If $ \ 1 \ $ is an eigen value of $ \ T \ $ , then find rank of $ \ T-I_4 \ $ . , $ I_4= 4 \times 4 \ \ $ identity matrix

Question

Let $ T:\mathbb R^4 \to \mathbb R^4 \ $ be a linear mapping such that $$\ker T= \{(x,y,z,w)| x+y+z+w=0 \} \ .$$

If $ \ 1 \ $ is an eigenvalue of $T$ , then find the rank of $T-I_4$. Here $ I_4$ is the $ 4 \times 4 $ identity matrix .

Answer:

$ \ker T= \{(x,y,z,w)| x+y+z+w=0 \},$ so $ \dim (\ker T)=3$.

Thus, $$\text{rank}\,T=4-\dim(\ker T) =4-3=1.$$

Then I can not find $ \text{rank}\,(T-I_4)$?

Help me doing this.

Answer 1

Because the kernel of $T$ has dimension $3$, the geometric multiplicity of $1$ is one. So the kernel of $T-I$ has dimension $1$, which tells you that the rank of $T-I$ is $3$.