## UsukiDoll Group Title Can anyone help me with this proof? Am I on the right track? Let A and B be equivalent square matrices. Prove that A is nonsingular if and only if B is nonsingular one year ago one year ago

1. UsukiDoll Group Title

Given: Matrix A and Matrix B are equivalent square matrices. Since we are given a biconditional statement, we need to prove two situations: If Matrix A is nonsingular then Matrix B is nonsingular If Matrix B is nonsingular then Matrix A is nonsingular

2. UsukiDoll Group Title

Let's see.... Theorem 2.13 states that two m x n matrices A and B are equivalent of and only if B = PAQ for some nonsingular matrices P and Q.

3. UsukiDoll Group Title

We have a product of nonsingular matrices P and Q. If B is nonsingular, B=PAQ, then A is nonsingular A=P^-1BQ^-1

4. UsukiDoll Group Title

|dw:1361430166380:dw|

5. UsukiDoll Group Title

Conversely, if A is nonsingular, A = PBQ, then B is nonsingular, B=P^-1AQ^-1 because P and Q are products of nonsingular matrices.

6. UsukiDoll Group Title

|dw:1361430248917:dw|

7. UsukiDoll Group Title

|dw:1361430265828:dw|

8. UsukiDoll Group Title

@UnkleRhaukus

9. UsukiDoll Group Title

?????????????

10. UnkleRhaukus Group Title

it looks like your on the right track, but i can not remember this stuff

11. UsukiDoll Group Title

@joemath314159 and @hartnn

12. UsukiDoll Group Title

@zepdrix

13. UsukiDoll Group Title

thank goodness! I was searching everywhere for someone to help me see if I'm on the right track.

14. zepdrix Group Title

I'm afraid this might be over my head :c

15. UsukiDoll Group Title

16. UsukiDoll Group Title

: (

17. UsukiDoll Group Title

have you done proof writing before or no?

18. zepdrix Group Title

mmmm not so much... and i don't have a very good grasp on matrices yet i'm afraid :P darn.

19. UsukiDoll Group Title

20. zepdrix Group Title

Poor lil pony :o

21. UsukiDoll Group Title

crud. This thing is like due in 11-12 days! I like to plan ahead and such. wahhh

22. UsukiDoll Group Title

@KingGeorge do you know how to write proofs in Linear Algebra? I've done some and I was wondering if you could tell me if I'm on the right track.

23. KingGeorge Group Title

This seems to be on the right track to me.

24. UsukiDoll Group Title

ok so what do I need to write next for this proof or is that it?

25. KingGeorge Group Title

have you learned that a matrix A is nonsingular if and only if it's invertible?

26. UsukiDoll Group Title

invertible?

27. UsukiDoll Group Title

oh that! yes square matrix and the AB=Ba=In. yeah I dealt with that before.

28. UsukiDoll Group Title

if a matrix is singular, there's no inverse, so it's impossible for AB=BA=In to work

29. KingGeorge Group Title

Alright. Suppose A is nonsingular, and $$A=PBQ$$ for some non-singular matrices P,Q. This means that A, P, Q are invertible. Then $B=P^{-1}AQ^{-1}$But since $$A^{-1}$$ exists, look at the product$P^{-1}AQ^{-1}QA^{-1}P =BQA^{-1}P=\text{Id}$So B has an inverse and must be non-singular. This is how I would do it. Just switch A and B to prove the other statement you need.

30. UsukiDoll Group Title

ok the switching is a bit nerve wrecking. omg... that's a lot of letters.

31. KingGeorge Group Title

Just work through it slowly. You had most of it yourself. You're only missing the last couple lines.

32. UsukiDoll Group Title

ok let me try ummm... B is nonsingular and B =PAQ for some non-singular matrices P,Q. This means that B, P, and Q are invertiable

33. KingGeorge Group Title

right.

34. UsukiDoll Group Title

ok so that means that |dw:1361502760161:dw|

35. UsukiDoll Group Title

ok so B inverse must exist

36. KingGeorge Group Title

You already know that B inverse exists. You need to show that A inverse exists.

37. KingGeorge Group Title

What can you multiply $P^{-1}BQ^{-1}$by to get the identity?

38. UsukiDoll Group Title

to get if B is nonsingular A is nonsingular. |dw:1361503034980:dw|

39. KingGeorge Group Title

I'm not understanding what you're trying to do there. Would you mind elaborating?

40. UsukiDoll Group Title

I'm trying to prove the other statement. I know there's two statements If Matrix A is nonsingular then Matrix B is nonsingular If Matrix B is nonsingular then Matrix A is nonsingular

41. UsukiDoll Group Title

Theorem 2.13 states that two m x n matrices A and B are equivalent of and only if B = PAQ for some nonsingular matrices P and Q.

42. UsukiDoll Group Title

so much letters going oN!

43. KingGeorge Group Title

So can you find some product of matrices such that $$P^{-1}BQ^{-1}$$ multiplied by that product of matrices gives you the identity?

44. UsukiDoll Group Title

hmmm... B = PAQ A = P^-1BQ^-1

45. UsukiDoll Group Title

(PAQ)(P^-1BQB^-1) = (P^-1BQB^-1)(PAQ)

46. UsukiDoll Group Title

:/

47. KingGeorge Group Title

I think you're on the right track. You want $P^{-1}BQ^{-1}X=\text{Id}$though. So $$BQ^{-1}X=P$$ and $$Q^{-1}X=B^{-1}P$$. If you continue this one more step, what do you get to be for $$X$$?

48. KingGeorge Group Title

Did that make sense?

49. UsukiDoll Group Title

omg no. I'm lost... :/

50. UsukiDoll Group Title

ok wait so A = P^-1BQ^-1 that must be Ax = Id

51. UsukiDoll Group Title

o.o wahhh this is torture.

52. KingGeorge Group Title

That's the right idea. We're finding an $$X$$ such that $$AX=\text{Id}$$. So if $AX=P^{-1}BQ^{-1}X=\text{Id},$you need to find $$X$$ in terms of $$P,Q,$$ and $$B$$ (and/or their inverses). Make more sense?

53. UsukiDoll Group Title

yeah more work and torture D:

54. UsukiDoll Group Title

but how do I find x in terms of P? Won't that be P = whatever it is X.

55. UsukiDoll Group Title

|dw:1361504529420:dw|

56. KingGeorge Group Title

That's what I was going to suggest next actually.

57. UsukiDoll Group Title

|dw:1361504588920:dw|

58. UsukiDoll Group Title

|dw:1361504640692:dw|

59. UsukiDoll Group Title

|dw:1361504663129:dw|

60. UsukiDoll Group Title

|dw:1361504710718:dw|

61. KingGeorge Group Title

Hold up. Look at it as an algebraic equation like this where you solve for x.$\frac{1}{p}b\frac{1}{q}x=1$

62. UsukiDoll Group Title

|dw:1361504883640:dw|

63. UsukiDoll Group Title

|dw:1361504957573:dw|

64. KingGeorge Group Title

Sorry, internet glitched out on me there. But yes. That right. Converting that back into matrix form, you get $X=PB^{-1}Q$That make sense?

65. UsukiDoll Group Title

yeah sorry...I feel a bit weak...need to eat dinner

66. KingGeorge Group Title

No worries. I'll probably be gone when you get back though. Good luck!

67. UsukiDoll Group Title

@KingGeorge are you going to be on openstudy tomorrow? I had a busy day today and feel tired.

68. UsukiDoll Group Title

Oh well can I email it to you tomorrow?

69. UsukiDoll Group Title

@ArkGoLucky

70. UsukiDoll Group Title

|dw:1361519596240:dw|

71. UsukiDoll Group Title

|dw:1361519710516:dw|

72. UsukiDoll Group Title

|dw:1361519769721:dw|

73. UsukiDoll Group Title

@AriPotta

74. UsukiDoll Group Title

ughhhh so close!!! X( I'm trying to prove the second statement

75. ArkGoLucky Group Title

Sorry don't know this stuff

76. UsukiDoll Group Title

Ok. I may have got it. Suppose B is nonsingular and B=PAQ for some non-singular matrices P and Q. That means B, P, and Q are invertible. |dw:1361526378870:dw| |dw:1361526405075:dw| |dw:1361526433824:dw| |dw:1361526465398:dw| Since B^-1 exist we look at the product of nonsingular matrices. |dw:1361526502682:dw| Therefore A has an inverse and it must be nonsingular

77. UsukiDoll Group Title

@KingGeorge I may have gotten it

78. UsukiDoll Group Title

@wio

79. UsukiDoll Group Title

can you please check if I got this right?

80. wio Group Title

Hold on.

81. experimentX Group Title

don't divide by matrices, division is not defined. Inverses and division are different.

82. UsukiDoll Group Title

ok so how do I prove the converse of this

83. UsukiDoll Group Title

Given: Matrix A and Matrix B are equivalent square matrices. Since we are given a biconditional statement, we need to prove two situations: If Matrix A is nonsingular then Matrix B is nonsingular If Matrix B is nonsingular then Matrix A is nonsingular

84. wio Group Title

The converse of what statement?

85. UsukiDoll Group Title

If Matrix A is nonsingular then Matrix B is nonsingular If Matrix B is nonsingular then Matrix A is nonsingular

86. UsukiDoll Group Title

these are the two situations

87. UsukiDoll Group Title

King George and I did the first one yesterday

88. wio Group Title

Okay, what is meant by A and B are equivalent?

89. UsukiDoll Group Title

just the second one. I attempted through the drawings last night

90. experimentX Group Title

use the same argument as you used on the first one, P and Q are invertible by definition.http://en.wikipedia.org/wiki/Matrix_equivalence

91. toxicsugar22 Group Title

Hi can you help me after you help this person please

92. UsukiDoll Group Title

but the letters are different on the second

93. UsukiDoll Group Title

I drew the attempt on it last night

94. UsukiDoll Group Title

I'm not sure if that's it or something

95. UsukiDoll Group Title

no and I can't use fancy stuff that my class didn't learn. I already got chewed once by the professor

96. UsukiDoll Group Title

like he will give points but taunt you in the comments

97. UsukiDoll Group Title

Theorem 2.13 states that two m x n matrices A and B are equivalent of and only if B = PAQ for some nonsingular matrices P and Q.

98. UsukiDoll Group Title

We have a product of nonsingular matrices P and Q. If B is nonsingular, B=PAQ, then A is nonsingular A=P^-1BQ^-1

99. UsukiDoll Group Title

Conversely, if A is nonsingular, A = PBQ, then B is nonsingular, B=P^-1AQ^-1 because P and Q are products of nonsingular matrices.

100. experimentX Group Title

I think you earlier asked the question AB is invertible iff both A and B are invertible. Isn't this same as proving it?

101. UsukiDoll Group Title

the second part. the one with conversely..... I don't know if I got it right even though it says to flip the letters A and B

102. UsukiDoll Group Title

? ? ? I think that's a different question that involves something different.

103. experimentX Group Title

this is same, B = PAQ what did you prove above and what do you want to prove next? write in short.

104. UsukiDoll Group Title

THere's two situations to this: If Matrix A is nonsingular then Matrix B is nonsingular If Matrix B is nonsingular then Matrix A is nonsingular

105. UsukiDoll Group Title

I want to prove through the usage of Theorem 2. 13 whatever I posted previously

106. UsukiDoll Group Title

I already have one done and I attempted the second one (it's drawn)

107. experimentX Group Title

$B = PAQ => P^{-1}BQ^{-1}=A$ Let $$P^{-1}=P'$$ and $$Q^{-1}=Q'$$ isn't this same as before?

108. UsukiDoll Group Title

THat's if B is nonsingular then A is nonsingular since we have the product of nonsingular matrices P and Q

109. experimentX Group Title

sorry, I didn't get you ... where do you have problem understanding it?

110. wio Group Title

If matrix equivalence is an equivalence relation, then it follows the transitive property.

111. experimentX Group Title

you don't have to go through the rigorous way like you did the first one, Both P^-1 and Q^-1 are invertible matrices. If you take P and Q on the side of B, you will get same expression as you did on the first one.

112. UsukiDoll Group Title

really? so A=PBQ then B=P^-1AQ^-1

113. experimentX Group Title

yes P and Q are invertible matrices by definition of Equivalent Matrices. So their inverses are also invertible matrices making the the relation symmetric.

114. UsukiDoll Group Title

omg I've been freaking out and stressing for nothing then! -________-

115. UsukiDoll Group Title

but ok I should put that for if A is nonsingular then B is nonsingular

116. experimentX Group Title

No ... use symmetry argument. A = PBQ B = P'AQ' <-- this is (similar) to first.

117. zzr0ck3r Group Title

you can do this with elementry matricies.

118. UsukiDoll Group Title

symmetry argument?

119. experimentX Group Title

Theorem 1: A = PBQ if B, P, and Q is non invertible then A is non invertible. B = P'AQ',P', Q' is non invertible, then B is non invertible from theorem 1. Guess not symmetry argument.

120. experimentX Group Title

write it up like that.

121. UsukiDoll Group Title

what about... A = PBQ. IF B, P, and Q are non invertible, then A is non invertible. B = P^-1AQ^-1. If P^-1 and Q^-1 are non invertible then B is non invertible.

122. experimentX Group Title

sure that works too.

123. UsukiDoll Group Title

so. If B=PAQ. If A, P, and Q are non invertible then B is non invertible A = P^-1BQ^-1. If P^-1 and Q^-1 are non invertible then A is non invertible

124. experimentX Group Title

yes

125. UsukiDoll Group Title

:D thank you