Verify the uniqueness of A in Theorem 10. Let \(T:{\mathbb{R}^n} \to {\mathbb{R}^m}\) be a linear transformation such that \(T\left( {\bf{x}} \right) = B{\bf{x}}\) for some \(m \times n\) matrix B. Show that if A is the standard matrix for T, then \(A = B\). (Hint:Show that A and B have the same columns.)

For matrix Aof the order\(m \times n\), if vector\({\bf{x}}\)is in\({\mathbb{R}^n}\), then transformation\(T\)of vector x is represented as\(T\left( x \right)\), and it is in the dimension\({\mathbb{R}^m}\).

It can also be written as\(T:{\mathbb{R}^n} \to {\mathbb{R}^m}\).

Here, the dimension \({\mathbb{R}^n}\) is the domain, and the dimension \({\mathbb{R}^m}\) is the codomain of transformation \(T\).

Standard matrix A can be represented as shown below:

\(A = \left( {\begin{array}{*{20}{c}}{T\left( {{{\bf{e}}_1}} \right)}& \cdots &{T\left( {{{\bf{e}}_n}} \right)}\end{array}} \right)\)

For the matrix of the order \(m \times n\), vector \(T\left( {{{\bf{e}}_j}} \right)\) is in the \(j{\rm{ th}}\) column. And the \(j{\rm{ th}}\)column of the identity matrix is represented by \({{\bf{e}}_j}\).

For matrix B, suppose\({{\bf{b}}_1},{{\bf{b}}_2},{{\bf{b}}_3},...\)are the columns.

The image of vector x is represented as\(T\left( {\bf{x}} \right) = A{\bf{x}}\).

So, for matrix B, the image of the\(j{\rm{ th}}\)column is

\(\begin{aligned} T\left( {{{\bf{e}}_j}} \right) &= B{{\bf{e}}_j}\\ &= {{\bf{b}}_j}.\end{aligned}\)

Thus,\(T\left( {{{\bf{e}}_n}} \right) = {{\bf{b}}_n}\).

Substitute the values in the standard matrix\(A = \left( {\begin{array}{*{20}{c}}{T\left( {{{\bf{e}}_1}} \right)}& \cdots &{T\left( {{{\bf{e}}_n}} \right)}\end{array}} \right)\), as shown below:

\(\begin{aligned} A &= \left( {\begin{array}{*{20}{c}}{T\left( {{{\bf{e}}_1}} \right)}& \cdots &{T\left( {{{\bf{e}}_n}} \right)}\end{array}} \right)\\ &= \left( {\begin{array}{*{20}{c}}{{{\bf{b}}_1}}& \cdots &{{{\bf{b}}_n}}\end{array}} \right)\\ &= B\end{aligned}\)

Hence, it is proved that \(A = B\).