# Using the rank-nullity theorem in a proof

• Nov 29th 2009, 08:59 AM
kevinlightman
Using the rank-nullity theorem in a proof
I'm stuck on this proof, can anyone help?

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• Nov 29th 2009, 06:54 PM
HallsofIvy
Quote:

Originally Posted by kevinlightman
I'm stuck on this proof, can anyone help?

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The first is pretty straight forward. In general, to prove "\$\displaystyle A\subset B\$" you show: "if y is a member of A then y is a member or B".

Here you want to prove: \$\displaystyle Im(S+ T)\subset Im(S)+ Im(T)\$
If y is in Im(S+ T) then y= (S+ T)x for some x in V. But "(S+ T)x" is defined as Sx+ Tx so we have y= Sx+ Tx. Let \$\displaystyle y_1= Sx\$ and \$\displaystyle y_2= Tx\$. Then \$\displaystyle y_1\$ is in Im(S), \$\displaystyle y_2\$ is in Im(T), and \$\displaystyle y= y_1+ y_2\$. Therefore, y is in Im(S)+ Im(T). The rest of that should be straight forward.

The second may be even simpler. Saying "ST= 0" mean (ST)v= 0 for all v. But that is the same as S(Tv)= 0 so that Tv is in the kernel of S for all v.