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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?

Beverlie
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  • While we can consider the problem here as resolved by the proposed duplicate (taking the minor step of algebra to move the dimension of the intersection to the other side of the equation), the present Question has a couple of quirks to be addressed by the OP. E.g. the use of "direct sum" to describe the last term in that equation. If this is to be an internal direct sum, then the intersection of ker $T$ and ker $T^t$ would need to be the trivial subspace. If it were to be an external direct sum, then the statement would be false unless again the intersection of subspaces is trivial. – hardmath Jun 03 '18 at 16:45

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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 $V+W$.

Bart Michels
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  • If the LHS is not $0$, then the sum is not direct by definition. You could still consider external direct sum and the formula would fail in the way you describe, but I doubt that was meant here. – Ennar Jun 03 '18 at 10:50