Intuition behind $\ker(T)=\ker(T^*)$ for $T$ a normal operator Let $T : V \to V$ be a normal operator and $V$ a finite-dimensional vector space. Show that $\ker(T)=  \ker(T^*)$ and $\text{im}(T) = \text{im}(T^*)$. 
I know how to rigorously show this, but I'm curious if anyone has an intuitive way of understanding why this has to be the case.
 A: We normally think of an operator or matrix as a mapping from vectors to vectors $x\mapsto Tx$. But there is a dual way of thinking of operators, as acting on co-vectors $y^*\mapsto y^*T$. Think of a row vector multiplied with a matrix to output another row vector. We usually think of this latter mapping as $y\mapsto T^*y$, bringing co-vectors to vectors, though in reality what is happening is $y\mapsto y^*\mapsto y^*T\mapsto (y^*T)^*=T^*y$.
Now $T$ acts very differently on vectors and on co-vectors. Some transformations you can see this directly: If $T$ rotates vectors in some way, it un-rotates co-vectors, i.e. it rotates in the reverse direction. And if $T$ shears in some direction, $T^*$ shears in some dual direction.
However, $T$ and $T^*$ do not act independently of each other, of course. For example, if $T$ annihilates a co-vector $y$, that is, $y^*T=0$, then $y^*(Tx)=0$, so $y$ is perpendicular to the image of $T$. More generally, if $y^*$ is a co-eigenvector of $T$ (or in familiar terms, $T^*y=\lambda y$) and $x$ is an eigenvector of $T$ then $$\mu y^*x=y^*Tx=\lambda^*y^*x$$ So, unless $\mu=\lambda^*$, each co-eigenvector $y$ must be perpendicular to each eigenvector $x$. For general matrices, we get a bunch of (generalized) eigenvectors and another bunch of co-eigenvectors, with one set perpendicular to most of the others, but not necessarily to themselves ...like this:

Each eigenvector in blue is perpendicular to the co-eigenvectors in red, except the one that has the same eigenvalue*, and vice-versa.
Normal transformations can be thought of as ones that have an orthonormal basis of eigenvectors. By the identity above, the co-eigenvectors are also orthogonal (or can be made so), and are forced to coincide with the eigenvectors. In particular, if $T^*y=0$, that is, $y^*$ is a zero co-eigenvector, then $y$ must be perpendicular to all eigenvectors $x$, except those that have zero eigenvalues -- hence $y$ must be a zero eigenvector as well. (And conversely.)
