(M) Let \({{\mathop{\rm a}\nolimits} _1},...,{{\mathop{\rm a}\nolimits} _5}\) denote the columns of the matrix \(A\), where \(A = \left( {\begin{array}{*{20}{c}}5&1&2&2&0\\3&3&2&{ - 1}&{ - 12}\\8&4&4&{ - 5}&{12}\\2&1&1&0&{ - 2}\end{array}} \right),B = \left( {\begin{array}{*{20}{c}}{{{\mathop{\rm a}\nolimits} _1}}&{{{\mathop{\rm a}\nolimits} _2}}&{{{\mathop{\rm a}\nolimits} _4}}\end{array}} \right)\)

1. Explain why \({{\mathop{\rm a}\nolimits} _3}\) and \({{\mathop{\rm a}\nolimits} _5}\) are in the column space of \(B\).
2. Find a set of vectors that spans \({\mathop{\rm Nul}\nolimits} A\).
3. Let \(T:{\mathbb{R}^5} \to {\mathbb{R}^4}\) be defined by \(T\left( x \right) = A{\mathop{\rm x}\nolimits} \). Explain why \(T\) is neither one-to-one nor onto.

It is given that \(B = \left( {\begin{array}{*{20}{c}}{{{\mathop{\rm a}\nolimits} _1}}&{{{\mathop{\rm a}\nolimits} _2}}&{{{\mathop{\rm a}\nolimits} _4}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}5&1&2\\3&3&{ - 1}\\8&4&{ - 5}\\2&1&0\end{array}} \right)\).

Consider the augmented matrix \(\left( {\begin{array}{*{20}{c}}B&{{{\mathop{\rm a}\nolimits} _3}}\end{array}} \right)\) as shown below:

\(\left( {\begin{array}{*{20}{c}}B&{{{\mathop{\rm a}\nolimits} _3}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}5&1&2&2\\3&3&{ - 1}&2\\8&4&{ - 5}&4\\2&1&0&1\end{array}} \right)\)

Consider the augmented matrix \(\left( {\begin{array}{*{20}{c}}B&{{{\mathop{\rm a}\nolimits} _5}}\end{array}} \right)\) as shown below:

\(\left( {\begin{array}{*{20}{c}}B&{{{\mathop{\rm a}\nolimits} _5}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}5&1&2&0\\3&3&{ - 1}&{ - 12}\\8&4&{ - 5}&{12}\\2&1&0&{ - 2}\end{array}} \right)\)

Consider the matrix \({\mathop{\rm C}\nolimits} = \left( {\begin{array}{*{20}{c}}5&1&2&2\\3&3&{ - 1}&2\\8&4&{ - 5}&4\\2&1&0&1\end{array}} \right)\).

Use the code in MATLAB to obtain the row-reduced echelon form as shown below:

\(\begin{array}{l} > > {\mathop{\rm C}\nolimits} = \left( {5\,\,1\,\,\,2\,\,\,2;\,3\,\,\,3\,\,\, - 1\,\,\,2;\,\,8\,\,\,4\,\,\, - 5\,\,\,4;\,\,2\,\,\,1\,\,\,0\,\,\,1} \right)\\ > > {\mathop{\rm U}\nolimits} = {\mathop{\rm rref}\nolimits} \left( {\mathop{\rm C}\nolimits} \right)\end{array}\)

\(\left( {\begin{array}{*{20}{c}}5&1&2&2\\3&3&{ - 1}&2\\8&4&{ - 5}&4\\2&1&0&1\end{array}} \right) \sim \left( {\begin{array}{*{20}{c}}1&0&0&{\frac{1}{3}}\\0&1&0&{\frac{1}{3}}\\0&0&1&0\\0&0&0&0\end{array}} \right)\)

Consider the matrix \({\mathop{\rm D}\nolimits} = \left( {\begin{array}{*{20}{c}}5&1&2&0\\3&3&{ - 1}&{ - 12}\\8&4&{ - 5}&{12}\\2&1&0&{ - 2}\end{array}} \right)\).

Use the code in MATLAB to obtain the row-reduced echelon form as shown below:

\(\begin{array}{l} > > {\mathop{\rm D}\nolimits} = \left( {5\,\,1\,\,\,2\,\,\,0;\,3\,\,\,3\,\,\, - 1\,\,\, - 12;\,\,8\,\,\,4\,\,\, - 5\,\,\,12;\,\,2\,\,\,1\,\,\,0\,\,\, - 2} \right)\\ > > {\mathop{\rm U}\nolimits} = {\mathop{\rm rref}\nolimits} \left( {\mathop{\rm D}\nolimits} \right)\end{array}\)

\(\left( {\begin{array}{*{20}{c}}5&1&2&0\\3&3&{ - 1}&{ - 12}\\8&4&{ - 5}&{12}\\2&1&0&{ - 2}\end{array}} \right) \sim \left( {\begin{array}{*{20}{c}}1&0&0&{\frac{{10}}{3}}\\0&1&0&{\frac{{ - 26}}{3}}\\0&0&1&{ - 4}\\0&0&0&0\end{array}} \right)\)

The same conclusions can be drawn when considering the row-reduced echelon form of A.

\(A \sim \left( {\begin{array}{*{20}{c}}1&0&{\frac{1}{3}}&0&{\frac{{10}}{3}}\\0&1&{\frac{1}{3}}&0&{\frac{{ - 26}}{3}}\\0&0&0&1&{ - 4}\\0&0&0&0&0\end{array}} \right)\)

A typical vector v in Col A has the property that the equation \(A{\mathop{\rm x}\nolimits} = {\mathop{\rm v}\nolimits} \) is consistent.

It means \({{\mathop{\rm a}\nolimits} _3}\)and \({a_5}\) are in the column space of \(B\), and both the systems of equations are consistent.

Thus, it is proved that \({{\mathop{\rm a}\nolimits} _3}\)and \({a_5}\) is in the column space of \(B\).

b)

The row-reduced echelon form of matrix A is shown below:

\(A \sim \left( {\begin{array}{*{20}{c}}1&0&{\frac{1}{3}}&0&{\frac{{10}}{3}}\\0&1&{\frac{1}{3}}&0&{\frac{{ - 26}}{3}}\\0&0&0&1&{ - 4}\\0&0&0&0&0\end{array}} \right)\)

Write the general solution of \(A{\mathop{\rm x}\nolimits} = 0\) using the row-reduced echelon form of matrix A.

\(\begin{array}{l}{{\mathop{\rm x}\nolimits} _1} = \frac{{ - 1}}{3}{{\mathop{\rm x}\nolimits} _3} - \frac{{10}}{3}{{\mathop{\rm x}\nolimits} _5}\\{{\mathop{\rm x}\nolimits} _2} = \frac{{ - 1}}{3}{{\mathop{\rm x}\nolimits} _3} + \frac{{26}}{3}{{\mathop{\rm x}\nolimits} _5}\\{{\mathop{\rm x}\nolimits} _4} = 4{{\mathop{\rm x}\nolimits} _5}\end{array}\)

\({{\mathop{\rm x}\nolimits} _3}\)and \({{\mathop{\rm x}\nolimits} _5}\)

The general solution of \(A{\mathop{\rm x}\nolimits} = 0\) in the parametric form is shown below:

\(\begin{array}{c}{\mathop{\rm x}\nolimits} = \left( {\begin{array}{*{20}{c}}{{{\mathop{\rm x}\nolimits} _1}}\\{{{\mathop{\rm x}\nolimits} _2}}\\{{{\mathop{\rm x}\nolimits} _3}}\\{{{\mathop{\rm x}\nolimits} _4}}\\{{{\mathop{\rm x}\nolimits} _5}}\end{array}} \right)\\ = {{\mathop{\rm x}\nolimits} _3}\left( {\begin{array}{*{20}{c}}{\frac{{ - 1}}{3}}\\{\frac{{ - 1}}{3}}\\1\\0\\0\end{array}} \right) + {{\mathop{\rm x}\nolimits} _5}\left( {\begin{array}{*{20}{c}}{\frac{{ - 10}}{3}}\\{\frac{{26}}{3}}\\0\\4\\1\end{array}} \right)\end{array}\)

Thus, the set of vectors that span \({\mathop{\rm Nul}\nolimits} A\) is \(\left\{ {\left( {\begin{array}{*{20}{c}}{\frac{{ - 1}}{3}}\\{\frac{{ - 1}}{3}}\\1\\0\\0\end{array}} \right),\left( {\begin{array}{*{20}{c}}{\frac{{ - 10}}{3}}\\{\frac{{26}}{3}}\\0\\4\\1\end{array}} \right)} \right\}\).

c)

Let \(T:{\mathbb{R}^n} \to {\mathbb{R}^m}\) be a linear transformationand A be the standard matrix for T. Then, according to Theorem 12in section 1.9,

a.\(T\) maps \({\mathbb{R}^n}\)onto \({\mathbb{R}^m}\) if and only if the columns of A span \({\mathbb{R}^m}\).

b.\(T\) is one-to-one if and only if the columns of A are linearly independent.

The row-reduced echelon form of A demonstrates that the columns of \(A\) are linearly dependent and do not span \({\mathbb{R}^4}\).

Therefore, according to theorem 12, \(T\) is neither one-to-one nor onto.