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.

Nullspaces. We start with the following result

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).

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