Question

Definition: The vector space $V$ is called the direct sum of $W_1$ and $W_2$ if $W_1$ and $W_2$ are subspaces of $V$ such that $W_1 \cap W_2=\{0\}$ and $W_1 + W_2=V$

We denote that $V$ is the direct sum of $W_1$ and $W_2$ by writing $V=W_1\oplus W_2$ .

Now, suppose that $V$ is a vector space over a field $\mathbb{F}$ and $T:V \longrightarrow V$ is a linear transformation with distinct eigenvalues $\lambda_1,...,\lambda_k$ . Show that $V=E{_\lambda __1}\oplus E{_\lambda __2}\oplus ...\oplus E{_\lambda __k}$ , where $E{_\lambda __i}$ is the eigenspace of $\lambda _i$, if and only if $T$ is diagonalizable.

Answer 1

$T:V\rightarrow V\ \ \ \ \ where,\\ \lambda_1,\lambda_2,\lambda_3,................,\lambda_k\text{ are distinct eigenvalues and }\\ E_{\lambda_1},E_{\lambda_2},E_{\lambda_3},..............,E_{\lambda_k}\text{ are eigenspaces corresponding to }\lambda_{i}\\\text{and }M_T(X)\text{ be the minimal polynomial of T}$

$Let,\\V=E_{\lambda_1}\oplus E_{\lambda_2}\oplus ...............\oplus E_{\lambda_k}\\\text{ then by primary decomposition theorem} \\M_T(X)=(X-\lambda_1)(X-\lambda_2)...................(X-\lambda_k)\\ \Righarrow V=E_{\lambda_1}\oplus E_{\lambda_2}\oplus ...............\oplus E_{\lambda_k}\\ \Rightarrow E_{\lambda_i}=Ker(T-\lambda_iI)\\ let, \ \ v\in E_{\lambda_i}\ \Rightarrow\ (T-\lambda_iI)(v)=0\ \ \ \ \ \ \ \ \ \ \ (\because )\\ \Rightarrow T(v)=\lambda_iv\\\Rightarrow v_i=E_{\lambda_i}\\\text{ Since, each eigenspace correspond to a basis. Let, }B_i\text{ be the basis corresponding to }E_{\lambda_i}\\\text{and }B\text{ be the basis for }V\text{ then, }\\B=B_1\cup B_2\cup B_3\cup ...............\cup B_k \\\text{ and each basis }B_i\text{ are characteristics vector of }T\text{ and the members of B are characteristics vectors}\\\Rightarrow T\text{ is diagonalizable.}\\\\\\\text{Conversely, suppose T is diagonalizable}\\\exists\text{ a basis B consists of characteristics vectors of T }$$\text{ Consider a polynomial }\\p(X)=(X-\lambda_1)(X-\lambda_2).................(X-\lambda_k)\ \ \ \in \mathbb F[X]\\\text{claim:}\ \ p(X)=M_T(X)\\p(T)=(T-\lambda_1I)(T-\lambda_2I)............................(T-\lambda_kI)\\v_i\in B\\\Rightarrow v_i\text{ is characteristics vector corresponding to }\lambda_i\\p(T)=(T-\lambda_1I)(T-\lambda_2I)............................(T-\lambda_iI)\\p(T)(v_i)=(T-\lambda_1I)(T-\lambda_2I)............................(T-\lambda_kI)(v_i)\\\Rightarrow p(T)(v_i)=0\\\text{since each eigen value }\lambda\text{is a root of }M_T(X),\\\Rightarrow p(X)=M_T(X)\\and\ \ as,\\M_T(X)=(X-\lambda_1)(X-\lambda_2)......................(X-\lambda_k)\\ \text{then by Primary Decomposition Theorem}\\V=E_{\lambda_1}\oplus E_{\lambda_2}\oplus ..........................\oplus E_{\lambda_k}$