Is $\ker T = \ker T^*$ for idempotent operators?

Question

Recall that if $T : V \to W$ is a linear map between finite-dimensional inner product spaces, then $T^* : W \to V$ is defined as the unique linear map satisfying $\langle Tv, w \rangle = \langle v, T^*w \rangle$ for all $v \in V$ and $w \in W$. Notice that $w \in \ker T^*$ iff $T^* w = 0$ iff $0 = \langle v, T^* w \rangle = \langle Tv, w \rangle$ for all $v \in V$, so $\ker T^* = T(V)^\perp$.

Consequently, if $S : V \to V$ is linear, then $V = S(V) \oplus S(V)^\perp = S(V) \oplus \ker S^*$. If $S : V \to V$ is also idempotent ($S^2 = S$), then $Sv = v$ iff $S(Sv) = v$ shows that $S(V)$ is the fixed points of $S$, and the only fixed point in the kernel is $0$, so $S(V) + \ker S$ is direct with dimension $\dim V$ by rank-nullity, so $S(V) \oplus \ker S = V = S(V) \oplus \ker S^*$.

Is $\ker S = \ker S^*$? In general, $A \oplus B = A \oplus B'$ does not imply $B = B'$, so I suspect not.

Answer 1

For an idempotent operator $S: V \to V$ let $K$ be its kernel and $P$ its image, so that $V = K \oplus P$. The adjoint operator $S^*$ is also idempotent, and we can see that if $v \in P^\perp$ then $\langle -, S^* v \rangle = \langle S(-), v \rangle = 0$. Therefore $P^\perp \subseteq \ker S^*$. Applying the same logic to $1 - S$ shows that $K^\perp \subseteq \operatorname{im} S$, so $S^*$ has kernel $P^\perp$ and image $K^\perp$.

Thinking of $S$ as a projection to $P$ by killing $K$, then $S^*$ is a projection to $K^\perp$ by killing $P^\perp$. A little thought should convince you that $K = P^\perp$ if and only if the operator $S$ is self-adjoint.

Answer 2

Say $S : V \to V$ has image $I$ and kernel $K$. $S$ is idempotent $\iff$ $I$ is a subspace of its fixed points $\iff$ its fixed points are precisely $I$. Thus $K \cap I = \{0 \}$ (there's only one fixed point in the kernel), so $K + I$ is a direct sum of subspaces, so $V = K \oplus I$ by rank-nullity. Construct orthormal bases $k_1, \dots, k_m$ for $K$ and $i_1, \dots, i_n$ for $I$, so that $\mathcal{B} = k_1, \dots, k_m, i_1, \dots, i_n$ is a basis for $V$ consisting of eigenvectors of $S$. Then for all $v \in V$, $$Sv = S\left( \sum_{i = 1}^m \alpha_i k_i + \sum_{j = 1}^n \beta_j i_j \right) = \sum_{j = 1}^n \beta_j i_j$$ In other words, $S$ maps $v = v_K + v_I$ to $v_I$, the unique part of $v$ in $I$. Conversely, if $S : V \to V$ satisfies $V = K \oplus I$ and $Sv = v_I$ for all $v$, then $S^2 v = S v_I = v_I = S v$ for all $v$, so $S$ is idempotent. In conclusion, $S$ is idempotent $\iff$ $S$ is a projection (meaning $V = K \oplus I$ and $Sv = v_I$).

The question asks if $K = \ker S^* = I^\perp$ for all projections $S$, which is false. For example, $$\begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} : \mathbb{R}^2 \to \mathbb{R}^2$$ is a horizontal shear followed by orthogonal projection onto the $x$-axis, which squashes all the points in each kernel coset $(x, y) + K = (x, y) + \mathrm{Span}(-1, 1)$ down to their intersection $(x + y, 0)$ with the $x$-axis (the image, $I = \mathrm{Span}(1, 0)$). It is idempotent (equivalently, it is a projection since $\mathbb{R}^2 = \mathrm{Span}(-1, 1) \oplus \mathrm{Span}(1, 0)$, and $(x + y, 0)$ is the $I$ part of $(x, y) = (-y, y) + (x + y, 0)$), but clearly $K \neq I^\perp = $ the $y$-axis.

However, $K = I^\perp$ seems like a property worth investigating. Indeed, it seems like a reasonable use of English to say a projection $S$ (as described above) is an orthogonal projection iff $K$ and $I$ are orthogonal, i.e. $K = I^\perp$. In fact, $K = I^\perp$ iff $S$ is self-adjoint. Recall that $S^* : V \to V$ has kernel $I^\perp$, which is also the kernel of $S$ if $S = S^*$. Conversely, if $K = I^\perp$, then the aforementioned $\mathcal{B} = k_1, \dots, i_n$ is an orthonormal eigenbasis, so the matrix $[S]_\mathcal{B}$ for $S$ with respect to the orthonormal basis $\mathcal{B}$ is diagonal with $1$s and $0$s along the diagonal, whence $[S]_\mathcal{B}$ is Hermitian; it follows that $S = S^*$, as $[S^*]_\mathcal{B}$ is the conjugate transpose of $[S]_\mathcal{B}$.

For a given subspace $U$ of $V$, define a projection onto $U$ as a projection with image $I = U$. There can be at most one orthogonal projection onto $U$: if $S_1, S_2$ are orthogonal projections onto $U$, then they have the same image $U$ and kernel $U^\perp$, so both transformations map $v = v_U + v_{U^\perp}$ to $v_U$ for all $v \in V$. And an orthogonal projection always exists, given by $$v \mapsto \sum_{i = 1}^k \langle v, u_i \rangle u_i$$ for any orthonormal basis $u_1, \dots, u_k$ of $U$. It is the unique idempotent, self-adjoint operator with image $U$.