# Ker(T) = Range(T)

Let $$V$$ be a vectorspace of dimension 6 over a field $$F$$. There exists $$T \in A(V)$$ such that $$Ker(T) = Range(T)$$ (True or false)

I know that Kernel is the subspace of domain and Range is the subspace of Codomain and $$x=0$$ is the only point in both spaces. My question is can we find a linear operator such that $$Ker T = Range T?$$

• Hint: Can you do it for dimension 2? Commented Jun 16, 2019 at 14:32
• Use the Dimensions Theorem Commented Jun 16, 2019 at 14:35

First "decipher" $$\DeclareMathOperator \im{im} \ker T = \im T$$. For $$v \in V$$, $$Tv = \im T = \ker T$$, thus $$T^2 v = 0$$. Therefore $$T^2 = 0$$.

Also by the Rank-Nullity theorem, $$\dim (\ker T) + \dim (\im T) = \dim V = 6$$, thus $$\DeclareMathOperator \rank {rank} \dim (\ker T) = 3$$.

For simplicity, we consider the Jordan canonical form of possible $$T$$. The argument above shows that $$T$$ is nilpotent, thus the Jordan form exists. Since $$\dim (\ker T) = 3$$, there should be $$3$$ Jordan blocks. Easy to see that $$X^2$$ is the minimal polynomial of $$T$$, thus the max size of the Jordan blocks is $$2$$. Since $$\dim V = 6$$, each Jordan blocks should be of size $$2$$. Therefore if such operator exists, then under certain basis the matrix of $$T$$ should be $$\mathrm {diag}(J_2(0),J_2(0), J_2(0)),$$ where $$J_2(0)$$ denotes the Jordan block of size $$2$$ with all diagonal entries $$0$$.

Now we check this $$T$$. By definition, if the basis is $$(e_j)_1^6$$, then $$T e_{2j-1} = 0, Te_{2j} = e_{2j-1}, j=1,2,3.$$ Direct verification shows $$\ker T = \mathrm {span} (e_1, e_3, e_5)$$ and $$\im T = \mathrm {span} (e_1, e_3, e_5)$$. In conclusion, such operator exists, whose matrix is described above.

Here's how to solve it and also the thought process.

The first thing you need to do when you see something that involves kernel image and dimensions is to apply the dimension theorem.

Theorem: If $$T:V\rightarrow V$$ a linear operator then $$\dim V = \dim \ker T + \dim \text{Range} T$$.

Since $$\ker T = \text{Range} T$$ we have that $$\dim V = 2\dim \ker T$$. Which is only possible if the dimension of $$V$$ is even.

But this is the case! We have that $$\dim V=6$$, so this means that the dimension theorem won't help us to disprove the claim. It also "hints" that there is probably such an example.

So we try to construct an example. We choose a $$6$$ dimensional vector space $$V$$, the obvious example is $$\mathbb{R}^6$$. Then we need that $$\ker T, \text{Range} T$$ are $$3$$ dimensionals (by the dimensions theorem). The best way to construct a linear operator is to describe where we send the basis $$e_1,e_2,...,e_6$$.

We need the dimension of the kernel to be $$3$$, so lets send $$e_1,e_2,e_3$$ to zero. Now we need the image to equal to kernel so we send $$e_4,e_5,e_6$$ to $$e_1,e_2,e_3$$.

In other words define $$T(x_1,x_2,...,x_6) = (x_4,x_5,x_6,0,0,0)$$. This is a linear operator from $$\mathbb{R}^6$$ to itself whose kernel equals to it's image (equals to $$\{(x,y,z,0,0,0):x,y,z\in\mathbb{R}\}$$).