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- Feb 5, 2012

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Here's a problem that I want to confirm my answer. Note that for the second part of the question it states, "prove that \(T\) is bonded by the above claim". I used a different method and couldn't find a method that relates the first part to prove the second.

**Problem:**

Suppose \(X\) is a n-dimensional linear vector space. Prove that any linear operator \(T\) on \(X\) is uniquely determined by \(\{Tx_i\}_{i=1}^{n}\) with \(\{x_i\}_{i=1}^{n}\) a basis for \(X\). Moreover, prove that \(T\) is bounded by the above claim.

**My Ideas:**

Suppose there are two representations of \(Tv\) where \(v\in V\). That is,

\[Tv=a_1 Tx_1+\cdots+a_n Tx_n=b_1 Tx_1+\cdots+b_n Tx_n\]

\[(a_1 Tx_1+\cdots+a_n Tx_n)-(b_1 Tx_1+\cdots+b_n Tx_n)=0\]

Since \(T\) is linear,

\[T((a_1-b_1) x_1+(a_2-b_2) x_2+\cdots+(a_n-b_n) x_n)=0\]

\[(a_1-b_1) x_1+(a_2-b_2) x_2+\cdots+(a_n-b_n) x_n=0\]

Since \(\{x_i\}_{i=1}^{n}\) is linearly independent,

\[a_i=b_i\mbox{ for all }i=1,\,2,\,\cdots,\,n\]

That is, \(T\) is uniquely determined by \(\{Tx\}_{i=1}^{n}\).

Now we shall show that \(T\) is bounded. Since \(X\) is n-dimensional, \(X\) is topologically isomorphic. That is there exist two positive constants \(c_1\) and \(c_2\) such that,

\[c_1\|x\|\leq \|Tx\|\leq c_2 \|x\|\]

for all \(x\in X\). Hence it's obvious that \(T\) is bounded.