Let \(T:V\to W\) be a linear map of vector spaces over a field, with \(V\) finite-dimensional. Prove using Jordan canonical form that \[ \dim V=\dim\ker T+\dim\operatorname{im}T. \]
Solution
Set \(U=\operatorname{im}T\), which is finite-dimensional because \(V\) is finite-dimensional.
1. Define an endomorphism \(S\) of the finite-dimensional space \(V\oplus U\) by\[ S(v,u)=(0,Tv). \]Then \(S^2=0\).
For \((v,u)\in V\oplus U\), \[ S^2(v,u)=S(0,Tv)=(0,T0)=(0,0). \] Thus the minimal polynomial of \(S\) divides \(x^2\). In particular it splits over the base field, so \(S\) has Jordan canonical form consisting only of nilpotent Jordan blocks of sizes \(1\) and \(2\).
2. For every Jordan block of \(S\), its rank plus its nullity equals its size.
A size-\(1\) block is \(J_1(0)=[0]\), which has rank \(0\) and nullity \(1\). A size-\(2\) block is \[ J_2(0)=\begin{pmatrix}0&1\\0&0\end{pmatrix}, \] which has rank \(1\) and nullity \(1\). Thus each block contributes its full dimension to rank plus nullity. Since rank and nullity add across block-diagonal direct sums, \[ \operatorname{rank}S+\operatorname{nullity}S=\dim(V\oplus U). \]
3. The image and kernel of \(S\) are\[ \operatorname{im}S=\{0\}\oplus U, \qquad \ker S=\ker T\oplus U. \]
By definition, \[ \operatorname{im}S=\{(0,Tv):v\in V\}=\{0\}\oplus\operatorname{im}T=\{0\}\oplus U. \] Also \(S(v,u)=0\) exactly when \(Tv=0\), with no restriction on \(u\in U\). Hence \[ \ker S=\ker T\oplus U. \]
4. Therefore\[ \dim V=\dim\ker T+\dim\operatorname{im}T. \]
By 3, \[ \operatorname{rank}S=\dim U, \qquad \operatorname{nullity}S=\dim\ker T+\dim U. \] Substituting these into 2 gives \[ \dim V+\dim U =\dim U+\dim\ker T+\dim U. \] Cancelling \(\dim U\) from both sides gives \[ \dim V=\dim\ker T+\dim U =\dim\ker T+\dim\operatorname{im}T. \]