# Thread: Is this onto or 1 to 1 ?

1. ## Is this onto or 1 to 1 ?

T: R^3 ----> W , where W is the vector space of all symmetric 2x2 matrices

T [ a b c ] =
[ a+b+c b-2c ]
[ b-2c a-c ]

I'm not sure how to show it is 1 to 1. After being transformed, I see that a = c and b = 2c and 2a = -b.

But I think I know it is not "onto" because T of R^3 cannot possibly cover all of the range of M22. A side-question on a little notation problem, is the space of 2x2 matrices M22 ?

2. [quote=crystal-maiden;285899]T: R^3 ----> W , where W is the vector space of all symmetric 2x2 matrices

T [ a b c ] =
[ a+b+c b-2c ]
[ b-2c a-c ]

I am new to abstract algebra, but I have a few thoughts/questions.

But I think I know it is not "onto" because T of R^3 cannot possibly cover all of the range of M22. A side-question on a little notation problem, is the space of 2x2 matrices M22 ?
1) A basis for the vector space of symmetric matrices is
$\left( \begin{array}{cc}
1 & 0 \\
0 & 0 \end{array} \right)$
, $\left( \begin{array}{cc}
0 & 1 \\
1 & 0 \end{array} \right)$
, $\left( \begin{array}{cc}
0 & 0 \\
0 & 1 \end{array} \right)$

Therefore it has dimension 3, so it is possible to create a 1-to-1 and onto map, T. For instance, $T([a b c]^T) = a*\left( \begin{array}{cc}
1 & 0 \\
0 & 0 \end{array} \right)+b*\left( \begin{array}{cc}
0 & 1 \\
1 & 0 \end{array} \right)+c*\left( \begin{array}{cc}
0 & 0 \\
0 & 1 \end{array} \right) = \left( \begin{array}{cc}
a & b \\
b & c \end{array} \right)$

This T will hit every symmetric matrix (onto), and every element in $R^3$ maps to a unique symmetric matrix.

I'm not sure how to show it is 1 to 1. After being transformed, I see that a = c and b = 2c and 2a = -b.
2) I do not know how you concluded your values above.

3) Consider the symmetric matrix:
$\left( \begin{array}{cc}
x & y \\
y & z \end{array} \right)$

Your transformation is going to take $[a b c]^T$ into the symmetric matrix:
$\left( \begin{array}{cc}
a+b+c & b-2c \\
b-2c & a-c \end{array} \right)$

This gives us three equations, namely:
$x = a+b+c$
$y = b-2c$
$z = a-c$
we can express this in terms of matrices
$\left( \begin{array}{ccc}
1 & 1 & 1 \\
0 & 1 & -2 \\
1 & 0 & -1\end{array} \right)\left( \begin{array}{c}
a\\
b\\
c \end{array} \right)=\left( \begin{array}{c}
x\\
y\\
z\end{array} \right)$

if you row-reduce this last matrix you will see that it has full rank.
First, this means that for any $(x,y,z)^T$ (any symmetric matrix) there is a unique solution for $(a,b,c)^T$ (an element in $R^3$). Thus, T is onto.
Second, the full rank shows that each $(a,b,c)^T$ in $R^3$ maps to a unique $(x,y,z)^T$ in the 2x2 symmetric matrix space. Thus, T is one-to-one.

Like I said, i am new to abstract algebra, I hope this at least gets you started.

3. interesting approach, that would work.
I was trying to find the kernal of the transformation such that a+b+c=0 and b-2c=0 and a-c = 0
and if I can prove the kernal is indeed zero, then I know for sure it is 1 to 1.
*edit* ooo.. I figured it out thanks so much!