Proof. Let \(T\in \mathcal {L}(A,B)\).

i.

\(T(0)=T(0+0)=T(0)+T(0)\), and subtracting \(T(0)\) from both sides gives \(T(0)=0\). So \(0\in \text {null}\,T\) and the null space is not empty.

ii.

If \(x,y\in \) null\(T\) then \(T(x+y)=T(x)+T(y)=0+0=0\), so \((x+y)\in \) null\(T\).

iii.

If \(x\in \) null\(T\) and \(\alpha \in \mathbb {F}\), then \(T(\alpha x)=\alpha T(x)=\alpha \cdot 0=0\), so \(\alpha x\in \) null\(T\).

Being a non-empty subset of \(A\) closed under addition and scalar multiplication, null\(T\) is a subspace of \(A\). □

Proof. Suppose \(T\) is injective and let \(x\in \) null\(T\). Then \(Tx=0=T0\), and injectivity gives \(x=0\). So null\(T=\{0\}\).

Conversely suppose null\(T=\{0\}\) and let \(Tx=Ty\). By linearity \(T(x-y)=Tx-Ty=0\), so \(x-y\in \) null\(T=\{0\}\), giving \(x-y=0\), that is \(x=y\). So \(T\) is injective.

This is a genuine economy: for a linear map, injectivity need only be checked at the single vector \(0\), not at every pair of vectors. □

Proof. \(0=T(0)\) lies in range\(T\), so it is not empty. If \(u,v\in \) range\(T\), say \(u=Tx\) and \(v=Ty\), then \(u+v=Tx+Ty=T(x+y)\in \) range\(T\); and for \(\alpha \in \mathbb {F}\), \(\alpha u=\alpha Tx=T(\alpha x)\in \) range\(T\). Hence range\(T\) is a subspace of \(B\). □

Proof.

i.

Let \(\dim A=n\) and let \(\{u_1,\dots ,u_k\}\) be a basis of null\(T\), so \(\dim \text {null}\,T=k\). Being a linearly independent set in \(A\), it extends to a basis \[\{u_1,\dots ,u_k,v_1,\dots ,v_m\}\] of \(A\), where \(k+m=n\). The claim is that \(\{Tv_1,\dots ,Tv_m\}\) is a basis of range\(T\); granting this, \(\dim \text {range}\,T=m=n-k\), which is the assertion.

It spans range\(T\)

Let \(y\in \) range\(T\), say \(y=Tx\). Write \(x\) in the basis above as \(x=\sum _{i=1}^{k}a_iu_i+\sum _{j=1}^{m}b_jv_j\). Applying \(T\) and using \(Tu_i=0\), \[y=Tx=\sum _{i=1}^{k}a_iTu_i+\sum _{j=1}^{m}b_jTv_j=\sum _{j=1}^{m}b_jTv_j.\]

It is linearly independent

Suppose \(\sum _{j=1}^{m}c_jTv_j=0\). By linearity \(T\left (\sum _{j}c_jv_j\right )=0\), so \(\sum _{j}c_jv_j\in \) null\(T\) and can therefore be written in the basis of the null space as \(\sum _{j}c_jv_j=\sum _{i}d_iu_i\). Rearranging, \[\sum _{j=1}^{m}c_jv_j-\sum _{i=1}^{k}d_iu_i=0,\] a linear relation among the vectors of a basis of \(A\). All its coefficients must therefore vanish; in particular every \(c_j=0\).

So \(\{Tv_1,\dots ,Tv_m\}\) is a basis of range\(T\), which is in particular finite dimensional, and \(\dim A=k+m=\dim \text {null}\,T+\dim \text {range}\,T\).

ii.

Since range\(T\) is a subspace of \(B\), \(\dim \text {range}\,T\leq \dim B\). By part (i), \[\dim \text {null}\,T=\dim A-\dim \text {range}\,T\geq \dim A-\dim B>0,\] so null\(T\neq \{0\}\) and by Proposition 6.2.4 \(T\) is not injective.

iii.

By part (i), \(\dim \text {range}\,T=\dim A-\dim \text {null}\,T\leq \dim A<\dim B\). A subspace of strictly smaller dimension cannot be all of \(B\), so range\(T\neq B\) and \(T\) is not surjective.

□