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For a complex $m$-by-$n$ matrix $T$, the products $T^{\ast }T$ and $T{T}^{\ast }$ are Gram products; the star denotes the conjugate transpose. For a real matrix, it reduces to the matrix transpose. If the columns of $T$ are $v_1,\dots ,v_n$, then $T^{\ast }T$ is their Gram matrix under the standard complex inner product:Thus a rectangular matrix gives square Hermitian positive semidefinite matrices. The identityproves nonnegativity and gives $\ker (T^{\ast }T)=\ker T$. Similarly, $\ker (T{T}^{\ast })=\ker T^{\ast }$. These are the kernel of a Gram product identities; combined with the rank-nullity theorem, they show that each Gram product has the same matrix rank as $T$.

For example, $T=(12)$ gives $T{T}^{\ast }=5$, while $T^{\ast }T=\left(\begin{array}{cc}1& 2\\ 2& 4\end{array}\right)$. Both have matrix rank one despite their different sizes. The second is singular because the two columns of $T$ are linearly dependent.

Fix one common eigenvector $v\ne 0$, and write $A{A}^{\top }v=\lambda v$ and $B{B}^{\top }v=\mu v$. By (d),so $\lambda \mu v=0$ and at least one of the two eigenvalues is zero.

To exclude both being zero, use the kernel of a Gram product identity:If both products annihilate $v$, then $A^{\top }v=B^{\top }v=0$. By (b), $v\in \mathrm{im}L$, and by (e), this is $\mathrm{im}B$. On the other hand, $B^{\top }v=0$ makes $v$ orthogonal to every vector in $\mathrm{im}B$. In particular $v$ is orthogonal to itself, so $\Vert v{\Vert }^2=0$ and $v=0$, a contradiction.

Therefore each specified common eigenvector belongs to exactly one of $\ker (A{A}^{\top })$ and $\ker (B{B}^{\top })$. Both eigenvalues are nonnegative by the displayed squared-norm identities, so the other eigenvalue is strictly positive. This is the spectral mechanism behind orthogonal null-space completion of a square matrix.