Eigenvalues and Eigenvectors
Invariant Subspaces
Suppose $ T \in \mathcal{L}(V) $ and $ U $ is a subspace of $ V $ invariant under $ T $. Then $ U $ is invariant under $ p(T) $ for every polynomial $ p \in \mathcal{P}(\mathbf{F}) $.
Every subspace of an eigenspace is invariant.
Suppose that $ V $ is finite-dimensional and $ k \in \{ 1, \dots \dim V - 1 \}$. Suppose $ T \in \mathcal{L}(V) $ is such that every subspace of $ V $ of dimension $ k $ is invariant under $ T $. Then
\[\bigcap \{U : \dim U = k,\ v \in U\} = \operatorname{span}(v).\]
The $\supseteq$ direction is trivial ($v$ is in each such $U$). For $\subseteq$, I show any $w \notin \operatorname{span}(v)$ can be excluded by some $k$-subspace through $v$ — so $w$ can’t be in the intersection.
Suppose $w \notin \operatorname{span}(v)$. Then $v, w$ are independent, so extend them to a basis
\[v,\ w,\ x_3,\ \dots,\ x_n \qquad (n = \dim V).\]Now set
\[U = \operatorname{span}(v,\ x_3,\ x_4,\ \dots,\ x_{k+1}) = \operatorname{span}(v) + \operatorname{span}(x_3,\dots,x_{k+1}).\]That’s $v$ together with $k-1$ of the $x_i$’s, so $\dim U = k$, and $v \in U$. But $w \notin U$: the vectors $v, x_3, \dots, x_{k+1}$ are part of a basis that also includes $w$, so $w$ is independent of them and hence not in their span. This $U$ contains $v$, has dimension $k$, and misses $w$. $\blacksquare$
“Every $k$-dimensional subspace is invariant” collapses to “every line is invariant”** — i.e. all the way down to $k=1$ — because a line is recoverable as the intersection of the $k$-subspaces sitting above it.
Eigenvectors
Every eigenvector for a nonzero eigenvalue lies in $\operatorname{range} T$.
Suppose $\lambda \ne 0$ is an eigenvalue with eigenvector $v$: $Tv = \lambda v$. Then divide by $\lambda$ (legal since $\lambda \ne 0$):
\[v = \tfrac{1}{\lambda}(\lambda v) = \tfrac{1}{\lambda} Tv = T\!\left(\tfrac{1}{\lambda}v\right) \in \operatorname{range} T.\]For $\lambda = 0$ the eigenvectors are in $\operatorname{null} T$, and there’s no reason they’d be in the range. $\blacksquare$
Suppose $ T \in \mathcal{L}(V) $. Then every list of eigenvectors of $ T $ corresponding to distinct eigenvalues of $ T $ is linearly independent.
Suppose $ V $ is finite-dimensional and $ v_1, \dots, v_m \in V $.
$ v_1, \dots, v_m \in V $ is linearly independent $ \iff \exists T \in \mathcal{L}(V) $ such that $ v_1, \dots, v_m \in V $ are eigenvectors of $ T $ corresponding to distinct eigenvalues.
Tight upper bounds of the number of distinct eigenvalues:
- $ \dim V $
- $ 1 + \dim \operatorname{range} T $
$\operatorname{range} T$ contains $m$ linearly independent vectors, which forces
\[m \le \dim \operatorname{range} T.\]The eigenvalues of $T$ are these $m$ nonzero ones, plus possibly $0$. Since $0$ is a single value, it adds at most $1$ to the count of distinct eigenvalues:
\[\#\{\text{distinct eigenvalues}\} \le m + 1 \le \dim \operatorname{range} T + 1. \qquad \blacksquare\]
Let $r = \dim \operatorname{range} T$ and $n = \dim V$. Rank-nullity: $\dim \operatorname{null} T = n - r$.
- If $0$ is not an eigenvalue: $T$ injective, $r = n$, so $1 + r = n + 1 > n = \dim V$. The range bound is the weaker (larger) one here — but it’s still valid, just not as good as $\dim V$.
- If $0$ is an eigenvalue: $r \le n - 1$, so $1 + r \le n = \dim V$. Now the range bound is the stronger (smaller, better) one.
So the range bound $1 + r$ is the better bound exactly when $0$ is an eigenvalue — which is the whole point of the problem: it improves on $\dim V$ precisely in the non-injective case, by using the range to corral the nonzero eigenvalues and spending only “+1” on zero.
Null space and range of $ p(T) $ are invariant under $ T $.
Suppose $ T \in \mathcal{L}(V) $ is such that every nonzero vector in $ V $ is an eigenvector of $ T $. Then $ T $ is a scalar multiple of the identity operator.
Suppose $ T \in \mathcal{L}(V) $. Suppose $ S \in \mathcal{L}(V) $ is invertible.
(a) $ p(STS^{-1}) = Sp(T)S^{-1} $.
(b) $ S:\; E(\lambda,\, S^{-1}TS)\;\xrightarrow{\ \sim\ }\; E(\lambda,\, T) $.
$S^{-1}TS$ (conjugation) is just $T$ “viewed in a different basis”, and $S$ is the dictionary translating vectors from the new coordinates back to the old.
(a) Polynomials of $T$ transform the same way — $p$ of the conjugate is the conjugate of $p(T)$. In particular, taking $p$ to be the minimal polynomial of $T$: $p(STS^{-1}) = Sp(T)S^{-1} = S\cdot 0\cdot S^{-1} = 0$, which reproves that $T$ and $STS^{-1}$ share the same minimal polynomial. So, eigenvalues are basis-independent facts about the operator.
(b) Eigenvaluees are untouched; eigenvectors are genuine vectors, so they get translated by the dictionary $S$.
Suppose $V$ is finite-dimensional, $ T \in \mathcal{L}(V) $, and $ \lambda \in \mathbf{F} $.
$\lambda$ is an eigenvalue of $T \iff \lambda$ is an eigenvalue of the dual operator $ T’ \in \mathcal{L}(V’). $
Suppose $ \mathbf{F} = \mathbb{R} $, $ T \in \mathcal{L}(V) $, and $ \lambda \in \mathbb{R} $.
$\lambda$ is an eigenvalue of $T \iff \lambda$ is an eigenvalue of the complexification $ T_{\mathbb{C}}. $
Suppose $ \mathbf{F} = \mathbb{R} $, $ T \in \mathcal{L}(V) $, and $ \lambda \in \mathbb{C} $.
$\lambda$ is an eigenvalue of the complexification $T_{\mathbb{C}} \iff \bar{\lambda} $ is an eigenvalue of $ T_{\mathbb{C}}. $
Recall $V_{\mathbb{C}} = \{ u + iv : u, v \in V \}$ with $T_{\mathbb{C}}(u+iv) = Tu + iTv$. Define conjugation $C : V_{\mathbb{C}} \to V_{\mathbb{C}}$ by
\[C(u + iv) = u - iv.\]Two properties, both routine to check:
- $C$ is a bijection, in fact an involution: $C\big(C(u+iv)\big) = C(u - iv) = u + iv$, so $C^{-1} = C$. In particular $C$ sends nonzero vectors to nonzero vectors.
- $C$ is conjugate-linear: $C(\alpha w) = \bar\alpha\, C(w)$ for $\alpha \in \mathbb{C}$. (Direct from the scalar-multiplication rule on $V_{\mathbb{C}}$.)
Claim: $T_{\mathbb{C}} \circ C = C \circ T_{\mathbb{C}}$, i.e. $T_{\mathbb{C}}$ commutes with conjugation.
\[T_{\mathbb{C}}\big(C(u+iv)\big) = T_{\mathbb{C}}(u - iv) = Tu - iTv = C(Tu + iTv) = C\big(T_{\mathbb{C}}(u+iv)\big).\ \checkmark\]Suppose $\lambda$ is an eigenvalue of $T_{\mathbb{C}}$: there is $w \neq 0$ in $V_{\mathbb{C}}$ with
\[T_{\mathbb{C}}\,w = \lambda w.\]Apply $C$ to both sides. On the right, conjugate-linearity gives $C(\lambda w) = \bar\lambda\, Cw$. On the left, the commuting relation gives $C(T_{\mathbb{C}} w) = T_{\mathbb{C}}(Cw)$. Hence
\[T_{\mathbb{C}}(Cw) = \bar\lambda\,(Cw).\]Since $C$ is a bijection and $w \neq 0$, we have $Cw \neq 0$. So $Cw$ is an eigenvector of $T_{\mathbb{C}}$ with eigenvalue $\bar\lambda$ — meaning $\bar\lambda$ is an eigenvalue of $T_{\mathbb{C}}$.
That proves the forward direction. The converse needs no new work: the statement is symmetric under $\lambda \leftrightarrow \bar\lambda$, since $\overline{\bar\lambda} = \lambda$. Concretely, apply what we just proved to $\bar\lambda$ in place of $\lambda$: if $\bar\lambda$ is an eigenvalue, so is $\overline{\bar\lambda} = \lambda$. $\blacksquare$
Suppose $ \mathbf{F} = \mathbb{R} $, $ T \in \mathcal{L}(V) $, and $ \lambda \in \mathbb{C} $.
The eigenspaces are swapped by conjugation:
\[C\big(E(\lambda, T_{\mathbb{C}})\big) = E(\bar\lambda, T_{\mathbb{C}}).\]
Suppose $ T \in \mathcal{L}(V) $ and $ \lambda $ is an eigenvalue of $ T $, then
\[\left\lvert \lambda \right\rvert \le n \max\{\left\lvert \mathcal{M}(T, (v_1,\dots,v_n))_{j,k} \right\rvert : 1 \le j, k \le n \}.\]
Sum of eigenspaces is a direct sum.
Eigenspace is invariant under commuting operator.
$ T $ is diagonalizable
$ \iff V = \operatorname{null} (T - \lambda I) \oplus \operatorname{range} (T - \lambda I) $.
In an upper-triangular matrix,
\[\{\text{distinct diagonal entries}\} = \{\text{zeros of min poly}\} = \{\text{eigenvalues}\}.\] \[1 \le (\text{min-poly exponent of } \lambda) \le (\text{times } \lambda \text{ appears on the diagonal}),\]
| $T \in \mathcal{L}(V)$ | Basis | Subspaces | Dimensions | Minimal polynomial ($ m = \deg p \le \dim V $) |
|---|---|---|---|---|
| Upper-triangularizable | $ Tv_k \in \operatorname{span}(v_1, \dots, v_k) $ for each $ k = 1, \dots, n $ | $ \operatorname{span}(v_1, \dots, v_k) $ is invariant under $T$ for each $ k = 1, \dots, n $ | $ (z - \lambda_1)\dots(z - \lambda_m) $ for some $ \lambda_1, \dots, \lambda_m \in \mathbf{F} $ (repetitions allowed) | |
| Diagonalizable | $\exists$ a basis of $V$ consisting of eigenvectors of $T$ | $V = E(\lambda_1,T)\oplus\cdots\oplus E(\lambda_m,T)$ | $\sum_k \dim E(\lambda_k,T) = \dim V$ | $ (z - \lambda_1)\dots(z - \lambda_m) $ for some list of distinct numbers $ \lambda_1, \dots, \lambda_m \in \mathbf{F} $ |
Diagonalizable means the eigenspaces are as big as they can be — big enough to fill $V$. Each column says “fill $V$” in a different dialect: enough eigenvectors for a basis, eigenspaces summing directly to $V$, dimensions adding to $\dim V$, and — the min poly one — no eigenvalue needing a repeated factor to be annihilated (a repeat is exactly the symptom of an eigenspace that came up short, like the $(0,1)$ vector that $(T-5I)$ couldn’t kill in one step).
Commutativity
If $AB = BA$, then $B$ maps each eigenspace of $A$ into itself. Every $E(\lambda, A)$ is invariant under $B$.
For $v \in E(\lambda, A)$, $A(Bv) = B(Av) = B(\lambda v) = \lambda(Bv)$, so $Bv \in E(\lambda, A)$. ∎
$A$ and $B$ commute iff $B$ preserves every generalized eigenspace of $A$ and, within each, is compatible with $A$’s Jordan structure.
- Diagonalizable pair: commute $\iff$ simultaneously diagonalizable (common eigenbasis). ← the one to know.
- General pair: no clean iff; the operative fact is the lemma — commuting means each preserves the other’s eigenspaces.