prove\disprove: $T$ is an isomorphism, given $T:\mathbb{R}^4 \to \mathbb{R}^4 $ a linear transformation. let dim(Im$(T+5I)=$dim(ker$(I-T))$.

Question

let $T:\mathbb{R}^4 \to \mathbb{R}^4$ be a linear transformation. given dim$(Im(T+5I)=$dim$(ker(I-T))$. prove/disprove $T$ is an isomorphism

My Attempt:

Let $A$ be the representative matrix for $T$. then: $rank(I-A)=3, rank(A+5I)=1$. this means that $rank(I-A)+rank(A+5I)=rank(I-A+A+5I)=rank(5I)=4$.

but rank of two matrices is not linear. what can you do using the linearity of $T$?

Answer 1

It holds $\dim(\ker(I-T))= \dim(Im(T+5I))=4-\dim (\ker(T+5I))$, whence $$4= \dim (\ker(T+5I))+\dim(\ker(I-T)).$$ This means that $\mathbb R^4$ decomposes as the direct sum of the eigenspaces $V_1$ and $V_{-5}$ (only $V_1$ if $T=Id$, and only $V_{-5}$ if $T=-5 Id$).

Therefore, $T$ is diagonalizable with eigenvalues $-5,1$ (possibly only $-5$ or only $1$), thus invertible.