Linear-Algebra-Done-Right · Unit 13 · Video 3 · Interactive Practice

Distinct Linear Factors Decide Diagonalizability

IKey Formulas

Formula	Name	What it says
$T$ diagonalizable $\iff p(z) = (z - \lambda_1)\cdots(z - \lambda_m)$ , with $\lambda_1, \dots, \lambda_m$ distinct	Necessary and sufficient condition	For $V$ finite-dimensional, $T \in \mathcal{L}(V)$ with minimal polynomial $p$ , and $\lambda_1, \dots, \lambda_m \in \mathbf{F}$
$q(T) = 0 \iff q$ is a polynomial multiple of $p$	Polynomials that send $T$ to $0$	So if $q(T) = 0$ and $q$ is a product of distinct factors $z - \lambda$ , then $p$ is a product of some of those factors, each once
$V = \operatorname{range}(T - \lambda_m I) \oplus \operatorname{null}(T - \lambda_m I)$	The splitting in the proof	The two subspaces meet only at $0$ because $\lambda_m - \lambda_k \neq 0$ for $k = 1, \dots, m - 1$
$T$ diagonalizable, $U$ invariant under $T$ $\implies T\|_U$ diagonalizable	Restriction to an invariant subspace	The minimal polynomial of $T\|_U$ divides that of $T$ , so its factors are distinct too

Key Insight: A repeated diagonal entry is allowed, a repeated factor of $p$ is not: the diagonal matrix with entries $8, 5, 5$ has $p(z) = (z - 8)(z - 5)$ , each factor once.

IIOnly Distinctness of the Zeros Matters

The operator on $\mathbf{C}^5$ with minimal polynomial $z^5 - 5z + c$ is diagonalizable exactly when its five zeros are distinct.

Constant term c 3

💡 For real $c$ , two zeros coincide only at $c = 4$ and $c = -4$ , so every other real $c$ gives a diagonalizable operator on $\mathbf{C}^5$ .

IIIThe Entry That Links the Two 6's

With diagonal $6, 6, 7$ fixed, the entries above the diagonal decide whether $p$ is $(z - 6)(z - 7)$ or $(z - 6)^2(z - 7)$ .

Entry (1, 2) 3 Entry (1, 3) 4 Entry (2, 3) 2

IVWhere Distinctness Is Used

For $T(x, y) = (2x + y, \mu y)$ the range and null space of $T - \mu I$ meet only at $0$ unless $\mu = 2$ .

μ 3

VQuiz Questions

Problem 1 · Read the Verdict off the Minimal Polynomial

Given: four operators on finite-dimensional vector spaces, each with the field and minimal polynomial listed — find the one that is diagonalizable.

✅ Correct!

z^3 - z = (z - 0)(z - 1)(z + 1)

: three degree-one factors with the distinct real numbers

0, 1, -1

❌ Not quite. The factor

z - 1

appears twice. Working over

\mathbf{C}

guarantees a product of degree-one factors, which gives an upper-triangular matrix, but not distinct ones.

❌ Not quite.

z^2 + 1

has no zeros in

\mathbf{R}

, so it is not a product of factors

z - \lambda

with

\lambda \in \mathbf{R}

. The operator has no eigenvalues at all.

❌ Not quite.

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

: the factor

z - 0

repeats, even though it is easy to miss without the

-0

❌ Not quite. Look for a product of degree-one factors over the given field with no factor repeated.

Show solution

Apply the test: $T$ is diagonalizable exactly when $p(z) = (z - \lambda_1)\cdots(z - \lambda_m)$ with $\lambda_1, \dots, \lambda_m \in \mathbf{F}$ distinct.

(a) $(z - 1)^2(z + 2)$ : the factor $z - 1$ repeats. Not diagonalizable.
(b) $z^3 - z = z(z - 1)(z + 1)$ : distinct $\lambda$ 's $0, 1, -1$ , all in $\mathbf{R}$ . Diagonalizable.
(c) $z^2 + 1$ has no real zeros, so it is not a product of degree-one factors over $\mathbf{R}$ . Not diagonalizable.
(d) $z^2(z - 3)$ : the factor $z$ repeats. Not diagonalizable.

Problem 2 · Differentiation on $\mathcal{P}_3(\mathbf{R})$

Given: $D \in \mathcal{L}(\mathcal{P}_3(\mathbf{R}))$ , $Dq = q'$ . Its only eigenvalue is $0$ , and its matrix in the basis $1, x, x^2, x^3$ is upper triangular with diagonal $0, 0, 0, 0$ — find the minimal polynomial of $D$ and decide whether $D$ is diagonalizable.

✅ Correct!

D^4 = 0

but

D^3 x^3 = 6 \neq 0

, so

p(z) = z^4

, and the factor

z

repeats four times.

❌ Not quite. One eigenvalue does not make

p

degree one:

p(z) = z

would mean

D = 0

, but

D x = 1 \neq 0

❌ Close, but… A repeated diagonal entry is allowed; a repeated factor of

p

is not. Here the factor

z

appears four times in

p

, which rules out a diagonal matrix.

❌ Not quite.

D^3 x^3 = 6 \neq 0

, so

z^3

does not send

D

0

; the verdict is right but the exponent is not.

Show solution

Each application of $D$ lowers the degree by one, so $D^4 q = 0$ for every $q \in \mathcal{P}_3(\mathbf{R})$ : the polynomial $z^4$ sends $D$ to $0$ , hence $p$ divides $z^4$ and $p(z) = z^k$ for some $k \leq 4$ .

But $D^3 x^3 = 6 \neq 0$ , so $D^3 \neq 0$ and $k = 4$ :

p(z) = z^4 = (z - 0)(z - 0)(z - 0)(z - 0).

The factor $z - 0$ repeats, so $D$ is not diagonalizable. A diagonal matrix of $D$ would have only the eigenvalue $0$ on its diagonal, making it the zero matrix — impossible since $D \neq 0$ .

Problem 3 · Polynomials That Send $T$ to $0$

Given: an operator $T$ on a finite-dimensional vector space $V$ with $T^3 = T$ ; and operators $R \in \mathcal{L}(\mathbf{R}^4)$ and $S \in \mathcal{L}(\mathbf{C}^4)$ with $R^2 = -I$ and $S^2 = -I$ — decide diagonalizability in each case.

Is $T$ (with $T^3 = T$ ) diagonalizable?

Which are diagonalizable: $R$ on $\mathbf{R}^4$ , $S$ on $\mathbf{C}^4$ ?

✅ Correct!

T

's minimal polynomial divides

z(z - 1)(z + 1)

, so it inherits distinct factors;

z^2 + 1

splits into

(z - i)(z + i)

only over

\mathbf{C}

❌ Not quite.

p

need not equal

z^3 - z

: it divides it, so it is a product of some of

z

z - 1

z + 1

, each at most once — distinct in every case.

❌ Not quite. Distinct eigenvalues filling the dimension is sufficient, not necessary.

T = I

satisfies

T^3 = T

, has one eigenvalue, and is diagonal.

❌ Not quite.

T^3 - T = 0

means

p

divides

z^3 - z = z(z - 1)(z + 1)

. What does that force about the factors of

p

❌ Not quite. Over

\mathbf{R}

z^2 + 1

has no zeros, so

R

has no eigenvalues and no eigenvector basis.

❌ Not quite. The field decides what counts as a degree-one factor: over

\mathbf{C}

z^2 + 1 = (z - i)(z + i)

with

i \neq -i

❌ Close, but… A repeated diagonal entry is allowed. Distinctness is required of the factors of

p = (z - i)(z + i)

, not of the diagonal entries.

❌ Not quite. Find

p

for each: it divides

z^2 + 1

. Over which field is that a product of distinct degree-one factors?

Show solution

$T^3 = T$ . The polynomial $z^3 - z = z(z - 1)(z + 1)$ sends $T$ to $0$ , so it is a polynomial multiple of $p$ . By unique factorization, the monic $p$ is a product of some of $z$ , $z - 1$ , $z + 1$ , each at most once. Those factors are distinct, so $T$ is always diagonalizable, whichever of $z$ , $z - 1$ , $z(z - 1)$ , …, $z^3 - z$ the minimal polynomial turns out to be.

$R^2 = -I$ on $\mathbf{R}^4$ . $p$ divides $z^2 + 1$ , which has no real zeros and no real degree-one factor, so $p(z) = z^2 + 1$ . It is not a product of factors $z - \lambda$ with $\lambda \in \mathbf{R}$ : $R$ is not diagonalizable.

$S^2 = -I$ on $\mathbf{C}^4$ . $p$ divides $z^2 + 1 = (z - i)(z + i)$ , so $p$ is $z - i$ , $z + i$ or $(z - i)(z + i)$ . Each is a product of distinct degree-one factors, so $S$ is diagonalizable; its diagonal may repeat $i$ and $-i$ .

Problem 4 · Restricting to an Invariant Subspace

Given: $T \in \mathcal{L}(\mathcal{P}_2(\mathbf{R}))$ , $(Tq)(x) = (x + 1)\,q'(x)$ , so $T1 = 0$ , $Tx = x + 1$ , $Tx^2 = 2x^2 + 2x$ ; and $U = \mathcal{P}_1(\mathbf{R}) = \operatorname{span}(1, x)$ , which is invariant under $T$ — find the minimal polynomial of $T$ and an eigenvector basis of $U$ for $T|_U$ .

What is the minimal polynomial of $T$ ?

Which list is a basis of $U$ consisting of eigenvectors of $T|_U$ ?

✅ Correct!

p_T(z) = z(z - 1)(z - 2)

has distinct factors, so

T|_U

is diagonalizable, with

p_{T|_U}(z) = z(z - 1)

and eigenvectors

1

(for

0

) and

1 + x

(for

1

❌ Not quite.

T

does not lower degrees:

T(1 + x) = 1 + x

, so

1

is an eigenvalue and

T^3 \neq 0

❌ Not quite.

T1 = 0

, so

0

is an eigenvalue and

z

must be a factor of

p

❌ Not quite. That is the minimal polynomial of

T|_U

. On all of

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

T(1 + x)^2 = 2(1 + x)^2

adds the eigenvalue

2

❌ Check the minimal polynomial. The matrix in the basis

1, x, x^2

is upper triangular with diagonal

0, 1, 2

❌ Not quite.

Tx = x + 1

is not a multiple of

x

, so

x

is not an eigenvector. An eigenvector basis of the whole space need not restrict to one of

U

❌ Not quite.

1 + x

is an eigenvector, but

Tx = x + 1

is not a multiple of

x

❌ Close, but…

T

is diagonalizable and

U

is invariant, so

T|_U

is diagonalizable: its minimal polynomial divides

z(z - 1)(z - 2)

❌ Check the basis of $U$ . Test each vector

q

: is

Tq

a scalar multiple of

q

Show solution

Minimal polynomial of $T$ . In the basis $1, x, x^2$ ,

\mathcal{M}(T) = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 1 & 2 \\ 0 & 0 & 2 \end{pmatrix},

upper triangular with diagonal $0, 1, 2$ , so these are the eigenvalues and, by the previous unit, $T(T - I)(T - 2I) = 0$ . Every eigenvalue is a zero of $p$ , so $p(z) = z(z - 1)(z - 2)$ . The factors are distinct: $T$ is diagonalizable, with eigenvectors $1$ , $1 + x$ , $(1 + x)^2$ .

The restriction. $T1 = 0 \in U$ and $Tx = 1 + x \in U$ , so $U$ is invariant. The minimal polynomial of $T|_U$ divides $z(z - 1)(z - 2)$ ; since $T|_U$ has eigenvalues $0$ (from $T1 = 0$ ) and $1$ (from $T(1 + x) = 1 + x$ ) and $\dim U = 2$ , it is $z(z - 1)$ . An eigenvector basis of $U$ is

1 \ (\text{eigenvalue } 0), \qquad 1 + x \ (\text{eigenvalue } 1).

The standard basis $1, x$ fails because $Tx = 1 + x$ is not a multiple of $x$ ; the minimal polynomial argument guarantees an eigenvector basis of $U$ before it is found.

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