Significance of $N(T^t)=R(T)^0$.

Question

I know that if $T:V\to W$ is a linear transformation where $V,W$ are finite dimensional.Then we have $Ker(T^t)=Im(T)^0$.But how to geometrically interpret this thing.What does it mean and why this has to be true?Can someone give me a clue?

Answer 1

Let $$T = \begin{pmatrix} a_1 & b_1 & ... & z_1 \\ a_2 & b_2 & ... & z_2\\ . & . &... & .\\ a_n & b_n & ... & z_n \end{pmatrix}$$

The range $R(T) = span \left\{ \begin{pmatrix} a_1 \\ a_2 \\ . \\ a_n \end{pmatrix} \begin{pmatrix} b_1 \\ b_2 \\ . \\ b_n \end{pmatrix} ... \begin{pmatrix} z_1 \\ z_2 \\ . \\ z_n \end{pmatrix} \right\}$.

A selection of linearly independent column vectors can form a basis for the range of $T$.

The annihilator $R(T)^0 = \{s : s \in W , r \in R(T), sr = 0\}$

The annihilator is the set of vectors $s$ that take every vector $r$ in the range to zero.

In this case $sr$ is a dot product.

$s.a = 0$, $s.b = 0$,...,$s.z = 0$.

$s$ are normal to the column vectors in $T$.

$$T^t = \begin{pmatrix} a_1 & a_2 & ... & a_n \\ b_1 & b_2 & ... & b_n\\ . & . &... & .\\ z_1 & z_2 & ... & z_n \end{pmatrix}$$

$$N(T^t) = \{w: T^tw = 0, w \in W\}$$

$T^tw = 0$ solves for $w.a = 0$, $w.b = 0$ ... $w.z = 0$.

$w$ are normal to the row vectors in $T^t$.

So the spans of $s$ and $w$ are identical, they solve the same equations.

$$N(T^t) = R(T)^0$$

Generalizing. The annihilators of $R(T)$ are normal to the basis of $R(T)$. The null space vectors of $T^t$ are normal to the same vectors.