{n \times m}} = \left( {\begin{array}{*{20}{c}} {{a_{11}}} & \cdots & \cdots & {{a_{1m}}} \\ \vdots & \ddots & \ddots & \vdots \\ {{a_{n1}}} & \cdots & \cdots & {{a_{nm}}} \\ \end{array} } \right)$.

The aim of this entry is to compare nullspace, column space and row space between $A$, $A^TA$ and $AA^T$. Obviously, $A^TA$ and $AA^T$ are $m\times m$ and $n \times n$ matrices respectively.

Proposition 1. The following $\displaystyle N(A)=N(A^TA)$

holds.

Proof. Pick an arbitrary element $x \in N(A)$, i.e. $Ax=0$, we can see that $\displaystyle A^TAx=A^T(Ax)=A^T0=0$

so $\displaystyle N(A) \subset N(A^TA)$.

Conversely, assume $x$ is such that $A^TAx=0$, as a consequence, $x^TA^TAx=0$. This gives us the fact $\displaystyle (Ax) \cdot (Ax)= (Ax)^T(Ax)=x^TA^TAx=0$.

Consequently, $Ax=0$ which proves $\displaystyle N(A^TA)\subset N(A)$.

In other words, $\displaystyle N(A)=N(A^TA)$.

Remark $\displaystyle N(A)=N(AA^T)$ is no longer true since these two matrices have different dimension.

Regarding to matrix $A$, one has $\displaystyle {\rm rank}(A)+\dim N(A)=m$.

Similarly, involving matrix $A^TA$, one gets $\displaystyle {\rm rank}(A^TA)+\dim N(A^TA)=m$.

It now follows from Proposition 1 that $\displaystyle {\rm rank}(A)={\rm rank}(A^TA)$.

In other words, column and row spaces associated to $A$ and $A^TA$ have the same dimension respectively.

Row spaces. We prove the following

Proposition 2. The following $\displaystyle RS(A)=RS(A^TA)$

holds.

Proof. The way to compare column spaces is to use the following facts $\displaystyle RS(A)=CS(A^T)$

and $\displaystyle CS(A^T)=A^T(\mathbb R^n)$.

Equivalently, from the first fact we need to show that $\displaystyle CS(A^T)=CS(A^TA)$.

In term of the second fact, once you have a suitable matrix $Q$ the column space of $AQ$ is indeed contained in the column space of $A$. Therefore $\displaystyle CS(A^TA) \subset CS(A^T)$

which turns out to be $\displaystyle RS(A^TA) \subset RS(A)$

since they have the same dimension, equality occurs.

Remark. This comes from the proof above. If you have a good matrix $Q$, the following is true $\displaystyle CS(AQ) \subset CS(A)$.

Column spaces. We prove the following

Proposition 3. The following $\displaystyle CS(A)=CS(AA^T)$

holds.

Proof. This is trivial by using Proposition 2.

Remark. If you have a good matrix $P$, the following is true $\displaystyle RS(PA) \subset RS(A)$.