Let $T$ be a self-adjoint operator on a real finite-dimensional inner product space $V$. And suppose that $b^2 − 4c < 0$. I want to show that the operator $T^2+bT+cI$ is invertible.
I'm wondering if the following argument is correct, especially the bolded part.
Let $\lambda$ be an eigenvalue of T with corresponding eigenvector $u$. Then $\lambda^2 + b\lambda + c$ is an eigenvalue of $T^2 + bT + cI$ with corresponding eigenvector $u$. Since $b^2 − 4c < 0$, we know that $\lambda^2 + b\lambda + c \ne 0$. And as $0$ isn't an eigenvalue of $T^2 + bT + cI$, it follows that that $T^2 + bT + cI$ is injective. Done.
No, this is wrong. At no point you used the fact that $T$ is self-adjoint. So, if your proof was correct, you would have proved taht $T^2+bT+cI$ is always invertible when $b^2-4c<0$. This is false; just take $T(x,y)=(-y,x)$, $b=0$ and $c=1$.
You proved correctly that if $\lambda$ is an eigenvalue of $T$, then $\lambda^2+b\lambda+c$ is an eigenvalue of $T^2+bT+cI$. But how do you know that every eigenvalue of $T^2+bT+cI$ can be obtained by this process?