Linear-Algebra-Done-Right · Unit 11 · Video 6 · Interactive Practice

Why Odd Dimension Forces a Real Eigenvalue

IKey Formulas

Statement	Name	Hypotheses and meaning
$\dim \operatorname{null}(T^2 + bT + cI)$ is even	Even-dimensional null space	$\mathbf{F} = \mathbf{R}$ , $V$ finite-dimensional, $T \in \mathcal{L}(V)$ , $b, c \in \mathbf{R}$ with $b^2 < 4c$ ; the dimension $0$ counts as even
$\lambda^2 + b\lambda + c = \left(\lambda + \frac{b}{2}\right)^2 + \left(c - \frac{b^2}{4}\right) > 0$	No eigenvalue when $T^2 + bT + cI = 0$	For every $\lambda \in \mathbf{R}$ when $b^2 < 4c$ ; then $Tv = \lambda v$ gives $0 = (\lambda^2 + b\lambda + c)v$ , so $v = 0$
$\dim V = \dim \operatorname{null}(T^2 + bT + cI) + \dim \operatorname{range}(T^2 + bT + cI)$	Parity count	Fundamental theorem of linear maps; with $\dim V$ odd and the null space even, the range is odd-dimensional and invariant under $T$
$\dim V$ odd $\implies$ $T$ has an eigenvalue	Odd dimension forces an eigenvalue	$V$ finite-dimensional, $T \in \mathcal{L}(V)$ ; over $\mathbf{C}$ every $V \neq \{0\}$ already works, so the new case is $\mathbf{F} = \mathbf{R}$

Key Insight: When the minimal polynomial $p$ has no real zero, it has a quadratic factor $x^2 + bx + c$ with $b^2 < 4c$ . That factor's null space uses up an even number of dimensions, so when $\dim V$ is odd, an odd-dimensional invariant range is left, and the induction continues there.

II $w$ and $Tw$ Span an Invariant Plane

If $T^2 + bT + cI = 0$ with $b^2 < 4c$ , then $Tw$ never lies on the line through $w$ .

T w₁ 2 w₂ 1

IIIOdd = Even + Odd: Descending to an Eigenvalue

A quadratic factor's null space takes an even number of dimensions, so an odd-dimensional invariant range remains.

dim V dim null 2

💡 At each stage, $b$ and $c$ come from a quadratic factor of the minimal polynomial of the current restriction; if that polynomial has a real zero, the zero is already an eigenvalue and the descent stops early.

IVQuiz Questions

Problem 1 · Which Dimension Is Impossible?

Given: $T \in \mathcal{L}(\mathbf{R}^5)$ and the quadratic $x^2 + 2x + 5$ , so $b = 2$ , $c = 5$ and $b^2 = 4 < 20 = 4c$ . Find the value that $\dim \operatorname{null}(T^2 + 2T + 5I)$ can never take.

✅ Correct! Since

b^2 < 4c

, the null space of

T^2 + 2T + 5I

has even dimension, and

3

is odd.

❌ Not quite.

0

is even, and it happens: for

T = I

T^2 + 2T + 5I = 8I

is injective, so its null space is

\{0\}

❌ Not quite.

4

is even and fits inside

\mathbf{R}^5

: two planes on which

T^2 + 2T + 5I = 0

, plus one line on which it is not, give dimension

4

❌ Not quite.

2

is even: one plane on which

T^2 + 2T + 5I = 0

and

T = I

on the other three coordinates give dimension

2

❌ Not quite. Check whether

b^2 < 4c

, then use the parity of the null space.

Show solution

The hypotheses of the theorem hold: $\mathbf{F} = \mathbf{R}$ , $V = \mathbf{R}^5$ is finite-dimensional, and $b^2 = 4 < 20 = 4c$ . So $\dim \operatorname{null}(T^2 + 2T + 5I)$ is even, and as a subspace of $\mathbf{R}^5$ its dimension is at most $5$ . The possible values are $0$ , $2$ , $4$ ; the odd values $1$ , $3$ , $5$ never occur.

All three even values happen. Let $A(x_1, x_2) = (-x_1 - 2x_2,\; 2x_1 - x_2)$ on $\mathbf{R}^2$ . Then $(A + I)(x_1, x_2) = (-2x_2, 2x_1)$ , so $(A + I)^2 = -4I$ , which says $A^2 + 2A + I = -4I$ , that is,

A^2 + 2A + 5I = 0.

$T = I$ : $T^2 + 2T + 5I = 8I$ , null space $\{0\}$ , dimension $0$ .
$T$ acts as $A$ on $(x_1, x_2)$ and as $I$ on $x_3, x_4, x_5$ : on those coordinates the operator is multiplication by $8$ , so the null space is the $(x_1, x_2)$ -plane, dimension $2$ .
$T$ acts as $A$ on $(x_1, x_2)$ , as $A$ on $(x_3, x_4)$ , and as $I$ on $x_5$ : dimension $4$ .

So $3$ is the impossible value.

Problem 2 · Which Quadratic Guarantees Even Parity?

Find the quadratic $x^2 + bx + c$ for which $\dim \operatorname{null}(T^2 + bT + cI)$ is even for every operator $T$ on every finite-dimensional real vector space $V$ .

✅ Correct!

b^2 = 36 < 40 = 4c

, and

x^2 + 6x + 10 = (x + 3)^2 + 1 > 0

for every real

x

❌ The boundary does not count. Here

b^2 = 16 = 4c

, so the strict inequality fails:

x^2 + 4x + 4 = (x + 2)^2

, and

T = -2I

\mathbf{R}

gives

T^2 + 4T + 4I = 0

, a null space of dimension

1

❌ Not quite. With

c = -1

b^2 = 1 > -4 = 4c

, so the quadratic has the real zeros

\frac{-1 \pm \sqrt{5}}{2}

;

T = \lambda I

\mathbf{R}

with

\lambda

one of them has a null space of dimension

1

❌ Close, but compare carefully.

b^2 = 25 > 24 = 4c

, and

x^2 - 5x + 6 = (x - 2)(x - 3)

;

T = 2I

\mathbf{R}

gives a null space of dimension

1

❌ Not quite. The theorem needs the strict inequality

b^2 < 4c

Show solution

The theorem guarantees an even dimension whenever $b^2 < 4c$ . Compute $b^2 - 4c$ :

$x^2 + 4x + 4$ : $16 - 16 = 0$ , not negative.
$x^2 + x - 1$ : $1 + 4 = 5 > 0$ .
$x^2 - 5x + 6$ : $25 - 24 = 1 > 0$ .
$x^2 + 6x + 10$ : $36 - 40 = -4 < 0$ .

For each of the first three, the quadratic has a real zero $\lambda$ , and on $V = \mathbf{R}$ the operator $T = \lambda I$ gives $T^2 + bT + cI = (\lambda^2 + b\lambda + c)I = 0$ , a null space of dimension $1$ , which is odd. So none of them guarantees even parity.

For $x^2 + 6x + 10$ , the hypothesis $b^2 < 4c$ holds, so $\dim \operatorname{null}(T^2 + 6T + 10I)$ is even for every such $T$ . Completing the square, $x^2 + 6x + 10 = (x + 3)^2 + 1$ , which is positive at every real $x$ .

Problem 3 · A Plane Block Plus a Real Line

Given: $T \in \mathcal{L}(\mathbf{R}^3)$ with $T(x, y, z) = (x - 2y,\; 2x + y,\; 3z)$ and the quadratic $x^2 - 2x + 5$ ( $b = -2$ , $c = 5$ , $b^2 = 4 < 20$ ). Find $\dim \operatorname{null}(T^2 - 2T + 5I)$ and the real eigenvalue of $T$ .

What is dim null(T² − 2T + 5I)?

What is the real eigenvalue of T?

✅ Correct! The null space is the

xy

-plane and the range is the

z

-axis, an odd-dimensional invariant subspace on which

T

is multiplication by

3

❌ Not quite. On the

z

-axis,

T^2 - 2T + 5I

sends

(0, 0, z)

(9 - 6 + 5)(0, 0, z) = 8(0, 0, z) \neq 0

for

z \neq 0

, so

e_3

is not in the null space.

❌ Not quite. Since

b^2 < 4c

, the null space has even dimension, so

1

is impossible before any computation.

❌ Not quite. Compute on

(1, 0, 0)

T^2(1, 0, 0) = (-3, 4, 0)

and

-2T(1, 0, 0) = (-2, -4, 0)

, so adding

5(1, 0, 0)

gives

0

❌ Not quite.

1

appears on the diagonal of the matrix of

T

on the

xy

-plane, but

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

is not a multiple of

(1, 0, 0)

; on the

xy

-plane

T

has no eigenvalues because

\lambda^2 - 2\lambda + 5 = (\lambda - 1)^2 + 4 > 0

❌ Not quite.

5

is the coefficient

c

, not an eigenvalue; check

T(0, 0, 1)

instead.

❌ Close, but that is the wrong operator.

8

is how

T^2 - 2T + 5I

acts on the

z

-axis, the value

3^2 - 2 \cdot 3 + 5

;

T

itself acts there by

3

❌ Check the dimension. Apply

T^2 - 2T + 5I

to each standard basis vector.

❌ Check the eigenvalue. Look for a vector that

T

sends to a multiple of itself.

Show solution

Apply $T$ twice:

T^2(x, y, z) = T(x - 2y,\; 2x + y,\; 3z) = (-3x - 4y,\; 4x - 3y,\; 9z).

Then

(T^2 - 2T + 5I)(x, y, z) = \big((-3x - 4y) - 2(x - 2y) + 5x,\; (4x - 3y) - 2(2x + y) + 5y,\; 9z - 6z + 5z\big) = (0, 0, 8z).

So $\operatorname{null}(T^2 - 2T + 5I) = \{(x, y, 0)\}$ , of dimension $2$ (even, as the theorem requires), and $\operatorname{range}(T^2 - 2T + 5I) = \{(0, 0, z)\}$ , of dimension $1$ . The count $3 = 2 + 1$ is odd $=$ even $+$ odd.

The range is invariant under $T$ , and on it $T(0, 0, z) = 3(0, 0, z)$ . So $\lambda = 3$ is an eigenvalue with eigenvector $(0, 0, 1)$ . It is the only real one: if $Tv = \lambda v$ with $v = (x, y, 0) \neq 0$ in the plane, then $0 = (T^2 - 2T + 5I)v = (\lambda^2 - 2\lambda + 5)v$ , but $\lambda^2 - 2\lambda + 5 = (\lambda - 1)^2 + 4 > 0$ ; and a vector with $z \neq 0$ forces $\lambda = 3$ and then $x = y = 0$ .

Problem 4 · Where Must an Eigenvalue Exist?

Find the real vector space $V$ on which every operator $T \in \mathcal{L}(V)$ has an eigenvalue.

✅ Correct!

\mathcal{P}_2(\mathbf{R})

has basis

1, x, x^2

, so its dimension

3

is odd, and the theorem gives every operator on it an eigenvalue.

❌ Not quite.

\dim \mathbf{R}^4 = 4

is even, and

T(x_1, x_2, x_3, x_4) = (-x_2, x_1, -x_4, x_3)

satisfies

T^2 + I = 0

, so

Tv = \lambda v

would need

\lambda^2 + 1 = 0

: no eigenvalue.

❌ The ambient $5$ is a trap. This subspace is the null space of a nonzero linear functional on

\mathbf{R}^5

, so its dimension is

5 - 1 = 4

, which is even.

❌ The index is a trap.

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

has basis

1, x, x^2, x^3

, so its dimension is

4

, which is even.

❌ Not quite. Compute each dimension and look for the odd one.

Show solution

The theorem applies to finite-dimensional real spaces of odd dimension. Compute the dimensions:

$\dim \mathbf{R}^4 = 4$ .
$\dim \mathcal{P}_2(\mathbf{R}) = 3$ (basis $1, x, x^2$ ).
The subspace $U = \{x \in \mathbf{R}^5 : x_1 + \cdots + x_5 = 0\}$ is the null space of the surjective map $\varphi(x) = x_1 + \cdots + x_5$ onto $\mathbf{R}$ , so $\dim U = 5 - 1 = 4$ .
$\dim \mathcal{P}_3(\mathbf{R}) = 4$ (basis $1, x, x^2, x^3$ ).

Only $\mathcal{P}_2(\mathbf{R})$ is odd-dimensional, so every operator on it has an eigenvalue.

Each $4$ -dimensional space carries an operator with none. Choose a basis $u_1, u_2, u_3, u_4$ and define $T$ by $Tu_1 = u_2$ , $Tu_2 = -u_1$ , $Tu_3 = u_4$ , $Tu_4 = -u_3$ . Then $T^2 = -I$ , so $Tv = \lambda v$ with $v \neq 0$ gives $T^2 v = \lambda^2 v = -v$ , that is $(\lambda^2 + 1)v = 0$ , impossible for real $\lambda$ .

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Why Odd Dimension Forces a Real Eigenvalue

IKey Formulas

IIwww and TwTwTw Span an Invariant Plane

IIIOdd = Even + Odd: Descending to an Eigenvalue

IVQuiz Questions

II $w$ and $Tw$ Span an Invariant Plane