1. ## invertable and isomorphic

Can someone please explain why this isn't invertable, I'm having trouble with the definitions.

T: M_2x2(R)->P(2) defined by T a b
C d = a+2bx+(c+d)x^2. (The a b c d is supposed to be a mtrix)

And then could someone explain why the following pairs of vector spaces are isomorphic?

M_2x2(R) and P_3(R)

And why the following pairs are not isomorphic?
V={A $\in$M_2x2(R):tr(A)=0} and R^4

2. The map T : M_2x2(R) -> P(2) is not invertible, because the two vector spaces (M_2x2(R) and P(2)) don't have the same dimension. Indeed, if the map had been invertible, then the two vector spaces would have been isomorphic, and then they would have had the same dimension. But M_2x2(R) has dimension 4 and P(2) has dimension 3, whereas they are non-isomorphic.

As for M_2x2(R) and P_3(R) there is a general theorem, which you may know: Two real vector spaces are isomorphic, if and only if they have the same dimension. These two have the same dimension.

Otherwise, map

1 0
0 0

to 1+0x+0x^2+0x^3,

and map

0 1
0 0

to

0+1x+0x^2+0x^3,

and so on. This gives rise to an isomorphism.

For the last question, they are not isomorphic, because their dimensions are unequal.

3. Originally Posted by tn11631
Can someone please explain why this isn't invertable, I'm having trouble with the definitions.

T: M_2x2(R)->P(2) defined by T a b
C d = a+2bx+(c+d)x^2. (The a b c d is supposed to be a mtrix)
What happens if you try to find its inverse?
Suppose $T\left(\begin{bmatrix}a & b \\ c & d\end{bmatrix}\right)= u+ 2vx+ wx^2$. Given u, v, and w, how would you solve for a, b, c, and d? Of course, you have a= u and 2b= v, but then c+ d= w. There exist an infinite number of c and values for any w. The function is not "one-to-one" and so does not have an inverse.

And then could someone explain why the following pairs of vector spaces are isomorphic?

M_2x2(R) and P_3(R)
$M_{2x2}(R)$ is the set of 2 by 2 matrices: $\begin{bmatrix}a & b \\c & d\end{bmatrix}$ and $P_3(R)$ is the set of polynomials of degree at most 3: $m+ nx+ px^2+ qx^3$. Since each depends upon 4 numbers, try $f\left(\begin{bmatrix}a & b \\ c & d\end{bmatrix}= a+ bx+ cx^2+ dx^3$. That is the same as the mapping of "basis" vectors HappyJoe gives.

And why the following pairs are not isomorphic?
V={A $\in$M_2x2(R):tr(A)=0} and R^4
A 2 by 2 matrix contains 4 numbers. The requirement that the trace be 0 means that the two diagonal numbers must have sum 0: any such matrix is of the form $\begin{bmatrix}a & b \\ c & -a\end{bmatrix}= a\begin{bmatrix}1 & 0 \\ 0 & -1\end{bmatrix}+ b\begin{bmatrix}0 & 1 \\ 0 & 0\end{bmatrix}+ c\begin{bmatrix}0 & 0 \\ 1 & 0\end{bmatrix}$.

That is, those three matrices form a basis for V so V has dimension 3 while $R^4$, of course, has dimension 4. Since they are of different dimensions, they cannot be isomorphic.