Calculating Dimension of Intersection of Kernels

Question

Let T be sqaure matrix and regarded as a linear operator on a finite dimensional vector space V such that $T^2 = 0$.

$\dim(\ker T \cap \ker T^t) =\dim \ker T+\dim\ker\ T^t−\dim(\ker T\dot+\ker T^t) $

($\dot+ $ denotes direct sum)

Is the abvoe equation true? If so, why is it so?

Answer 1

No, but it is true if you take the ordinary subspace sum:

$\dim(\ker T \cap \ker T^t) =\dim \ker T+\dim\ker\ T^t−\dim(\ker T+\ker T^t)$

Note that with direct sum it can't be true, since $\dim(\ker T\dot+\ker T^t)= \dim \ker T+\dim\ker\ T^t$ and your RHS is zero, while the LHS need not be, for example for $T=0$.

So I assume the formula you want is with the ordinary sum. In that case, more generally, for subspaces $V,W$ of a vector space we have $$\dim(V \cap W) = \dim V + \dim W - \dim(V + W)$$ To prove this, we can construct a basis of $V+W$ by taking a basis $(e_i)$ of $V \cap W$, extending it to a basis $(f_i)$ of $V$ and $(g_i)$ of $W$. Then you can show that $(e_i) \cup (f_i) \cup (g_i)$ is a basis of $V+W$.