5.6 Axler's Linear Algebra Done Right

Question

Consider the following theorem:

Theorem 5.6 Equivalent conditions to be an eigenvalue: Suppose V is finite-dimensional, $T \in \mathcal{L}(V),$ and $\lambda \in \mathbb{F}.$ Then the following are equivalent:

(a) $\lambda$ is an eigenvalue of T ;

(b) $T-\lambda I$ is not injective;

(c) $T- \lambda I$ is not surjective;

(d)$T- \lambda I$ is not invertible

Proof: Conditions (a) and (b) are equivalent because the equation $Tv = \lambda v$ is equivalent to the equation $(T - \lambda I)v = 0$.

From here I understand that $v \neq 0$, so that $(T - \lambda I)$ is a linear transformation that sends v to $Tv - \lambda v$. But from here I don't see why this implies that $(T - \lambda I)v = 0$ is not injective.

Can someone please explain? Thank you!

Answer 1

The false statement

If $(T-\lambda I) v = 0$, then it no way means that $T - \lambda I = 0$. Counterexample: suppose $T$ is a linear operator on $\mathbb F^2$ whose matrix w.r.t. some basis $e_1,e_2$ is $$ \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}, $$ then for $\lambda = 1$ and $v = [1, 0]^{\mathsf T}$, you could verify that $$ (T-I)v = 0, $$ but clearly $T - I \neq 0$ in this case, and $v \neq 0$, either.

Preparations

By definition, $T \in \mathcal L (V, W)$ is injective $\iff$

• whenever $v,w \in V$ satisfy $Tv = Tw$, $v = w$.

Equivalently, i.e. take the contrapositive statement,

• whenever $v,w \in V$ that $v \neq w$, $Tv \neq Tw$.

Therefore, $T$ is not injective $\iff$

• there exists $v, w \in V$ that $v \neq w$ but $T v = Tw$.

Since $T$ is a linear mapping, $Tv = Tw \iff T(v - w) = 0$. Let $u = v -w$, then we actually have

• $T$ is not injective $\iff$ there exists $u \neq 0$ that $Tu = 0$ $\iff$ $\mathrm{null}\,T \neq \{0\}(\bigstar)$.

Solve the problem

Now get back to the question. By definition,

• (a) holds $\iff$ there exists $ v \neq 0$ that $T v = \lambda v$,

equivalently, $(T-\lambda I)v = 0$. As a linear mapping, by applying $(\bigstar)$ to $T-\lambda I$, we get that equivalently $T -\lambda I$ is not injective.

Answer 2

Not injective means that $\text{null}\ (T-\lambda I) \neq \{0\}$ (see 3.16). Thus $(T-\lambda I)$ is not injective if and only if there exists some $v \in \text{null}\ (T-\lambda I), v \neq 0$, so if and only if for some $v \neq 0$, $(T-\lambda I)v = 0$.

Answer 3

In order to tackle your question, we need the following theorems:

1- A linear map $A$ is injective, iff it has a trivial kernel. (That is, Ker($A$)={0})

2- Let $V,W$ be vector spaces, such that dim$V$ = dim$W$, $\varphi:V\rightarrow W$ a linear map. Then, the following statements are equivalent:

a. $\varphi$ is injective;

b. $\varphi$ is surjective;

c. $\varphi$ is bijective.

3- A map $f:X\rightarrow Y$ is bijective, iff $f$ is invertible.

Do you see how the rest of your theorem follows? (Hint: $T-\lambda L$ is a linear transformation)