## UsukiDoll 3 years ago 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

1. UsukiDoll

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

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

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

|dw:1361430166380:dw|

5. UsukiDoll

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

|dw:1361430248917:dw|

7. UsukiDoll

|dw:1361430265828:dw|

8. UsukiDoll

@UnkleRhaukus

9. UsukiDoll

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

10. UnkleRhaukus

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

11. UsukiDoll

@joemath314159 and @hartnn

12. UsukiDoll

@zepdrix

13. UsukiDoll

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

14. zepdrix

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

15. UsukiDoll

16. UsukiDoll

: (

17. UsukiDoll

have you done proof writing before or no?

18. zepdrix

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

19. UsukiDoll

20. zepdrix

Poor lil pony :o

21. UsukiDoll

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

22. UsukiDoll

@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

This seems to be on the right track to me.

24. UsukiDoll

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

25. KingGeorge

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

26. UsukiDoll

invertible?

27. UsukiDoll

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

28. UsukiDoll

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

29. KingGeorge

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

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

31. KingGeorge

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

32. UsukiDoll

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

right.

34. UsukiDoll

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

35. UsukiDoll

ok so B inverse must exist

36. KingGeorge

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

37. KingGeorge

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

38. UsukiDoll

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

39. KingGeorge

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

40. UsukiDoll

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

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

so much letters going oN!

43. KingGeorge

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

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

45. UsukiDoll

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

46. UsukiDoll

:/

47. KingGeorge

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

Did that make sense?

49. UsukiDoll

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

50. UsukiDoll

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

51. UsukiDoll

o.o wahhh this is torture.

52. KingGeorge

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

yeah more work and torture D:

54. UsukiDoll

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

55. UsukiDoll

|dw:1361504529420:dw|

56. KingGeorge

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

57. UsukiDoll

|dw:1361504588920:dw|

58. UsukiDoll

|dw:1361504640692:dw|

59. UsukiDoll

|dw:1361504663129:dw|

60. UsukiDoll

|dw:1361504710718:dw|

61. KingGeorge

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

|dw:1361504883640:dw|

63. UsukiDoll

|dw:1361504957573:dw|

64. KingGeorge

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

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

66. KingGeorge

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

67. UsukiDoll

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

68. UsukiDoll

Oh well can I email it to you tomorrow?

69. UsukiDoll

@ArkGoLucky

70. UsukiDoll

|dw:1361519596240:dw|

71. UsukiDoll

|dw:1361519710516:dw|

72. UsukiDoll

|dw:1361519769721:dw|

73. UsukiDoll

@AriPotta

74. UsukiDoll

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

75. anonymous

Sorry don't know this stuff

76. UsukiDoll

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

@KingGeorge I may have gotten it

78. UsukiDoll

@wio

79. UsukiDoll

can you please check if I got this right?

80. anonymous

Hold on.

81. experimentX

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

82. UsukiDoll

ok so how do I prove the converse of this

83. UsukiDoll

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. anonymous

The converse of what statement?

85. UsukiDoll

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

86. UsukiDoll

these are the two situations

87. UsukiDoll

King George and I did the first one yesterday

88. anonymous

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

89. UsukiDoll

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

90. experimentX

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

Hi can you help me after you help this person please

92. UsukiDoll

but the letters are different on the second

93. UsukiDoll

I drew the attempt on it last night

94. UsukiDoll

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

95. UsukiDoll

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

96. UsukiDoll

like he will give points but taunt you in the comments

97. UsukiDoll

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

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

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

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

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

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

103. experimentX

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

104. UsukiDoll

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

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

106. UsukiDoll

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

107. experimentX

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

108. UsukiDoll

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

109. experimentX

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

110. anonymous

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

111. experimentX

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

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

113. experimentX

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

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

115. UsukiDoll

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

116. experimentX

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

117. zzr0ck3r

you can do this with elementry matricies.

118. UsukiDoll

symmetry argument?

119. experimentX

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

write it up like that.

121. UsukiDoll

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

sure that works too.

123. UsukiDoll

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

yes

125. UsukiDoll

:D thank you