Suppose \(A = QR\), where \(R\) is an invertible matrix. Showthat \(A\) and \(Q\) have the same column space.

A matrix with order \(m \times n\) can be written as the multiplication of an upper triangular matrix \(R\) and a matrix \(Q\) which is formed by applying the Gram–Schmidt orthogonalization processto the \({\rm{col}}\left( A \right)\).

The matrix \(R\) can be found by the formula \({Q^T}A = R\).

Let \({\bf{y}} \in {\rm{Col}}A\). Then we have\(A{\bf{x}} = {\bf{y}}\) for any vector \({\bf{x}}\).

Then

\(\begin{aligned}{}A{\bf{x}} &= QR{\bf{x}}\\{\bf{y}} &= Q\left( {R{\bf{x}}} \right)\\{\bf{y}} &= Q{\bf{v}}\end{aligned}\)

Where \({\bf{v}} = R{\bf{x}}\).

Hence it is shown that \({\bf{y}}\) is also in \({\rm{Col}}Q\).

So \({\rm{Col}}A \subseteq {\rm{Col}}Q\).

Conversely let \({\bf{y}} \in {\rm{Col}}Q\), then for some vector \({\bf{x}}\) we have

\(Q{\bf{x}} = {\bf{y}}\).

Now it is given that \(R\) is invertible, hence

\(\begin{aligned}{}A &= QR\\A{R^{ - 1}} &= Q\\A{R^{ - 1}}{\bf{x}} &= Q{\bf{x}}\\A\left( {{R^{ - 1}}{\bf{x}}} \right) &= {\bf{y}}\\Aw &= {\bf{y}}\end{aligned}\)

Where \(w = {R^{ - 1}}{\bf{x}}\).

Hence \(y\) is also in the column space of \(A\).

So \({\rm{Col}}A \supseteq {\rm{Col}}Q\).

Thus \({\rm{Col}}A = {\rm{Col}}Q\). Hence it is showed.