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Problem of the Week #5 - April 30th, 2012

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Chris L T521

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Jan 26, 2012
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Thanks to those who participated in last week's POTW. I'll just say it's no fun unless more people participate!

This week's problem was again proposed by yours truly (it would be nice if more people proposed some problems! (Smile)).

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Problem: Let $\phi:V\rightarrow V$ be a linear operator on a finite-dimensional vector space $V$. Let $k=\text{dim}(\ker \phi)$, and let $d=\text{dim}(V)$. Let $\phi^2:V\otimes V\rightarrow V\otimes V$ be the unique linear operator which satisfies $\phi^2(v_1\otimes v_2) = \phi(v_1)\otimes \phi(v_2)$ for all $v_1, v_2\in V$. Prove that $\text{dim}(\ker\phi^2) = 2dk-k^2$.

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I will provide some hints for this week's problem:

Apply the rank-nullity theorem. Also note that if $V$ and $W$ are finite-dimensional vector spaces and $V\otimes W$ is the tensor product of $V$ with $W$, then $\text{dim}(V\otimes W) = \text{dim}(V)\cdot\text{dim}(W)$.

Remember to read the POTW submission guidlines to find out how to submit your answers!
 
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Chris L T521

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Jan 26, 2012
995
The week 5 POTW was correctly answered Sudharaka. Here's my solution:

Proof: By the rank-nullity theory, we know that

\[\text{dim}(\ker\phi) + \text{dim}(\text{Im}\,\phi) = \text{dim}(V).\]


Thus, $\text{dim}(\text{Im}\,\phi) = \text{dim}(V) - \text{dim}(\ker\phi) = d-k$. Now, if we consider the map $\phi^2:V\otimes V\rightarrow V\otimes V$, we apply rank-nullity theorem again to obtain that


\[\text{dim}(\ker\phi^2) + \text{dim}(\text{Im}\,\phi^2) = \text{dim}(V\otimes V)\]


implying that


\[\text{dim}(\ker\phi^2) = \text{dim}(V\otimes V) - \text{dim}(\text{Im}\,\phi^2).\]


By properties of tensor product, we see that for a finite-dimensional vector space, $\text{dim}(V\otimes V) = (\text{dim}(V))^2$. Thus, we have in our case that


\[\text{dim}(\ker\phi^2) = d^2 - \text{dim}(\text{Im}\,\phi^2).\]


Now, observe that under $\phi^2$, $v_1\otimes v_2 \mapsto \phi(v_1)\otimes \phi(v_2)$. Since $\phi(v_1),\phi(v_2)\in\text{Im}\,\phi$, we have $\text{Im}(\phi^2)\subseteq\text{Im}(\phi)\otimes \text{Im}(\phi)$. In a similar fashion, one can show that $\text{Im}(\phi)\otimes \text{Im}(\phi)\subseteq\text{Im}(\phi^2)$, implying that $\text{Im}(\phi)\otimes \text{Im}(\phi)=\text{Im}(\phi^2)$.


Thus, $\text{dim}(\text{Im}\,\phi^2) = (\text{dim}(\text{Im},\phi))^2=(d-k)^2$. Therefore, $\text{dim}(\ker\phi^2)= d^2 - (d-k)^2 = (d-(d-k))(d+(d-k)) = k(2d-k)=2dk-k^2$. Q.E.D.

and here's Sudharaka's solution:

By the Rank-Nullity Theorem,

\[\text{dim}(V\otimes V)=\text{dim}(\ker\phi^2)+\text{dim}(\text{im }\phi^2)\]


Since, \(\text{dim}(V\otimes V) = \text{dim}(V)\cdot\text{dim}(V)=d^2\)


\[\text{dim}(\ker\phi^2)=d^2-\text{dim}(\text{im }\phi^2)~~~~~~~~~~(1)\]


\[\text{im }\phi^2=\{\phi^2(v_1\otimes v_2)~|~v_1\otimes v_2\in V\otimes V\}\]


\[\Rightarrow \text{im }\phi^2=\{\phi(v_1)\otimes\phi(v_2)~|~v_1\otimes v_2\in V\otimes V\}=\phi(V)\otimes\phi(V)=(\text{im }\phi)\otimes(\text{im }\phi)\]


\[\Rightarrow \text{dim}(\text{im }\phi^2)=\{\text{dim}(\text{im }\phi)\}^2~~~~~~~~(2)\]


Considering the linear operator \(\phi\) and using the Rank-Nullity theorem we get,


\[\text{dim}(V)=\text{dim}(\ker\phi)+\text{dim} (\text{im }\phi)\]


\[\Rightarrow \text{dim} (\text{im }\phi)=d-k\]


\[\Rightarrow \{\text{dim} (\text{im }\phi)\}^2=(d-k)^2~~~~~~~~~(3)\]


By (2) and (3),


\[\text{dim}(\text{im }\phi^2)=(d-k)^2\]


Now by (1),


\[\text{dim}(\ker\phi^2)=d^2-(d-k)^2=2dk-k^2\]
 
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