11.29 Isomorphism

1. Let \(V\) and \(W\) be \(n\)-dimensional vector spaces and \(T: V \rightarrow W\) linear map. Suppose \(\{v_1, \ldots, v_n\}\) is a basis for \(V\). Prove that \(T\) is an isomorphism \(\iff \{T(v_1), \ldots, T(v_n)\}\) is a basis for \(W\).

   We will do this in 2 steps:

   • (i) \(T\) is injective \(\iff \{T(v_1), \ldots, T(v_n)\}\) is L.I.

   • (ii) \(T\) is surjective \(\iff \{T(v_1), \ldots, T(v_n)\}\) spans \(W\).

   • (i) We have seen this before but the basic idea is:

   \(T\) is injective \(\iff \text{Null}(T) = \{0\}\)​\(\iff\) if \(T(v) = 0\) then \(v = 0\) for all \(v \in V\)

   \(\iff\) if \(T(\lambda_1v_1+\lambda_2v_2+\ldots+\lambda_{n}v_{n})=0\) then \(\lambda_1 = \lambda_2 = \ldots = \lambda_n = 0\)​\(\iff\) if \(T(v_1),\ldots,T(v_{n})\) is L.I. * (ii) \(\Rightarrow\))

   Suppose \(T\) is surjective. If \(w \in W\), \(\exists v \in V\) such that \(T(v) = w\).

   Since \(\{v_1, \ldots, v_n\}\) is a basis for \(V\), we can write \(v = \lambda_1 v_1 + \lambda_2 v_2 + \ldots + \lambda_n v_n\).

   So \(w = T(v) = T(\lambda_1 v_1 + \lambda_2 v_2 + \ldots + \lambda_n v_n)\).

   \(\Rightarrow \exists \lambda_1, \ldots, \lambda_n \in F\) such that \(w=\lambda_1T(v_1)+\lambda_2T(v_2)+\ldots+\lambda_{n}T(v_{n})\).

   So \(\{T(v_1), \ldots, T(v_n)\}\) spans \(W\).

   \(\Leftarrow\)) Let \(w\in W\), since \(\{T(v_1), \ldots, T(v_n)\}\) spans \(W\), \(\exists \lambda_1, \ldots, \lambda_n \in F\) such that \(w=\lambda_1T(v_1)+\lambda_2T(v_2)+\ldots+\lambda_{n}T(v_{n})=T\left(\lambda_1v_1\right)+T(\lambda_2v_2)+\ldots+T\left(\lambda_{n}v_{n}\right)\).

   So if we take \(v = \lambda_1 v_1 + \lambda_2 v_2 + \ldots + \lambda_n v_n \in V\), \(T(v) = w\).

   Since \(w\) was arbitrary, \(T\) is surjective.

2. Suppose \(V\) and \(W\) are finite dimensional.

   Let \(n = \dim V\) and define \(E_v = \{T \in L(V, W) : T(v) = 0\}\).

   1. Show that \(E_v\) is a subspace of \(L(V, W)\)?

      \(T,S\in E_{v}\Rightarrow T(v)=0\) and \(S(v) = 0\).

      \((T+S)(v)=T(v)+S(v)=0+0\Rightarrow T+S\in E_{v}\).

      If \(T\in E_{v},\lambda\in F\) \(\Rightarrow\left(\lambda T)(v\right)=\lambda T(v)=0\) \(\Rightarrow\lambda T\in E_{v}\)

      So \(E_v\) is a subspace of \(L(V,W)\). 2. Suppose \(v \in V\), What is \(\dim(E_v)\)?

      Note that if \(v=0\), \(E_{v}=L(V,W)\) \(\Rightarrow\dim(E_{v})=\dim(L\left(V,W\right))=\dim\left(V\right)\dim(W).\)

      So let's consider the case \(v\neq0\).

      Since \(v \neq 0\), we can extend it to a basis of \(V\)
      We'll call \(B=\{v_{},v_2,\ldots,v_{n}\}\). Now consider any basis of \(W\): \(B' = \{w_1, w_2, \ldots, w_m\}\).

      So now by theorem we have an isomorphism \(\Phi:\mathcal{L}(V,W)\rightarrow M_{m\times n}\left(F\right),\Phi\left(T\right)=\left\lbrack T\right\rbrack_{BB^{\prime}}\)

      Note that \(T\in E_{V}\;\iff\;T(v)=0\Rightarrow\text{the first column of }[T]_{BB^{\prime}}\) is \(0\).

      (Since by definition of \([T]_{BB^{\prime}}\), the first column is the coordinates of \(T(v)\) in my basis \(B'\) and so, since \(T(v)=0\), the coordinates are all \(0\).)

      So \(\text{dim}(E_{V})=\text{dim}(M)=m\left(n-1\right)\)

      Therefore \(\text{dim}(E_V) = m(n - 1) = \text{dim}(W)(\text{dim}(V) - 1)\)

3. Show that \(V\) (vector space) and \(\mathcal{L}(F,V)\) are isomorphic.

   Note that if \(V\) is finite-dimensional, we know an isomorphism exists since \(V\) and \(\mathcal{L}(F,V)\) have the same dimension: \(\text{dim}(\mathcal{L}(F,V))=\text{dim}(F)\cdot\text{dim}(V)=\text{dim}(V)\)

   Let's define: \(\Phi:\mathcal{L}(F,V)\rightarrow V,\,\Phi(T)=T(1)\) We'll see that \(\Phi\) is an isomorphism!

   • \(\Phi\) is a linear map. \(\Phi(T+\lambda S)=(T+\lambda S)(1)=T(1)+\lambda S(1)=\Phi\left(T)+\Phi\lambda(S\right)\)​\(\Rightarrow\) \(\Phi\) is a linear map.

   • \(\Phi\) is injective: it is enough to check that \(\text{Null}(\Phi)=\{0\}\).

   Let \(T\in\text{Null}(\Phi)=\Phi(T)=0\Rightarrow T\left(1\right)=0\)​\(\Rightarrow\) for any \(\lambda\in F\), \(T(\lambda)=0\Rightarrow\lambda T(1)=0\) \(\Rightarrow\) \(T=0\) (since \(T: F \rightarrow V\)) Therefore, \(\text{Null}(\Phi)=\{0\}\) \(\Rightarrow\) \(\Phi\) is injective. * \(T\) is surjective: \(\Phi:\mathcal{L}(F,V)\rightarrow V\)​

   Let \(v \in V\), we want \(T\in\mathcal{L}(F,V)\) such that \(\left.\Phi(T\right)=v\). So let's take \(T:F\rightarrow V\), \(T(\lambda)=\lambda v\) So \(\Phi\left(T\right)=T(1)=v\) So \(T\) is surjective.

   Thus \(\Phi\) is an isomorphism, then \(V\) (vector space) and \(\mathcal{L}(F,V)\) are isomorphic.