QR decomposition proof

Question

Let $A\in\mathbb{M}_{m\times n}(\mathbb{R})$ with $m>n$ and $rank(A)=n$ and take the decomposition $A=QR$ with $Q\in\mathbb{M}_{m\times n}(\mathbb{R})$ a orthogonal matrix and $R\in\mathbb{M}_n(\mathbb{R})$ a upper triangular matrix with diagonal elements positive. Assume that $R^*(A)$ is a column space of $A$ and $N^*(A)$ is a null space of $A$ * is just a notation. Show that

i)$R^*(A)=R^*(Q)$

ii)$N^*(A^t)=N^*(Q^t)$

iii)$R^*(A^t)=R^*(Q^t)=\mathbb{R}^n$

iv)$N^*(A)=N^*(Q)=0_{\mathbb{R}^n}$

Can anyone give me help?

Answer 1

you have a matrix with full column rank meaning that the columns are linearly independent. we also have $$A = QR, Q^\top Q= I_n, R \text{ upper triangular.} $$

we will try to prove that column space $C(A), C(Q)$ of $A$ and $Q$ are the same. i wil do the case $n = 3.$ the general case should be similar and you can do that. we will write $A = QR$ as $$[a_1, a_2, a_3] = [q_1, q_2, q_3]R $$ this means each of the columns of $A$ are a linear combinations of the columns of $Q$. in particuls $a_3 = r_{13}q_1 + r_{23}q_2 + r_{33}q_3.$ that implies $C(A) \subset C(Q).$

to establish $C(Q) \subset C(A),$ we can use the fact that $R$ is invertible and write $A = QR$ as $A R^{-1} = Q$ which implies the columns of $Q$ are linear combinations of the columns of $A.$ and we are done.

next i will do part(iv). we will show that null spaces and $A$ and $R$ are the same. suppose $x \in \ker(A).$ then $Ax = 0 \to QRx =0 \to Q^\top Q R x = 0\to Rx = 0.$ this shows $\ker A \subset \ker R.$ the other way is much easier and i will let you do it.

Answer 2

Whenever $A=BF$, where $F$ is an invertible matrix, we have $R^*(A)=R^*(B)$ (using your notation).

Indeed, if $v\in R^*(A)$, we have $v=Ax$ for some $x$, so $v=(BF)x=B(Fx)\in R^*(B)$. The converse inclusion follows from $B=AF^{-1}$.

Under the same hypotheses, we have $A^t=F^tB^t$; if $w\in N^*(B)$, then $A^tw=F^tB^tw=0$, so $w\in N^*(A^t)$. The converse inclusion follows from the fact that $B^t=(F^{-1})^tA^t$.

In case $A=FB$, with invertible $F$, it's true that $A$ and $B$ have the same rank, because multiplication by $F$ induces a (bijective) linear map from the column space of $B$ to the column space of $A$. Thus $$ n=\operatorname{rank}Q^t=\operatorname{rank}A^t $$ The rank-nullity theorem says that both $Q^t$ and $A^t$ have zero null space.

As you see, it's not really relevant that $Q$ has orthogonal columns.