# Prove that TU = T0

• Nov 5th 2009, 10:04 AM
Nona
Prove that TU = T0
Hello,
I would like help in:
Assume T and U are in L(V,V ) and that V = N(T)+N(U). Prove that TU = T0 . (T0 is the zero linear transformation)
Thank you
• Nov 5th 2009, 10:44 AM
tonio
Quote:

Originally Posted by Nona
Hello,
I would like help in:
Assume T and U are in L(V,V ) and that V = N(T)+N(U). Prove that TU = T0 . (T0 is the zero linear transformation)
Thank you

Let's see if I succeed in decoding the above: T,U are linear operators and N(T), N(U) are the corresponding null spaces, or kernel, of these operators.

So, if $\displaystyle V = N(T)+N(U)$ then $\displaystyle TU=T_0$.

Assuming I guessed correctly your symbols, the claim is false: as example take $\displaystyle T,\,U\in L(\mathbb{R}^2,\mathbb{R}^2)$ , $\displaystyle T\left(\begin{array}{c}x\\y\end{array}\right)=\lef t(\begin{array}{c}x-y\\x-y\end{array}\right)\,,\,\,U\left(\begin{array}{c}x \\y\end{array}\right)=\left(\begin{array}{c}0\\y\e nd{array}\right)$

It's easy to check that $\displaystyle \left(\begin{array}{c}a\\b\end{array}\right)=\left (\begin{array}{c}b\\b\end{array}\right)+\left(\beg in{array}{c}a-b\\0\end{array}\right)\in\,N(T)+N(U)$ , but $\displaystyle TU\neq T_0$ , as you can easily check.

If by the above symbols you meant something else then the above is worthless (perhaps the above's worthless EVEN if I guessed correctly your symbols).

Tonio
• Nov 5th 2009, 11:11 AM
Nona
Yes, you guessed write. And this is the question that I have to prove that TU=T_0.
I wonder if Prof. quotation is to prove TU not equal T_0. maybe typo
Thank you very much