Linear Algebra - Prove Isomorphism. Let $T : \Bbb R^n \rightarrow \Bbb R^n$ Linear transformation.
Prove that there is a real number $\alpha$ that the transformation $\alpha I-T$ is isomorphism.
isomorphism is only if $\ker T={0}$ or $\text{Im } T=\Bbb R^n$ because the transformation is from the same dimension.
If $0$ is an eigenvalue then $T$ isn't isomorphism, so I just need to choose $\alpha \neq 0$?
 A: Note that $\alpha I - T$ is an isomorphism if and only if $\alpha$ is not an eiegenvalue of $T$.  Noting that $T$ has at most $n$ distinct eigenvalues, it is of course possible to choose a suitable $\alpha$.
Alternatively, it is sufficient to select $\alpha > \|T\|$ for any matrix-norm $\|\cdot\|$.
A: We know by the Jordan block form of $T$ that $\alpha I-T$ is similar to an upper-triangular matrix with values on the diagonal equal to $\alpha-\lambda_i$ with $\lambda_i$ the eigenvalues of $T$. So long as $|\alpha|>|\lambda_i|$ for all $i$, the determinant is non-zero, hence the transformation $\alpha I-T$ is invertible.
A: Let $T_\alpha=\alpha I -T:\mathbb{R}^n \to \mathbb{R}^n$
Assume $T_\alpha$ is not an isomorphism, so $ker(T_a) \ne \{ \vec{0}\}$
Thus the set $S_\alpha:=\{\vec{v}\in \mathbb{R}^n: T(\vec{v})=\alpha \vec{v}\}= ker(T_\alpha) \subset \mathbb{R}^n$ has $\dim(S_\alpha)\ge 1$
For $\alpha \ne \beta$, if $T_\beta$ is not an isomorphism, then $S_\beta \cap S_\alpha = \{\vec{0}\}$
This is because if $\vec{v}\in S_\alpha \cap S_\beta$ we must have $T(\vec{v})=\alpha \vec{v}=\beta \vec{v} \Rightarrow \vec{v}=\vec{0}$
This means there are at most $n$ values $\alpha_1,\alpha_2,\dots, \alpha _n\in \mathbb{R}$ with $\dim(S_\alpha)\ge 1$, which means there must exists some (infinitely many of them actually) $\alpha \in \mathbb{R}$ with $S_\alpha = \{\vec{0}\}$ which is equivalent to saying there exists some $T_\alpha$ which is an isomorphism
A: Here's a bit more general/topological perspective for why this is true:
The set of all invertible $n$-by-$n$ matrices is an open set, so for any invertible matrix there exists a small neighborhood of invertible matrices surrounding it. 
In particular, there is a small neighborhood of invertible matrices surrounding the identity, so that
$$I + \frac{1}{\alpha}T$$
is invertible for perturbation $\frac{1}{\alpha}T$, so long as $\frac{1}{\alpha}$ is sufficiently small.

Ok, so why is the set of invertible matrices open? Well, the determinant is a continuous function, so under the determinant map the inverse image of open sets is open. In particular, the set of invertible matrices,
$$\{\text{invertible matrices}\} = \text{det}^{-1}(\mathbb{R}\backslash 0),$$
is the inverse image of an open set under the determinant, so it is open.
For more details about the topology, see this answer here
