Help with Answer Checking: Can you check my answers and help me?

  • Thread starter Ted123
  • Start date
In summary, the conversation involves solving multiple mathematical problems with various images and notations. The answers to each question are given and discussed, with clarification needed for some questions. The final question involves finding the dimension of the image of a given map, which is found to be 9.
  • #1
Ted123
446
0
Can anyone check my answers/help me?

1
[PLAIN]http://img689.imageshack.us/img689/3912/59048711.jpg

My answer: 3

2
[PLAIN]http://img171.imageshack.us/img171/812/10085212.jpg

My answer: b,c

3
[PLAIN]http://img252.imageshack.us/img252/7706/81927762.jpg

My answer: False

4
[PLAIN]http://img573.imageshack.us/img573/5450/70607877.jpg

My answer: a

5
[PLAIN]http://img51.imageshack.us/img51/6527/12780304.jpg

My answer: ?

6
[PLAIN]http://img715.imageshack.us/img715/8680/30828400.jpg

My answer: ?

7
[PLAIN]http://img251.imageshack.us/img251/1006/48454960.jpg

My answer: a,e and ...?
 
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  • #2
1,2,3 are correct.

Your answer for 4 is partially correct: there are two statements that are definitely true, not one.

For 5, use this theorem:

dim(V) = dim(kernel(phi)) + dim(image(phi))

For 6, I am not sure what this notation means:

[tex]\mathbb{R}^1 0[/tex]

Can you please clarify?

For 7, a) and e) are correct, but there is one more. What is the dimension of the quotient space V/U?
 
  • #3
jbunniii said:
1,2,3 are correct.

Your answer for 4 is partially correct: there are two statements that are definitely true, not one.

For 5, use this theorem:

dim(V) = dim(kernel(phi)) + dim(image(phi))

For 6, I am not sure what this notation means:

[tex]\mathbb{R}^1 0[/tex]

Can you please clarify?

For 7, a) and e) are correct, but there is one more. What is the dimension of the quotient space V/U?

For 4 is c) correct too?

And for 5 how do I find the dimension of image([itex]\phi[/itex]) ?

In question 6, I don't know what that notation means either, so I've sent an email and queried it.

For 7, d) is correct too (12 - 3 = 9)
 
  • #4
Ted123 said:
For 4 is c) correct too?

Yes. Can you explain why?

And for 5 how do I find the dimension of image([itex]\phi[/itex]) ?

This is given, implicitly. The map is a surjection (onto). So image([itex]\phi[/itex]) = ?

For 7, d) is correct too (12 - 3 = 9)

Right.
 
  • #5
jbunniii said:
Yes. Can you explain why?

For [itex]\phi[/itex] to be an isomorphism it must be one-to-one which implies the kernel is trivial.

jbunniii said:
This is given, implicitly. The map is a surjection (onto). So image([itex]\phi[/itex]) = ?

[itex]\text{image}(\phi ) = W[/itex] and [itex]\text{dim}(W)=4[/itex]

so [itex]\text{dim}[\text{kernel}(\phi )] = 8-4=4[/itex]

I've had an answer back for question 6 and apparently this should read [itex]\phi : \mathbb{R}^{10} \to \mathcal{C}[0,1][/itex] (ie. [itex]\mathbb{R}^{10}[/itex] is the space of real column vectors of length 10).
 
Last edited:
  • #6
Yes, those answers are correct.

OK, so for 6, you can again use

dim(V) = dim(kernel(phi)) + dim(image(phi))

Hint: the target space is irrelevant.
 
  • #7
jbunniii said:
Yes, those answers are correct.

OK, so for 6, you can again use

dim(V) = dim(kernel(phi)) + dim(image(phi))

Hint: the target space is irrelevant.

[itex]\text{dim}[\text{image}(\phi)] = 10-1 = 9[/itex]
 
  • #8
Ted123 said:
[itex]\text{dim}[\text{image}(\phi)] = 10-1 = 9[/itex]

Correct.
 

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