Show that if B is obtained from A by performing one or more row operations, then R(B) = R(A).
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Textbook Solutions for Linear Algebra: A Geometric Approach
Question
Given each matrix A, find matrices X and Y so that C(A) = N(X) and N(A) = C(Y).
a. \(A=\left[\begin{array}{rr} 3 & -1 \\6 & -2 \\ -9 & 3 \end{array}\right]\)
b. \(A=\left[\begin{array}{rrr} 1 & 1 & 0 \\ 2 & 1 & 1 \\ 1 & -1 & 2 \end{array}\right]\)
c. \(A=\left[\begin{array}{lll} 1 & 1 & 1 \\ 1 & 2 & 0 \\ 1 & 1 & 1 \\ 1 & 0 & 2 \end{array}\right]\)
Solution
Step 1 of 3
(a)
Let \(A= \left[ {\begin{array}{*{20}{c}}3&{ - 1}\\6&{ - 2}\\{ - 9}&3\end{array}} \right]\)
By applying elementary column operations
\(\left[ {\begin{array}{*{20}{c}}3&{ - 1}\\6&{ - 2}\\{ - 9}&3\end{array}} \right]\)
\({c_2} \Leftrightarrow {c_1}\)
\(\left[ {\begin{array}{*{20}{c}}{ - 1}&3\\{ - 2}&6\\3&{ - 9}\end{array}} \right]\)
\({c_2} \to {c_2} + 3{c_1}\)
\(\left[ {\begin{array}{*{20}{c}}{ - 1}&0\\{ - 2}&0\\3&0\end{array}} \right]\)
Since A has only one Linearly Independent column, basis of C(A) is
\(\left\{ {\left[ {\begin{array}{*{20}{c}}{ - 1}\\{ - 2}\\3\end{array}} \right]} \right\}\)
Thus \(C(A) = 1 \to (a)\)
Now we have to find Matrix X such that,
\(X\left[ {\begin{array}{*{20}{c}}{ - 1}\\{ - 2}\\3\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}0\\0\\0\end{array}} \right]\)
Choosing \(X = \left[ {\begin{array}{*{20}{c}}1&1&1\\{ - 2}&{ - 2}&{ - 2}\\0&0&0\end{array}} \right]\) which having rank(X)=2 so that null-space of \(X = 3 - 2 = 1\)
Thus \(N(X) = 1 \to (b)\)
\(\left[ {\begin{array}{*{20}{c}}1&1&1\\{ - 2}&{ - 2}&{ - 2}\\0&0&0\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{ - 1}\\{ - 2}\\3\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}0\\0\\0\end{array}} \right]\)
From (a) and (b)
Thus \(C(A) = N(X)\)
Let N(A) be the Null-space of A and \(\left[ {\begin{array}{*{20}{c}}a\\b\end{array}} \right] \in N(A)\)
\(\left[ {\begin{array}{*{20}{c}}3&{ - 1}\\6&{ - 2}\\{ - 9}&3\end{array}} \right]\left[ {\begin{array}{*{20}{c}}a\\b\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}0\\0\\0\end{array}} \right]\)
\( \to 3a - b = 0 \to (1)\)
\( \to 6a - 2b = 0 \to (2)\)
\( \to - 9a + 3b = 0 \to (3)\)
from (1)
\(3a = b\)
\(\left[ {\begin{array}{*{20}{c}}a\\b\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}a\\{3a}\end{array}} \right] = a\left[ {\begin{array}{*{20}{c}}1\\3\end{array}} \right]\)
Basis for \(N(A) = \left[ {\begin{array}{*{20}{c}}1\\3\end{array}} \right]\)
Now we have to find matrix Y such that basis of column space of Y is \(\left\{ {\left[ {\begin{array}{*{20}{c}}1\\3\end{array}} \right]} \right\}\)
Let \(Y = \left[ {\begin{array}{*{20}{c}}1&2\\3&6\end{array}} \right]\) having column 2 of Y is scalar multiple of column 1\( \to \)No: of linearly independent columns of Y is 1.
Rank(Y) =1
By elementary column operations column space of \(Y = \left\{ {\left[ {\begin{array}{*{20}{c}}1\\3\end{array}} \right]} \right\}\)
Hence \(N(A) = C(Y)\)
full solution