# Prove isomorphism

Show 40 post(s) from this thread on one page
Page 1 of 2 12 Last
• May 17th 2010, 12:42 AM
wattkow
Prove isomorphism
I have dosed off in class way too many times and I need help with a few problems from my pre-test:

#3
Prove that M_22 is isomorphic to P_3.

A simple hint in the right direction would be greatly appreciated.

• May 17th 2010, 01:26 AM
tonio
Quote:

Originally Posted by wattkow
I have dosed off in class way too many times and I need help with a few problems from my pre-test:

#3
Prove that M_22 is isomorphic to P_3.

A simple hint in the right direction would be greatly appreciated.

Hint and very important theorem: any two vector spaces of the same dimension defined over the same field are isomorphic.

Proof: very simple and exists in every decent text book in linear algebra. Look for it.

Tonio
• May 17th 2010, 07:05 AM
Chris L T521
Quote:

Originally Posted by wattkow
I have dosed off in class way too many times and I need help with a few problems from my pre-test:

#3
Prove that M_22 is isomorphic to P_3.

A simple hint in the right direction would be greatly appreciated.

Let $A\in M_{22}$ with $A=\begin{bmatrix}a&b\\c&d\end{bmatrix}$. What is the most basic mapping $L$ we can define such that $L:M_{22}\mapsto P_3$? (What do elements look like in $P_3$?)
• May 17th 2010, 07:19 AM
wattkow
What do elements look like in $P_3$ ?
$ax^3+bx^2+cx+d$?

As you can tell from my answer, I am rather lost in class :o
• May 17th 2010, 07:25 AM
Chris L T521
Quote:

Originally Posted by wattkow
What do elements look like in $P_3$ ?
$ax^3+bx^2+cx+d$?

As you can tell from my answer, I am rather lost in class :o

Be a little more confident! (That is correct, by the way :) )

With this, how would you define the mapping $L$ such that $L:M_{22}\mapsto P_3$?
• May 17th 2010, 07:33 AM
wattkow
Quote:

Originally Posted by Chris L T521
Be a little more confident! (That is correct, by the way :) )

With this, how would you define the mapping $L$ such that $L:M_{22}\mapsto P_3$?

I'm not sure how to relate
$
A=\begin{bmatrix}a&b\\c&d\end{bmatrix}
$
with $
ax^3+bx^2+cx+d
$
(Speechless)

Is it just $
T(\begin{bmatrix}a&b\\c&d\end{bmatrix})
$
= $
ax^3+bx^2+cx+d
$
?
• May 17th 2010, 07:41 AM
Chris L T521
Quote:

Originally Posted by wattkow
I'm not sure how to relate
$
A=\begin{bmatrix}a&b\\c&d\end{bmatrix}
$
with $
ax^3+bx^2+cx+d
$
(Speechless)

That's how you define the mapping! $L:M_{22}\mapsto P_3$ where $L\left(\begin{bmatrix}a&b\\c& d\end{bmatrix}\right)=ax^3+bx^2+cx+d$.

Can you justify that this mapping is an isomorphism?
• May 17th 2010, 07:49 AM
wattkow
Quote:

Originally Posted by Chris L T521
That's how you define the mapping! $L:M_{22}\mapsto P_3$ where $L\left(\begin{bmatrix}a&b\\c& d\end{bmatrix}\right)=ax^3+bx^2+cx+d$.

Can you justify that this mapping is an isomorphism?

Somehow show that it is 1-to-1?
• May 17th 2010, 07:50 AM
Swlabr
Quote:

Originally Posted by wattkow
Somehow show that it is 1-to-1?

Find the kernel. A linear map is 1-1 if and only if the kernel is trivial.

(That is, find every element which is mapped to zero.)
• May 17th 2010, 08:04 AM
wattkow
$

L\left(\begin{bmatrix}a&b\\c& d\end{bmatrix}\right)$
$

\left(\begin{bmatrix}x\\y\\z\end{bmatrix}\right)
$
= 0?

I'm lost because I know it's not possible to multiply a 2x2 matrix by a 3x2 matrix :o
• May 17th 2010, 08:09 AM
Swlabr
Quote:

Originally Posted by wattkow
$

L\left(\begin{bmatrix}a&b\\c& d\end{bmatrix}\right)$
$

\left(\begin{bmatrix}x\\y\\z\end{bmatrix}\right)
$
= 0?

I'm lost because I know it's not possible to multiply a 2x2 matrix by a 3x2 matrix :o

This isn't what you are trying to do. You want to find the kernel of the map L, which will be a 4-by-4 matrix but you don't want to think of it that way. You need to solve,

$L(M)=0 \Rightarrow \ldots$.

This is basically asking you if $ax^3+bx^2+cx+d = 0$ then what are $a, b, c$ and $d$?
• May 17th 2010, 08:12 AM
Chris L T521
Quote:

Originally Posted by Swlabr
This isn't what you are trying to do. You want to find the kernel of the map T, which will be a 4-by-4 matrix but you don't want to think of it that way. You need to solve,

$T(M)=0 \Rightarrow \ldots$.

This is basically asking you if $ax^3+bx^2+cx+d = 0$ then what are $a, b, c$ and $d$?

You mean 2x2 matrix...
• May 17th 2010, 08:15 AM
wattkow
Quote:

Originally Posted by Swlabr
This isn't what you are trying to do. You want to find the kernel of the map T, which will be a 4-by-4 matrix but you don't want to think of it that way. You need to solve,

$T(M)=0 \Rightarrow \ldots$.

This is basically asking you if $ax^3+bx^2+cx+d = 0$ then what are $a, b, c$ and $d$?

Is it suppose to look like this:
$

L\left(\begin{bmatrix}a&b&0\\c&d&0\end{bmatrix}\ri ght)
$
?
I assume $a, b, c, d = 0$?
• May 17th 2010, 08:16 AM
Swlabr
Quote:

Originally Posted by Chris L T521
You mean 2x2 matrix...

No, 4-by-4. You are mapping from a 4-dimensional vector space into a 4-dimensional vector space...

The matrix you are mapping from is 2-by-2, and I was presuming the OP was getting confused by the linear map ~ matrix equivalence (it seemed like he was trying to find the kernel of the 2-by-2 matrix). That is why I wanted to point out that T is 4-by-4.
• May 17th 2010, 08:16 AM
Swlabr
Quote:

Originally Posted by wattkow
Is it suppose to look like this:
$

L\left(\begin{bmatrix}a&b&0\\c&d&0\end{bmatrix}\ri ght)
$
?
I assume $a, b, c, d = 0$?

Is what supposed to look like that?

(The map T in my earlier post is the same as the L in the rest of the thread. I have changed this to reflect that.)
Show 40 post(s) from this thread on one page
Page 1 of 2 12 Last