Given each matrix A, find matrices X and Y so that C(A) = N(X) and N(A) = C(Y ). a. A = | StudySoup

Textbook Solutions for Linear Algebra: A Geometric Approach

Chapter 3.2 Problem 3

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 \(N(A) = C(Y)\)

Subscribe to view the
full solution

Title Linear Algebra: A Geometric Approach 2 
Author Ted Shifrin, Malcolm Adams
ISBN 9781429215213

Given each matrix A, find matrices X and Y so that C(A) = N(X) and N(A) = C(Y ). a. A =

Chapter 3.2 textbook questions

×

Login

Organize all study tools for free

Or continue with
×

Register

Sign up for access to all content on our site!

Or continue with

Or login if you already have an account

×

Reset password

If you have an active account we’ll send you an e-mail for password recovery

Or login if you have your password back