Question

Let \({{\bf{u}}_1},......,{{\bf{u}}_p}\) be an orthogonal basis for a subspace \(W\) of \({\mathbb{R}^n}\), and let \(T:{\mathbb{R}^n} \to {\mathbb{R}^n}\) be defined by \(T\left( x \right) = {\rm{pro}}{{\rm{j}}_W}x\). Show that \(T\) is a linear transformation.

Short answer

It is verified that \(T\) is a linear transformation.

Step-by-step solution

\(QR\) factorization of a Matrix and Linear Transformation

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 \({Q^T}A = R\).

A transformation \(T:X \to Y\), is said to be linear if it satisfies the following properties:

1. For any \({{\bf{x}}_1},{{\bf{x}}_2} \in X\), \(T\left( {{{\bf{x}}_1} + {{\bf{x}}_2}} \right) = T\left( {{{\bf{x}}_1}} \right) + T\left( {{{\bf{x}}_2}} \right)\).

2. For any \({\bf{x}} \in X\) and \(c \in F\), \(T\left( {c{\bf{x}}} \right) = cT\left( {\bf{x}} \right)\).

Showing the Transformation is Linear

Given mapping is \(T:{\mathbb{R}^n} \to {\mathbb{R}^n}\) such that

\(T\left( x \right) = {\rm{pro}}{{\rm{j}}_W}x\)

It is given that \(\left\{ {{{\bf{u}}_1},......,{{\bf{u}}_p}} \right\}\) is an orthonormal basis for \(W\), then lets us construct a matrix with \({{\bf{u}}_1},......,{{\bf{u}}_p}\)as column of the matrix and name it as \(U\).

Then we have

\(U{U^T} = {\rm{pro}}{{\rm{j}}_W}\)by the Theorem 10(from 6.3).

Now,

\(\begin{aligned}{}T\left( x \right) &= {\rm{pro}}{{\rm{j}}_W}x\\T\left( x \right) &= \left( {U{U^T}} \right)x\end{aligned}\)

Hence, we get that \(T\) is a matrix transformation. Also, we know that every matrix transformation is linear so \(T\) is a linear transformation.