Linear-Algebra-Done-Right · Unit 11 · Video 1 · Interactive Practice

Invariant Subspaces and Eigenvalues: Tv = λv

IKey Formulas

Formula	Name	What it says
$\mathcal{L}(V) = \mathcal{L}(V, V)$	Operators	An operator is a linear map from $V$ to itself, so its outputs can be fed back in: $Tv$ , $T(Tv)$ , …
$Tu \in U$ for every $u \in U$	$U$ invariant under $T$	For $T \in \mathcal{L}(V)$ and a subspace $U$ of $V$ ; equivalently $T\|_U \in \mathcal{L}(U)$ . Always invariant: $\{0\}$ , $V$ , $\operatorname{null} T$ , $\operatorname{range} T$
$\operatorname{span}(v)$ invariant $\iff Tv = \lambda v$ for some $\lambda \in \mathbf{F}$	One-dimensional invariant subspaces	For $T \in \mathcal{L}(V)$ and $v \in V$ with $v \ne 0$ ; every one-dimensional subspace of $V$ is $\operatorname{span}(v)$ for such a $v$
$Tv = \lambda v$ for some $v \in V$ , $v \ne 0$	$\lambda \in \mathbf{F}$ is an eigenvalue of $T$	$v \ne 0$ is essential, since $T0 = \lambda 0$ for every $\lambda$ ; on a real vector space only real $\lambda$ count

Key Insight: $V$ has a one-dimensional subspace invariant under $T$ exactly when $T$ has an eigenvalue $\lambda$ , and on that line $T$ is multiplication by $\lambda$ : applying $T$ $n$ times multiplies every vector on it by $\lambda^n$ .

IIVisualization 1 — Invariant Lines in the Plane

In $\mathbf{R}^2$ , the line $\operatorname{span}(v)$ with $v \ne 0$ is invariant under $T$ exactly when $Tv$ lands on it: $Tv = \lambda v$ .

Operator

IIIVisualization 2 — Applying the Opening Operator Again and Again

For $T(x, y, z) = (7x + 3z,\; 3x + 6y + 9z,\; -6y)$ , applying $T$ $n$ times multiplies every vector of $\operatorname{span}(3, 1, -1)$ by $6^n$ .

Start n 0

IVVisualization 3 — Differentiation and Its Invariant Subspaces

Which spans of the monomials $1$ , $x$ , …, $x^4$ does differentiation, $Dp = p'$ , map into themselves?

VQuiz Questions

Problem 1 · Find the Invariant Line

Given: $T \in \mathcal{L}(\mathbf{R}^3)$ defined by $T(x, y, z) = (5x + y - 2z,\; x + 4y + z,\; 2x + y + z)$ — find which of the lines below is invariant under $T$ , and the eigenvalue it gives.

Which line is invariant under $T$ ?

What does $T$ multiply that line by?

✅ Correct!

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

with

(1, 2, 1) \ne 0

, so

\operatorname{span}(1, 2, 1)

is invariant and

5

is an eigenvalue of

T

❌ Not a multiple.

T(1, 0, 0) = (5, 1, 2)

; a multiple of

(1, 0, 0)

must have second and third coordinates

0

❌ Not a multiple.

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

, but every multiple of

(1, 1, 1)

has three equal coordinates.

❌ Not a multiple.

T(0, 1, 0) = (1, 4, 1)

; a multiple of

(0, 1, 0)

must have first and third coordinates

0

❌ That is one coordinate of $Tv$ .

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

;

\lambda

is the ratio of

Tv

v

, coordinate by coordinate:

10 / 2 = 5

❌ That adds the coordinates of $Tv$ .

Tv = \lambda v

compares

Tv

with

v

coordinate by coordinate:

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

❌ A coefficient of the formula is not the eigenvalue. Compute

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

and compare it with

(1, 2, 1)

❌ Not quite. Compute

T

of each spanning vector and test whether the result is a multiple of that vector.

❌ Not quite. Find the

\lambda

with

T(1, 2, 1) = \lambda (1, 2, 1)

Show solution

For $v \ne 0$ , the line $\operatorname{span}(v)$ is invariant under $T$ exactly when $Tv = \lambda v$ for some $\lambda \in \mathbf{R}$ . Apply $T$ to each spanning vector:

T(1, 0, 0) = (5, 1, 2), \qquad T(0, 1, 0) = (1, 4, 1), \qquad T(1, 1, 1) = (4, 6, 4)

None of these is a multiple of the vector it came from. For $(1, 2, 1)$ :

T(1, 2, 1) = (5 + 2 - 2,\; 1 + 8 + 1,\; 2 + 2 + 1) = (5, 10, 5) = 5(1, 2, 1)

So $\operatorname{span}(1, 2, 1)$ is invariant under $T$ , on it $T$ is multiplication by $5$ , and since $(1, 2, 1) \ne 0$ , the number $5$ is an eigenvalue of $T$ .

Problem 2 · An Invariant Plane With No Invariant Line in It

Given: $T \in \mathcal{L}(\mathbf{R}^3)$ defined by $T(x, y, z) = (-y,\; x,\; 2z)$ , with $\mathbf{F} = \mathbf{R}$ — find the true statement.

✅ Correct!

T(x, y, 0) = (-y, x, 0)

stays in the plane, but

(-y, x, 0) = \lambda(x, y, 0)

forces

x = y = 0

; the only real eigenvalue is

2

, from

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

❌ The definition needs $v \ne 0$ .

T0 = \lambda 0

holds for every operator and every

\lambda

, which is exactly why the zero vector is excluded.

❌ That tests a line, not the plane.

(0, 1, 0)

lies in the plane

z = 0

; a plane is invariant when

T

keeps all of its vectors in it, and

T(x, y, 0) = (-y, x, 0)

does.

❌ Eigenvalues must lie in $\mathbf{F}$ . The computation is right in

\mathbf{C}^3

, but here

T

acts on

\mathbf{R}^3

i \notin \mathbf{R}

, and

(1, -i, 0)

is not even in

\mathbf{R}^3

❌ Not quite. Test the plane

z = 0

and the lines inside it separately.

Show solution

The plane. Every vector of $U = \{(x, y, 0) : x, y \in \mathbf{R}\}$ goes to $T(x, y, 0) = (-y, x, 0) \in U$ , so $U$ is invariant under $T$ .

Lines inside it. Suppose $(x, y, 0) \ne 0$ and $T(x, y, 0) = \lambda (x, y, 0)$ with $\lambda \in \mathbf{R}$ . Then $-y = \lambda x$ and $x = \lambda y$ , so $x = \lambda y = -\lambda^2 x$ , that is $(1 + \lambda^2) x = 0$ . Since $1 + \lambda^2 > 0$ , this gives $x = 0$ , and then $-y = \lambda x = 0$ gives $y = 0$ , a contradiction. No line in $U$ is invariant.

All eigenvalues. If $T(x, y, z) = \lambda(x, y, z)$ with $(x, y, z) \ne 0$ , the same argument gives $x = y = 0$ , so $z \ne 0$ and $2z = \lambda z$ gives $\lambda = 2$ . Indeed $T(0, 0, 1) = 2(0, 0, 1)$ , so $2$ is the only eigenvalue of $T$ .

Option (a) ignores the requirement $v \ne 0$ ; option (d) uses a scalar and a vector from $\mathbf{C}$ , which do not count for an operator on the real space $\mathbf{R}^3$ .

Problem 3 · Invariant Subspaces of Differentiation

Given: $D \in \mathcal{L}(\mathcal{P}_3(\mathbf{R}))$ defined by $Dp = p'$ — find an invariant span and the invariant $\mathcal{P}_m(\mathbf{R})$ .

Which subspace of $\mathcal{P}_3(\mathbf{R})$ is invariant under $D$ ?

Which of $\mathcal{P}_0(\mathbf{R})$ , $\mathcal{P}_1(\mathbf{R})$ , $\mathcal{P}_2(\mathbf{R})$ , $\mathcal{P}_3(\mathbf{R})$ are invariant under $D$ ?

✅ Correct!

D

sends

1

x + 3

x^2 - 2x

0

1

2x - 2 = 2(x + 3) - 8

, all in the span; and

\deg p \le m

gives

\deg p' \le m - 1

, so every

\mathcal{P}_m(\mathbf{R})

is invariant.

❌ Check every spanning vector.

D(x^2 + 1) = 2x

is in the span, but

D(x) = 1

is not:

a(x^2 + 1) + bx = 1

forces

a = b = 0

❌ Check $x^2$ too.

D(x^3) = 3x^2

stays, but

D(x^2) = 2x

is not a combination of

x^3

and

x^2

❌ Not a multiple. A one-dimensional span is invariant only if

D(x^3 - x) = 3x^2 - 1

is a multiple of

x^3 - x

, and the degrees already differ.

❌ Lowering the degree keeps you inside. If

\deg p \le m

then

\deg p' \le m

as well, so

p' \in \mathcal{P}_m(\mathbf{R})

❌ Those two are invariant, but not the only ones.

\operatorname{null} D

and

\operatorname{range} D

are always invariant;

\mathcal{P}_1(\mathbf{R})

and

\mathcal{P}_3(\mathbf{R})

are invariant too.

❌ Lowering the degree does not leave $\mathcal{P}_m(\mathbf{R})$ .

\mathcal{P}_m(\mathbf{R})

holds every polynomial of degree at most

m

, including those of smaller degree.

❌ Not quite. Apply

D

to each spanning polynomial and test whether the result is in the span.

❌ Not quite. Compare

\deg p'

with

\deg p

Show solution

To show $U = \operatorname{span}(u_1, \dots, u_k)$ is invariant it is enough to check each $Du_j \in U$ : by linearity, $D(c_1 u_1 + \cdots + c_k u_k) = c_1 Du_1 + \cdots + c_k Du_k$ is then a linear combination of vectors in $U$ .

(a) $D1 = 0$ , $D(x + 3) = 1$ , $D(x^2 - 2x) = 2x - 2 = 2(x + 3) - 8 \cdot 1$ , all in $U$ . So $U$ is invariant (in fact $U = \mathcal{P}_2(\mathbf{R})$ , since its three vectors have degrees $0$ , $1$ , $2$ ).

(b) $D(x) = 1$ ; if $a(x^2 + 1) + bx = 1$ , the $x^2$ and $x$ coefficients give $a = b = 0$ , and then the constant term is $0 \ne 1$ . Not invariant.

(c) $D(x^2) = 2x$ , and $ax^3 + bx^2$ never equals $2x$ . Not invariant.

(d) $D(x^3 - x) = 3x^2 - 1$ has degree $2$ , while nonzero multiples of $x^3 - x$ have degree $3$ . Not invariant.

The $\mathcal{P}_m(\mathbf{R})$ . If $\deg p \le m$ then $p'$ has degree at most $m - 1$ (or $p' = 0$ ), so $p' \in \mathcal{P}_m(\mathbf{R})$ . All four are invariant; $\mathcal{P}_0(\mathbf{R}) = \operatorname{null} D$ , $\mathcal{P}_2(\mathbf{R}) = \operatorname{range} D$ , and $\mathcal{P}_3(\mathbf{R})$ is the whole space.

Problem 4 · The Eigenvalues of Differentiation

Given: $D \in \mathcal{L}(\mathcal{P}_3(\mathbf{R}))$ defined by $Dp = p'$ — find all eigenvalues of $D$ .

✅ Correct! For

\lambda \ne 0

and

p \ne 0

\lambda p

has degree

\deg p

while

p'

has smaller degree or is

0

; for

\lambda = 0

, the nonzero polynomial

1

has

D1 = 0

, which is

\lambda \cdot 1

❌ Those are the factors in $D(x^k) = kx^{k-1}$ .

kx^{k-1}

is not a multiple of

x^k

when

k \ge 1

; an eigenvalue needs

Dp = \lambda p

for the same

p

❌ $e^{\lambda x}$ is not a polynomial. The operator acts on

\mathcal{P}_3(\mathbf{R})

, so the nonzero vector

p

Dp = \lambda p

must lie in

\mathcal{P}_3(\mathbf{R})

❌ Constants are the exception.

p = 1

is a nonzero polynomial with

D1 = 0

, which is

\lambda p

for

\lambda = 0

, so

0

is an eigenvalue; nothing requires

\lambda \ne 0

❌ Not quite. Compare the degrees of

p'

and

\lambda p

Show solution

Suppose $Dp = \lambda p$ with $p \in \mathcal{P}_3(\mathbf{R})$ , $p \ne 0$ , and let $n = \deg p$ .

If $\lambda \ne 0$ : $\lambda p$ is nonzero of degree $n$ . But $p'$ is $0$ when $n = 0$ and has degree $n - 1$ when $n \ge 1$ . Either way $p' \ne \lambda p$ , so no $\lambda \ne 0$ is an eigenvalue.

If $\lambda = 0$ : the equation is $p' = 0$ , which holds for every nonzero constant, e.g. $p = 1$ . So $0$ is an eigenvalue.

The only eigenvalue of $D$ is $0$ , and the invariant line it gives is $\operatorname{span}(1) = \mathcal{P}_0(\mathbf{R}) = \operatorname{null} D$ .

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