linear algebra 16 and 9 In this exercise you will prove Theorem 3 on page 233. Let our assumptions be as in the statement of this theorem and let the notation be as in Exercise 15a where dim V = dim W = n. Assume that the nullspace of L = {0}. Explain how it follows from Exercise 15a that the nullspace of M is {0}? How does it follow that the equation L(X) = Y has at least one solution for all Y E V? Hint: First explain why M is an n x n matrix. Then apply Theorem 9 on page 146. Assume that the equation L(X) = Y has at least one solution for all Y V. Prove that then this equation has only one solution. Hint: First prove that for each B in Rn the system MX = B has at least one solution. Then apply Theorem 9 on page 146. Let V and W be finite-dimensional vector spaces with dim V = dim W. Let L : V rightarrow V be a linear transformation. Then the following statements are equivalent: The nullspace of L is {0}. The equation L(X) = Y has at least one solution for all Y V. L is invertible. Use the Solution 16...YOU NEED TO GIVE WHAT IS 15 A TO GIVE PROPER RESPONSE. 9.....LET A=[N X N ] DETERMINANT .... LET THE 2 TWO EQUAL ROWS BE ....SAY I TH . AND J TH .ROW . SO WE HAVE ORIGINAL DETERMINANT A AS GIVEN ...LET |A|= D ..........1 NOW LET US EXCHANGE I TH . AND J TH. ROWS TO GET A DETERMINANT A\'.... LET |A\'| = D\' .........................................2 SINCE THE ROWS EXCHANGED ARE IDENTICAL AS GIVEN WE SHALL HAVE A AND A\' AS IDENTICAL ......SO .... |A|=D=|A\'=D\' .................D=D\'..................................3 BUT AS PER PRPERTY OF ROW EXCHANGE WE SHALL HAVE ...... D \' = - D .............................................4 PUTTING EQN.3 IN EQN.4 WE GET....... D = -D ..............2D=0................OR ................D=0... THAT IS ...|A| = 0 .............PROVED .
linear algebra 16 and 9 In this exercise you will prove Theorem 3 on page 233. Let our assumptions be as in the statement of this theorem and let the notation be as in Exercise 15a where dim V = dim W = n. Assume that the nullspace of L = {0}. Explain how it follows from Exercise 15a that the nullspace of M is {0}? How does it follow that the equation L(X) = Y has at least one solution for all Y E V? Hint: First explain why M is an n x n matrix. Then apply Theorem 9 on page 146. Assume that the equation L(X) = Y has at least one solution for all Y V. Prove that then this equation has only one solution. Hint: First prove that for each B in Rn the system MX = B has at least one solution. Then apply Theorem 9 on page 146. Let V and W be finite-dimensional vector spaces with dim V = dim W. Let L : V rightarrow V be a linear transformation. Then the following statements are equivalent: The nullspace of L is {0}. The equation L(X) = Y has at least one solution for all Y V. L is invertible. Use the Solution 16...YOU NEED TO GIVE WHAT IS 15 A TO GIVE PROPER RESPONSE. 9.....LET A=[N X N ] DETERMINANT .... LET THE 2 TWO EQUAL ROWS BE ....SAY I TH . AND J TH .ROW . SO WE HAVE ORIGINAL DETERMINANT A AS GIVEN ...LET |A|= D ..........1 NOW LET US EXCHANGE I TH . AND J TH. ROWS TO GET A DETERMINANT A\'.... LET |A\'| = D\' .........................................2 SINCE THE ROWS EXCHANGED ARE IDENTICAL AS GIVEN WE SHALL HAVE A AND A\' AS IDENTICAL ......SO .... |A|=D=|A\'=D\' .................D=D\'..................................3 BUT AS PER PRPERTY OF ROW EXCHANGE WE SHALL HAVE ...... D \' = - D .............................................4 PUTTING EQN.3 IN EQN.4 WE GET....... D = -D ..............2D=0................OR ................D=0... THAT IS ...|A| = 0 .............PROVED .