Prove that TU = T0

Nona · #1 $Old$ November 5th, 2009, 12:04 PM

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

tonio · #2 $Old$ November 5th, 2009, 12:44 PM

Quote:

Originally Posted by Nona $View Post$

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 $V = N(T)+N(U)$ then $TU=T_0$ .

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

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

Nona · #3 $Old$ November 5th, 2009, 01:11 PM

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