91Ó°ÊÓ

The zero vector is a condition for the subspace but it is not the only one. Therefore, the statement is false.

\(x = \sum\limits_{i = 1}^p {{c_i}{v_i}} \)

\(y = \sum\limits_{i = 1}^p {{d_i}{v_i}} \)

Check the vector addition:

\(\begin{array}{c}x + y = \sum\limits_{i = 1}^p {{c_i}{v_i}} + \sum\limits_{i = 1}^p {{d_i}{v_i}} \\c = \sum\limits_{i = 1}^p {\left( {{c_i} + {d_i}} \right){v_i} \in H} \end{array}\)

Multiply the vector by x:

\(cx = \sum\limits_{i = 1}^p {{{\left( {cc} \right)}_i}{v_i} \in H} \)

As both properties are satisfied, the given statement is true.

A non-empty subset H of \({\mathbb{R}^n}\) if an only if \(x + y\), \(cx \in H\)for all \(x,y \in H\), \(c \in \mathbb{R}\).

Check for zero vector.

As \(A\left( 0 \right) = 0\), the zero-vector condition is satisfied.

Similarly, \(A\left( {x + y} \right) = Ax + Ay\). Therefore, vector additionis satisfied.

For \(A\left( {cx} \right) = cAx\). Therefore, scalar multiplicationis satisfied.

Thus, the statement is true.

Suppose an identity matrix of orders n and \(b\) is

\(b = \left( {\begin{array}{*{20}{c}}1\\1\\ \vdots \\1\end{array}} \right)\).

The column vectors of A are

\({e_1} = \left( {\begin{array}{*{20}{c}}1\\0\\0\\ \vdots \\0\end{array}} \right)\), \({e_2} = \left( {\begin{array}{*{20}{c}}0\\1\\0\\0\\ \vdots \\0\end{array}} \right)\), \[{e_n} = \left( {\begin{array}{*{20}{c}}0\\0\\ \vdots \\0\\1\end{array}} \right)\].

Since span \(\left\{ {{e_1},{e_2},\,....,{e_n}} \right\} = {\mathbb{R}^n}\), every vector in \({\mathbb{R}^n}\) is an element of the column space of A. The solution set of \(A{\bf{x}} = {\bf{b}}\) is \(\left\{ b \right\}\) but \({\mathbb{R}^n}\) is not \(\left\{ b \right\}\). So, the given statement is false.

The pivot column of matrix A forms a basis and so the columns of B. As the columns of A and B are the same, the pivot columns of B from a basis of Col B. But the column of echelon form are often not in the column space A.

Thus, the statement is False.