Linear-Algebra-Done-Right · Unit 11 · Video 2 · Interactive Practice

Eigenvectors, T − λI and At Most dim V Eigenvalues

IKey Formulas

Statement	Name	Hypotheses
$\lambda$ eigenvalue $\iff T - \lambda I$ not injective $\iff$ not surjective $\iff$ not invertible	Eigenvalue test	$V$ finite-dimensional, $T \in \mathcal{L}(V)$ , $\lambda \in \mathbf{F}$
eigenvectors for $\lambda$ $= \operatorname{null}(T - \lambda I) \setminus \{0\}$	Eigenvectors: $v \neq 0$ with $Tv = \lambda v$	$T \in \mathcal{L}(V)$ , $\lambda$ an eigenvalue of $T$
$v_1, \ldots, v_m$ eigenvectors for distinct $\lambda_1, \ldots, \lambda_m$ $\Rightarrow$ linearly independent	Independence of eigenvectors	$T \in \mathcal{L}(V)$ ; $V$ need not be finite-dimensional
number of distinct eigenvalues of $T$ $\leq \dim V$	Eigenvalue bound	$V$ finite-dimensional, $T \in \mathcal{L}(V)$

Key Insight: The field decides existence, the dimension caps the count: $T(w, z) = (-z, w)$ has no eigenvalues over $\mathbf{R}$ and the two eigenvalues $\pm i$ over $\mathbf{C}$ , the most any operator on a $2$ -dimensional space can have.

IIIs $Tv$ on the Line Through $v$ ?

A nonzero $v$ is an eigenvector exactly when $Tv$ lands on the line $\operatorname{span}(v)$ .

Operator on R²

IIIWhere $T - \lambda I$ Stops Being Injective

$\lambda$ is an eigenvalue exactly when $T - \lambda I$ flattens the unit square, sending some $v \neq 0$ to $0$ .

Operator on R² λ 0

💡 The equivalence of the four conditions needs $\dim V < \infty$ : the backward shift on $\mathbf{F}^\infty$ is surjective, yet $0$ is an eigenvalue with eigenvector $(1, 0, 0, \ldots)$ .

IVWhy Distinct Eigenvalues Force Independence

Applying $T - \lambda_k I$ deletes the $v_k$ term and scales each other $v_j$ term by $\lambda_j - \lambda_k$ , nonzero when the eigenvalues are distinct.

λ₁ 4 λ₂ 2 λ₃ −1 Step 0

VQuiz Questions

Problem 1 · Test $T - \lambda I$ for Injectivity

Given: $T \in \mathcal{L}(\mathbf{R}^2)$ , $T(w, z) = (4w + 2z,\ w + 3z)$ . Find a $\lambda$ for which $T - \lambda I$ is not injective, and an eigenvector for it.

For which $\lambda$ is $T - \lambda I$ not injective?

Which vector is an eigenvector for that $\lambda$ ?

✅ Correct!

(T - 5I)(w, z) = (-w + 2z,\ w - 2z)

sends

(2, 1)

0

, so

5

is an eigenvalue and

T(2, 1) = (10, 5) = 5(2, 1)

❌ Not quite. The coefficients

4

and

3

in the formula are not automatically eigenvalues. For this

\lambda

, solving

(T - \lambda I)(w, z) = 0

gives only

(0, 0)

, so

T - \lambda I

is injective.

❌ Not quite.

(T + 2I)(w, z) = (6w + 2z,\ w + 5z)

: the second coordinate gives

w = -5z

, then the first gives

-28z = 0

, so only

(0, 0)

goes to

0

— injective.

❌ Not quite. Solve

(T - \lambda I)(w, z) = 0

for each candidate;

\lambda

qualifies only if a nonzero solution exists.

❌ Coordinates swapped.

T(1, 2) = (8, 7)

, which is not a multiple of

(1, 2)

. The equation

w - 2z = 0

gives

w = 2z

, so the first coordinate is twice the second.

❌ That eigenvector belongs to the other eigenvalue.

T(1, -1) = (2, -2) = 2(1, -1)

: it is an eigenvector for

2

, not for

5

❌ Not quite.

T(1, 1) = (6, 4)

is not a multiple of

(1, 1)

, so

(1, 1)

is not an eigenvector of

T

at all.

❌ Not quite. An eigenvector for

5

is a nonzero vector in

\operatorname{null}(T - 5I)

Show solution

Subtract $\lambda$ times each coordinate:

(T - \lambda I)(w, z) = \big((4 - \lambda)w + 2z,\ w + (3 - \lambda)z\big).

Check the candidates. For $\lambda = 3$ : $(w + 2z,\ w) = 0$ forces $w = 0$ , then $z = 0$ . For $\lambda = 4$ : $(2z,\ w - z) = 0$ forces $z = 0$ , then $w = 0$ . For $\lambda = -2$ : $(6w + 2z,\ w + 5z) = 0$ gives $w = -5z$ and $-28z = 0$ , so $z = w = 0$ . Each of these is injective, so none of $3, 4, -2$ is an eigenvalue.

For $\lambda = 5$ :

(T - 5I)(w, z) = (-w + 2z,\ w - 2z) = 0 \iff w = 2z.

So $\operatorname{null}(T - 5I) = \operatorname{span}\big((2, 1)\big) \neq \{0\}$ : $T - 5I$ is not injective, and $5$ is an eigenvalue. The eigenvectors for $5$ are $t(2, 1)$ with $t \neq 0$ . Check: $T(2, 1) = (8 + 2,\ 2 + 3) = (10, 5) = 5(2, 1)$ .

(The other eigenvalue is $2$ , with eigenvectors $t(1, -1)$ , $t \neq 0$ ; two eigenvalues on $\mathbf{R}^2$ is the maximum.)

Problem 2 · A Scaled Rotation, Two Fields

Given: $T(w, z) = (-2z,\ 2w)$ on $\mathbf{F}^2$ . Find its eigenvalues when $\mathbf{F} = \mathbf{R}$ and when $\mathbf{F} = \mathbf{C}$ .

Eigenvalues over $\mathbf{R}$ :

Eigenvalues over $\mathbf{C}$ :

✅ Correct! The coordinate equations force

\lambda^2 = -4

: no real solution, and

\lambda = \pm 2i

over

\mathbf{C}

❌ Not quite.

T

scales by

2

but also turns every arrow a quarter turn, so in

\mathbf{R}^2

Tv

with

v \neq 0

lies on

\operatorname{span}(v)

. Algebraically, over either field:

\lambda = 2

would need

-2z = 2w

and

2w = 2z

, and

\lambda = -2

would need

-2z = -2w

and

2w = -2z

; each pair forces

w = z = 0

❌ Not quite.

T - 0I = T

, and

T(w, z) = (-2z, 2w) = 0

only for

(0, 0)

, so

T

is injective and

0

is not an eigenvalue.

❌ The scale factor matters.

T(1, -i) = (2i, 2)

, while

i(1, -i) = (i, 1)

. Solving gives

\lambda^2 = -4

, not

\lambda^2 = -1

❌ The field changed. Over

\mathbf{C}

the equation

\lambda^2 = -4

has solutions, so

T

does have eigenvalues.

❌ Not quite. Write

T(w, z) = \lambda(w, z)

coordinate by coordinate and eliminate

w

❌ Not quite. Write

T(w, z) = \lambda(w, z)

coordinate by coordinate and eliminate

w

Show solution

$\lambda$ is an eigenvalue when $T(w, z) = \lambda(w, z)$ has a solution with $(w, z) \neq (0, 0)$ . Comparing coordinates:

-2z = \lambda w, \qquad 2w = \lambda z.

If $z = 0$ , the second equation gives $w = 0$ , which is excluded; so $z \neq 0$ . Substituting $w = \tfrac{\lambda}{2} z$ into the first equation:

-2z = \lambda \cdot \tfrac{\lambda}{2} z \;\Rightarrow\; \lambda^2 z = -4z \;\Rightarrow\; \lambda^2 = -4.

Over $\mathbf{R}$ : $\lambda^2 = -4$ has no solution, so $T$ has no eigenvalues (geometrically, $T$ rotates by $90\degree$ and doubles lengths, so $Tv$ is perpendicular to $v$ ).

Over $\mathbf{C}$ : $\lambda = 2i$ or $\lambda = -2i$ . For $\lambda = 2i$ : $z = \tfrac{2w}{2i} = -iw$ , eigenvector $(1, -i)$ , and $T(1, -i) = (2i, 2) = 2i(1, -i)$ . For $\lambda = -2i$ : eigenvector $(1, i)$ , and $T(1, i) = (-2i, 2) = -2i(1, i)$ .

Problem 3 · Eigenvalues and an Eigenvector over $\mathbf{C}$

Given: $T(w, z) = (w - z,\ w + z)$ on $\mathbf{C}^2$ . Find its eigenvalues and an eigenvector for $\lambda = 1 + i$ .

The eigenvalues of $T$ :

An eigenvector for $\lambda = 1 + i$ :

✅ Correct!

T(1, -i) = (1 + i,\ 1 - i)

and

(1 + i)(1, -i) = (1 + i,\ -i - i^2) = (1 + i,\ 1 - i)

❌ Not quite. Unlike

(w, z) \mapsto (-z, w)

, this

T

also contains

w \mapsto w

in the first coordinate. The equations give

(1 - \lambda)w = z

and

w = (\lambda - 1)z

, so

\lambda - 1

, not

\lambda

, squares to

-1

❌ Not quite.

T

stretches lengths by

\sqrt{2}

, but the eigenvalue equation is

(\lambda - 1)^2 = -1

, whose solutions are

1 \pm i

(each of absolute value

\sqrt{2}

❌ That is the answer over $\mathbf{R}$ . Here

\mathbf{F} = \mathbf{C}

, and

(\lambda - 1)^2 = -1

has two solutions.

❌ Not quite. Compare coordinates in

T(w, z) = \lambda(w, z)

and eliminate one variable.

❌ That eigenvector belongs to $1 - i$ .

T(1, i) = (1 - i,\ 1 + i) = (1 - i)(1, i)

❌ Not an eigenvector. Its image under

T

is not a scalar multiple of it:

T(1, 1) = (0, 2)

and

T(1 + i, 1) = (i,\ 2 + i)

❌ Not quite. Solve the first coordinate of

(T - (1 + i)I)(w, z) = 0

for

z

Show solution

Compare coordinates in $T(w, z) = \lambda(w, z)$ with $(w, z) \neq (0, 0)$ :

w - z = \lambda w, \qquad w + z = \lambda z,

that is, $z = (1 - \lambda)w$ and $w = (\lambda - 1)z$ . If $z = 0$ then $w = 0$ , excluded, so $z \neq 0$ . Substituting the second into the first:

z = (1 - \lambda)(\lambda - 1)z = -(\lambda - 1)^2 z \;\Rightarrow\; (\lambda - 1)^2 = -1 \;\Rightarrow\; \lambda = 1 \pm i.

For $\lambda = 1 + i$ : $z = (1 - \lambda)w = -iw$ , so the eigenvectors are $(w, -iw)$ with $w \neq 0$ ; taking $w = 1$ gives $(1, -i)$ . Check:

T(1, -i) = (1 + i,\ 1 - i), \qquad (1 + i)(1, -i) = (1 + i,\ -i - i^2) = (1 + i,\ 1 - i).\ \checkmark

For $\lambda = 1 - i$ the same steps give $z = iw$ and the eigenvector $(1, i)$ . Since $1 + i \neq 1 - i$ , the list $(1, -i), (1, i)$ is linearly independent, a basis of $\mathbf{C}^2$ .

Problem 4 · What Distinctness Buys

Given: $V$ is a $3$ -dimensional vector space over $\mathbf{F}$ and $T \in \mathcal{L}(V)$ . Which statement must be true?

✅ Correct! Eigenvectors for distinct eigenvalues are linearly independent, and a linearly independent list of length

\dim V = 3

is a basis.

❌ The word "distinct" matters. If

v

is an eigenvector for

\lambda

, so is

2v

, and the list

v, 2v

is linearly dependent.

❌ The bound is an upper bound. At most

\dim V = 3

; the identity operator

I

has the single eigenvalue

1

❌ That is the converse, and it fails. For

T = I

, any basis

v_1, v_2, v_3

is an independent list of eigenvectors, all for the eigenvalue

1

Show solution

(a) True. By the theorem, eigenvectors $v_1, v_2, v_3$ for distinct eigenvalues form a linearly independent list. A linearly independent list in $V$ whose length equals $\dim V$ is a basis of $V$ , and here the length is $3 = \dim V$ .

(b) False. If $Tv = \lambda v$ with $v \neq 0$ , then $T(2v) = \lambda(2v)$ , so $2v$ is also an eigenvector. The relation $2 \cdot v + (-1) \cdot (2v) = 0$ has the nonzero coefficients $2$ and $-1$ , so the list $v, 2v$ is linearly dependent.

(c) False. The theorem gives at most $\dim V = 3$ distinct eigenvalues, not exactly $3$ . The identity operator has only the eigenvalue $1$ , since $Iv = v$ for every $v$ and $Iv = \lambda v$ with $v \neq 0$ forces $\lambda = 1$ .

(d) False. This reverses the theorem. With $T = I$ , every nonzero vector is an eigenvector for $1$ , so a basis $v_1, v_2, v_3$ is a linearly independent list of eigenvectors whose eigenvalues are all equal.

Solved: 0 / 4

Eigenvectors, T − λI and At Most dim V Eigenvalues

IKey Formulas

IIIs TvTvTv on the Line Through vvv?

IIIWhere T−λIT - \lambda IT−λI Stops Being Injective

IVWhy Distinct Eigenvalues Force Independence

VQuiz Questions

IIIs $Tv$ on the Line Through $v$ ?

IIIWhere $T - \lambda I$ Stops Being Injective