MA20216: Algebra 2A

5.3 Transposes

There is a duality construction for linear maps also: let \(V,W\) be vector spaces, \(\phi \in L(V,W)\) and \(\alpha \in W^{*}\). Then \(\alpha \circ \phi :V\to \F \) is also linear, so that \(\alpha \circ \phi \in V^{*}\). This prompts:

Definition. Let \(\phi \in L(V,W)\) be a linear map of vector spaces. The transpose \(\phi ^T\) of \(\phi \) is the map \(\phi ^T:W^{*}\to V^{*}\) given by
\(\seteqnumber{0}{5.}{1}\)
\begin{equation*} \phi ^T(\alpha ):=\alpha \circ \phi , \end{equation*}

for all \(\alpha \in W^{*}\).

Lemma 5.13. \(\phi ^T:W^{*}\to V^{*}\) is also a linear map.

Proof. Let \(\alpha ,\beta \in W^{*}\) and \(\lambda \in \F \). We must show that
\(\seteqnumber{0}{5.}{1}\)
\begin{equation*} \phi ^T(\alpha +\lambda \beta )=\phi ^T(\alpha )+\lambda \phi ^T(\beta ). \end{equation*}

Unravelling the definition, this means
\(\seteqnumber{0}{5.}{1}\)
\begin{equation*} (\alpha +\lambda \beta )\circ \phi =\alpha \circ \phi +\lambda \beta \circ \phi . \end{equation*}

This is an equality of functions and so holds exactly when
\(\seteqnumber{0}{5.}{1}\)
\begin{equation*} (\alpha +\lambda \beta )(\phi (v))=\alpha (\phi (v))+\lambda (\beta (\phi (v))), \end{equation*}

for all \(v\in V\). However, this last is true by the very definition of addition and scalar multiplication in \(W^{*}\). □

Examples.
- (1) \(\id _V^T=\id _{V^{*}}\). Indeed, \(\id _V^T(\alpha )=\alpha \circ \id _V=\alpha \), for all \(\alpha \in V^{*}\).
- (2) \((\psi \circ \phi )^T=\phi ^T\circ \psi ^T\). Indeed, \((\psi \circ \phi )^T(\alpha )=\alpha \circ \psi \circ \phi = \phi ^T(\alpha \circ \psi )=\phi ^T(\psi ^T(\alpha ))\).

Here is why \(\phi ^T\) is called the transpose of \(\phi \):

Proposition 5.14. Let \(V,W\) be finite-dimensional vector spaces and \(\phi \in L(V,W)\) with matrix \(A\in M_{m\times n}(\F )\) with respect to bases \(\lst {v}1n\) and \(\lst {w}1m\) of \(V\) and \(W\).

Then \(\phi ^T\) has matrix \(A^T\) with respect to the dual bases \(\dlst {w}1m\) and \(\dlst {v}1n\) of \(W^{*}\) and \(V^{*}\).

Proof. Let \(\phi ^T\) have matrix \(B\) so that
\(\seteqnumber{0}{5.}{1}\)
\begin{equation*} \phi ^T(w^{*}_j)=\sum _{i=1}^nB_{ij}v^{*}_i. \end{equation*}

Evaluate both sides of this at \(v_k\) to get
\(\seteqnumber{0}{5.}{1}\)
\begin{align*} \phi ^T(w^{*}_j)(v_k)&=B_{kj}\\ \end{align*} or, unravelling the definition of \(\phi ^T\),
\(\seteqnumber{0}{5.}{1}\)
\begin{align*} w^{*}_j(\phi (v_k))&=B_{kj}. \end{align*} Now
\(\seteqnumber{0}{5.}{1}\)
\begin{equation*} \phi (v_k)=\sum _{i=1}^mA_{ik}w_i \end{equation*}

so that we also get
\(\seteqnumber{0}{5.}{1}\)
\begin{equation*} w_j^{*}(\phi (v_k))=A_{jk}. \end{equation*}

Comparing these we get \(B_{kj}=A_{jk}\) whence \(B=A^T\). □

The kernels and images of \(\phi \) and \(\phi ^T\) are intimately related via the annihilators and solution sets of §5.2:

Theorem 5.15. Let \(\phi \in L(V,W)\) be a linear map of vector spaces. Then
- (1)
  \(\seteqnumber{0}{5.}{1}\)
  \begin{align*} \ker \phi &=\sol (\im \phi ^T)\\ \im \phi &\leq \sol (\ker \phi ^T) \end{align*} with equality if \(V,W\) are finite-dimensional.
- (2)
  \(\seteqnumber{0}{5.}{1}\)
  \begin{align*} \ker \phi ^T&=\ann (\im \phi )\\ \im \phi ^T&\leq \ann (\ker \phi ) \end{align*} with equality if \(V,W\) are finite-dimensional.

Proof. We will prove (1) and leave (2) as an exercise⁸.

For the first equality, observe that \(v\in \ker \phi \) if and only if \(\phi (v)=0\) or, equivalently, by Theorem 5.3, \(\alpha (\phi (v))=0\), for all \(\alpha \in W^{*}\), which is the same as \(\phi ^T(\alpha )(v)=0\), for all \(\alpha \in W^{*}\), that is, \(v\in \sol (\im \phi ^T)\).

If \(V,W\) are finite-dimensional we now use this, along with rank-nullity and Proposition 5.6, to get
\(\seteqnumber{0}{5.}{1}\)
\begin{equation*} \dim V-\dim \im \phi = \dim \ker \phi =\dim \sol (\im \phi ^T)=\dim V-\dim \im \phi ^T \end{equation*}

so that
\(\seteqnumber{0}{5.}{1}\)
\begin{equation} \label {eq:26} \rank \phi =\dim \im \phi =\dim \im \phi ^T=\rank \phi ^T. \end{equation}

For \(\im \phi \leq \sol (\ker \phi ^T)\), let \(w\in \im \phi \) and \(\alpha \in \ker \phi ^T\) so that \(\alpha \circ \phi =0\) and \(w=\phi (v)\), for some \(v\in V\). Then \(\alpha (w)=\alpha (\phi (v))=(\alpha \circ \phi )(v)=0\) so that \(w\in \sol (\ker \phi ^T)\). Thus \(\im \phi \leq \sol (\ker \phi ^T)\).

Moreover, if \(V,W\) are finite-dimensional, use (5.2), rank-nullity and Proposition 5.6 to get
\(\seteqnumber{0}{5.}{2}\)
\begin{equation*} \dim \im \phi =\dim \im \phi ^T =\dim W-\dim \ker \phi ^T=\dim \sol (\ker \phi ^T). \end{equation*}

We conclude that \(\im \phi \) and \(\dim \sol (\ker \phi ^{T})\) have the same dimension and so coincide. □

⁸ Question 4 on sheet 9.

Along the way, we got (5.2):

Corollary 5.16. Let \(\phi \in L(V,W)\) be a linear map of finite-dimensional vector spaces. Then
\(\seteqnumber{0}{5.}{2}\)
\begin{equation*} \rank \phi =\rank \phi ^T. \end{equation*}

Remark. This gives us a new take on an old result⁹from Algebra 1B. Let \(A\in M_{m\times n}(\F )\) be the matrix of \(\phi \) with respect to bases of \(V\) and \(W\) so that, by Proposition 5.14, \(A^T\) is the matrix of \(\phi ^T\) with respect to the dual bases. Then the rank of \(\phi \) is the column rank of \(A\) while the rank of \(\phi ^T\) is the column rank of \(A^T\) which is the row rank of \(A\). Thus row rank and column rank coincide.

⁹ Algebra 1B, Proposition 1.7.7.

The punchline of Theorem 5.15 is that \(\phi \) and \(\phi ^T\) have “opposite” properties. For example:

Proposition 5.17. Let \(\phi \in L(V,W)\) be a linear map of finite-dimensional vector spaces. Then
- (1) \(\phi \) injects if and only if \(\phi ^T\) surjects.
- (2) \(\phi ^T\) injects if and only if \(\phi \) surjects.

Proof. For (1), \(\phi \) injects if and only if \(\ker \phi =\set 0\) while \(\phi ^T\) surjects if and only if \(\dim \im \phi ^T=\dim V\). By Theorem 5.15, the first happens if and only if \(\sol (\im \phi ^T)=\set 0\) but, by Proposition 5.6, this is equivalent to the \(\dim \im \phi ^T=\dim V\).

Item (2) is similar. □

Remarks.
- (1) This result is useful as it is sometimes easier to prove injectivity than surjectivity.
- (2) With a bit more effort, we can do better than Proposition 5.17: for example, using Theorem 5.3, we can prove that Proposition 5.17(2) holds even in infinite dimensions.