Linear-Algebra-Done-Right · Unit 11 · Video 3 · Interactive Practice

p(T) and Why Every Complex Operator Has an Eigenvalue

IKey Formulas

Formula	Name	What it says
$p(T) = a_0 I + a_1 T + a_2 T^2 + \cdots + a_m T^m$	A polynomial applied to an operator	For $T \in \mathcal{L}(V)$ and $p(z) = a_0 + a_1 z + \cdots + a_m z^m \in \mathcal{P}(\mathbf{F})$ , replace $z$ by $T$ ; the constant term $a_0 = a_0 z^0$ becomes $a_0 T^0 = a_0 I$
$(pq)(T) = p(T)\,q(T) = q(T)\,p(T)$	Multiplicative property	For $p, q \in \mathcal{P}(\mathbf{F})$ and $T \in \mathcal{L}(V)$ : products of polynomials become products of operators, and two polynomials in the same $T$ commute
$\text{null}\, p(T)$ and $\text{range}\, p(T)$ are invariant under $T$	Invariant subspaces from polynomials	For $T \in \mathcal{L}(V)$ and $p \in \mathcal{P}(\mathbf{F})$ ; with $p(z) = z - \lambda$ this says $\text{null}\,(T - \lambda I)$ is invariant
$T\big(q(T)v\big) = \lambda\, q(T)v$ , $\ q(T)v \ne 0$	Existence of eigenvalues	If $\mathbf{F} = \mathbf{C}$ , $V$ is finite-dimensional, $V \ne \{0\}$ and $T \in \mathcal{L}(V)$ , then $T$ has an eigenvalue: for $v \ne 0$ , $v, Tv, \dots, T^n v$ ( $n = \dim V$ ) is dependent, a nonconstant $p$ of smallest degree with $p(T)v = 0$ factors as $p(z) = (z - \lambda)q(z)$ , and $q(T)v$ is an eigenvector for $\lambda$

Key Insight: Both hypotheses are needed. The rotation $T(w, z) = (-z, w)$ on $\mathbf{R}^2$ gives $p(z) = z^2 + 1$ , which has no real zero; multiplication by $z$ on the infinite-dimensional $\mathcal{P}(\mathbf{C})$ never produces a dependent list $1, z, z^2, \dots$ , so no $p$ is found at all.

IITwo Routes to $(pq)(T)v$

Applying $q(T)$ then $p(T)$ , or $p(T)$ then $q(T)$ , reaches the same vector $(pq)(T)v$ ; an operator $S$ outside the polynomials in $T$ need not.

Polynomials Second operator

III $\text{null}\, p(T)$ and $\text{range}\, p(T)$ Are Invariant

For $T(x, y) = (y, 2x + y)$ , $T$ maps $\text{null}\, p(T)$ into itself and $\text{range}\, p(T)$ into itself.

Polynomial

💡 In $\mathbf{R}^2$ a line invariant under $T$ is spanned by an eigenvector, so these lines are eigenvector lines: $\text{null}\,(T - 2I)$ for $2$ , and $\text{range}\,(T - 2I) = \text{null}\,(T + I)$ for $-1$ .

IVRunning the Existence Proof

From a nonzero $v$ , the dependent list $v, Tv, T^2 v$ yields $p$ , a zero $\lambda$ of $p$ , and the eigenvector $q(T)v$ .

Operator

VQuiz Questions

Problem 1 · Compute $p(T)$

Given: $T \in \mathcal{L}(\mathbf{R}^2)$ , $T(x, y) = (2x + y,\ 3y)$ , and $p(z) = z^2 - 4z + 3$ — find $p(T)$ .

✅ Correct!

T^2(x, y) = (4x + 5y,\ 9y)

, so

T^2 - 4T + 3I

sends

(x, y)

(-x + y,\ 0)

❌ Not quite. The constant

3

was added to every entry of the matrix of

T^2 - 4T

, which is

\begin{pmatrix} -4 & 1 \\ 0 & -3 \end{pmatrix}

. The constant term of

p

becomes

3I

, which adds

3x

to the first coordinate and

3y

to the second, nothing else.

❌ Not quite.

T^2

T

composed with itself, not the matrix with each entry squared:

T^2(x, y) = T(2x + y,\ 3y) = (4x + 5y,\ 9y)

❌ Not quite. Using only

p(2) = -1

and

p(3) = 0

drops the

y

-term in the first coordinate:

T^2

contributes

5y

and

-4T

contributes

-4y

, leaving

+y

Show solution

Compose $T$ with itself:

T^2(x, y) = T(2x + y,\ 3y) = \big(2(2x + y) + 3y,\ 9y\big) = (4x + 5y,\ 9y).

Replace $z$ by $T$ and the constant $3$ by $3I$ :

p(T)(x, y) = (4x + 5y,\ 9y) - 4(2x + y,\ 3y) + 3(x, y) = (-x + y,\ 0).

In matrix form, $p(T) = \begin{pmatrix} -1 & 1 \\ 0 & 0 \end{pmatrix}$ .

Problem 2 · Which Operators Commute?

Given: $T, S \in \mathcal{L}(\mathbf{R}^2)$ with $T(x, y) = (y,\ -4x)$ and $S(x, y) = (x,\ 0)$ — which statement is true?

✅ Correct!

T

and

T^2 + 3I

are both polynomials in

T

(

z

and

z^2 + 3

), so they commute;

S

is not a polynomial in

T

, and here it does not commute with

T

❌ Not quite. Operators on the same space need not commute, just as

DM \ne MD

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

. Test

(1, 0)

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

, but

ST(1, 0) = S(0, -4) = (0, 0)

❌ Not quite. With

p(z) = z

and

q(z) = z^2 + 3

, the multiplicative property gives

p(T)q(T) = (pq)(T) = (qp)(T) = q(T)p(T)

: polynomials in the same operator always commute.

❌ Not quite. The order of composition is reversed.

TS

applies

S

first:

TS(1, 0) = T\big(S(1, 0)\big) = T(1, 0) = (0, -4)

Show solution

Polynomials in $T$ . $T = p(T)$ with $p(z) = z$ , and $T^2 + 3I = q(T)$ with $q(z) = z^2 + 3$ . By the multiplicative property, $p(T)q(T) = (pq)(T) = (qp)(T) = q(T)p(T)$ ; both sides equal $T^3 + 3T$ .

$T$ and $S$ . Apply each composition to $(1, 0)$ , rightmost operator first:

TS(1, 0) = T(1, 0) = (0, -4), \qquad ST(1, 0) = S(0, -4) = (0, 0).

The outputs differ, so $TS \ne ST$ .

Problem 3 · Null Space and Range of $p(T)$

Given: $T \in \mathcal{L}(\mathbf{R}^2)$ , $T(x, y) = (2x + y,\ 3y)$ , and $p(z) = z - 3$ — find $\text{null}\, p(T)$ and $\text{range}\, p(T)$ , both of which are invariant under $T$ .

What is $\text{null}\, p(T)$ ?

What is $\text{range}\, p(T)$ ?

✅ Correct!

p(T)(x, y) = (-x + y,\ 0)

. Invariance checks out:

T(1, 1) = (3, 3)

stays in

\text{null}\, p(T)

, and

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

stays in

\text{range}\, p(T)

❌ Check the null space. Solve

p(T)(x, y) = (-x + y,\ 0) = (0, 0)

: the only condition is

y = x

❌ Check the range. Every output

(-x + y,\ 0)

has second coordinate

0

, and every first coordinate occurs.

❌ Not quite. The null space and the range have been swapped.

\text{null}\, p(T)

is where

p(T)

gives

0

;

\text{range}\, p(T)

is the set of outputs of

p(T)

❌ Not quite.

(-1, 1)

is the first row of the matrix of

p(T)

. The range is spanned by the columns,

(-1, 0)

and

(1, 0)

Show solution

p(T)(x, y) = T(x, y) - 3(x, y) = (2x + y - 3x,\ 3y - 3y) = (-x + y,\ 0).

Null space. $(-x + y,\ 0) = (0, 0)$ exactly when $y = x$ , so $\text{null}\, p(T) = \text{span}\{(1, 1)\}$ .

Range. The outputs are $(t, 0)$ with $t = -x + y$ arbitrary, so $\text{range}\, p(T) = \text{span}\{(1, 0)\}$ .

Invariance. $T(1, 1) = (3, 3) = 3(1, 1)$ and $T(1, 0) = (2, 0) = 2(1, 0)$ , so $T$ maps each line into itself, as the theorem guarantees. Each line is spanned by an eigenvector, with eigenvalues $3$ and $2$ .

Problem 4 · The Proof on $\mathcal{P}_2(\mathbf{C})$ , and Where It Fails

Given: $D \in \mathcal{L}\big(\mathcal{P}_2(\mathbf{C})\big)$ , $Df = f'$ , and the starting vector $v = z^2$ . Run the existence proof: find the nonconstant $p$ of smallest degree with $p(D)v = 0$ and the eigenvector $q(D)v$ it produces. Then let $M \in \mathcal{L}\big(\mathcal{P}(\mathbf{C})\big)$ , $(Mf)(z) = z f(z)$ , with $v = 1$ .

Which $p$ does the proof find for $D$ ?

Which eigenvector $q(D)v$ results?

Why does the proof fail for $M$ on $\mathcal{P}(\mathbf{C})$ ?

✅ Correct! For

D

p(z) = z^3 = (z - 0)z^2

, so

\lambda = 0

and

q(D)v = D^2 z^2 = 2

, a nonzero constant with

D(2) = 0

. For

M

the list

1, z, z^2, \dots

never becomes dependent, which is possible only because

\mathcal{P}(\mathbf{C})

is infinite-dimensional.

❌ Check $p$ . For

p

of degree at most

2

p(D)z^2 = a_0 z^2 + 2a_1 z + 2a_2

, which is

0

only when

p = 0

❌ Not quite.

D^2 z^2 = 2 \ne 0

. Two derivatives leave a constant; a third is needed to reach

0

❌ Not quite.

Dz^2 = 2z

is not

2z^2

, so

(D - 2I)z^2 = 2z - 2z^2 \ne 0

;

z^2

is not an eigenvector.

❌ Check $q$ .

p(z) = z^3

has the zero

\lambda = 0

and splits as

(z - 0)\,q(z)

with

q(z) = z^2

; compute

q(D)z^2 = D^2 z^2

❌ Not quite. An eigenvector is nonzero by definition, and the proof shows

q(D)v \ne 0

because

\deg q < \deg p

❌ Check the failure for $M$ . The proof's first step needs the list

v, Mv, M^2 v, \dots

to become linearly dependent.

❌ Not quite.

M1 = z

, which is not

0 \cdot 1 = 0

. In fact

\deg(Mf) = \deg f + 1

for

f \ne 0

, so

Mf

is never a scalar multiple of

f

M

has no eigenvalues.

❌ Not quite. Over

\mathbf{C}

every nonconstant polynomial has a zero. Here no

p

is found at all:

p(M)1 = p

, which is

0

only when

p = 0

❌ Not quite.

M

is linear:

z(f + g) = zf + zg

and

z(cf) = c(zf)

. The obstacle is dimension, not linearity.

Show solution

The list. $\dim \mathcal{P}_2(\mathbf{C}) = 3$ , so take $v, Dv, D^2 v, D^3 v$ :

z^2,\quad 2z,\quad 2,\quad 0.

Smallest degree. If $p(z) = a_0 + a_1 z + a_2 z^2$ , then $p(D)z^2 = a_0 z^2 + 2a_1 z + 2a_2$ , which is $0$ only if $a_0 = a_1 = a_2 = 0$ . Degree $3$ works: $D^3 z^2 = 0$ . So $p(z) = z^3$ (any nonzero multiple gives the same result).

Factor. The zero is $\lambda = 0$ , and $p(z) = (z - 0)\,q(z)$ with $q(z) = z^2$ . Then $q(D)v = D^2 z^2 = 2 \ne 0$ and $D(2) = 0 = 0 \cdot 2$ , so $2$ is an eigenvector of $D$ with eigenvalue $0$ .

Multiplication by $z$ . Starting from $v = 1$ : $M^k 1 = z^k$ , and $1, z, \dots, z^n$ is linearly independent for every $n$ . For any nonzero polynomial $p$ , $p(M)1 = p \ne 0$ , so the first step never produces a polynomial. This does not contradict the theorem, because $\mathcal{P}(\mathbf{C})$ is infinite-dimensional — and $M$ indeed has no eigenvalues.

Solved: 0 / 4

p(T) and Why Every Complex Operator Has an Eigenvalue

IKey Formulas

IITwo Routes to (pq)(T)v(pq)(T)v(pq)(T)v

IIInull p(T)\text{null}\, p(T)nullp(T) and range p(T)\text{range}\, p(T)rangep(T) Are Invariant

IVRunning the Existence Proof

VQuiz Questions

IITwo Routes to $(pq)(T)v$

III $\text{null}\, p(T)$ and $\text{range}\, p(T)$ Are Invariant