Linear-Algebra-Done-Right · Unit 12 · Video 3 · Interactive Practice

The Minimal Polynomial Decides Upper-Triangularity

IKey Formulas

Formula	Name	What it says
$\mathcal{M}(T)$ upper triangular for some basis $\iff p(z) = (z - \lambda_1)\cdots(z - \lambda_m)$	Necessary and sufficient condition	For $V$ finite-dimensional and $T \in \mathcal{L}(V)$ with minimal polynomial $p$ , where $\lambda_1, \dots, \lambda_m \in \mathbf{F}$ and repeats are allowed
$\{\lambda_1, \dots, \lambda_m\} =$ eigenvalues of $T$	The $\lambda$ 's are the eigenvalues	The zeros of the minimal polynomial are exactly the eigenvalues of $T$
$\mathbf{F} = \mathbf{C} \implies$ every $T \in \mathcal{L}(V)$ has an upper-triangular matrix	The complex case	For $V$ a finite-dimensional complex vector space: by the fundamental theorem of algebra, $p$ is a product of factors $z - \lambda$
$v_k$ eigenvector $\iff$ column $k$ of $\mathcal{M}(T)$ is $0$ off the diagonal	Which basis vectors are eigenvectors	For $\mathcal{M}(T)$ upper triangular with respect to $v_1, \dots, v_n$ , $Tv_1 = \lambda_1 v_1$ , so $v_1$ is always an eigenvector; the later $v_k$ need not be

Key Insight: One formula can fail over $\mathbf{R}$ and pass over $\mathbf{C}$ : the factor $z^2 + 1$ becomes $(z - i)(z + i)$ only once $i$ and $-i$ are available scalars.

IIThe Zeros of $p$ Must Lie in $\mathbf{F}$

An operator on $\mathbf{F}^4$ with minimal polynomial $p$ has an upper-triangular matrix exactly when every complex zero of $p$ lies in $\mathbf{F}$ .

Field Re λ 0 Im λ 1

IIIColumn by Column in the Video's Basis

Each $Tv_k$ lies in $\operatorname{span}(v_1, \dots, v_k)$ , yet only the vectors whose column is $0$ off the diagonal are eigenvectors.

Column k 1

IVThe Induction Step as a Block Matrix

With $U = \operatorname{range}(T - \lambda_m I)$ , the basis $u_1, u_2, u_3$ of $U$ followed by $w$ splits $\mathcal{M}(T)$ into blocks.

Last factor z − λₘ

💡 With $\lambda_m = 3$ the vectors $u_1, u_2, u_3, w$ are the video's $v_1, v_2, v_3, v_4$ , so the video's matrix is one this induction step builds.

VQuiz Questions

Problem 1 · Which Real Operator Passes?

Given: four operators, each on a finite-dimensional real vector space, with the minimal polynomials below — find the one that has an upper-triangular matrix with respect to some basis.

✅ Correct!

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

is a product of factors

z - \lambda

with

\lambda \in \{0, 1, -1\} \subseteq \mathbf{R}

❌ An eigenvalue is not enough.

2

is a real eigenvalue, but the factor

z^2 + 1

has no real zeros, so

p

is not a product of real factors

z - \lambda

❌ Factor it first.

z^3 + z = z(z^2 + 1)

, and

z^2 + 1

has no real zeros:

p(i) = 0

with

i \notin \mathbf{R}

❌ Check the quadratic factor.

z^3 - 1 = (z - 1)(z^2 + z + 1)

, and

z^2 + z + 1 = (z + \tfrac12)^2 + \tfrac34 > 0

for every real

z

Show solution

By the theorem, an operator on a finite-dimensional real space has an upper-triangular matrix with respect to some basis exactly when its minimal polynomial is $(z - \lambda_1)\cdots(z - \lambda_m)$ with every $\lambda_k \in \mathbf{R}$ . A product of real factors $z - \lambda$ vanishes only at real numbers, so a minimal polynomial with a non-real zero fails.

$(z^2 + 1)(z - 2)$ vanishes at $i$ : fails.
$z^3 + z = z(z^2 + 1)$ vanishes at $i$ : fails.
$z^3 - 1 = (z - 1)(z^2 + z + 1)$ vanishes at $-\tfrac12 + \tfrac{\sqrt3}{2}i$ : fails.
$z^3 - z = z(z - 1)(z + 1) = (z - 0)(z - 1)(z - (-1))$ : passes.

Only the operator with minimal polynomial $z^3 - z$ has an upper-triangular matrix, and its diagonal entries are its eigenvalues $0$ , $1$ , $-1$ .

Problem 2 · The Diagonal over $\mathbf{C}$

Given: $T \in \mathcal{L}(\mathbf{C}^3)$ has minimal polynomial $(z^2 + 1)(z - 2)$ — find a list that can be the diagonal of an upper-triangular matrix of $T$ with respect to some basis of $\mathbf{C}^3$ .

✅ Correct! Over

\mathbf{C}

p(z) = (z - i)(z + i)(z - 2)

, and the diagonal entries are exactly the eigenvalues

i

-i

2

❌ That is the value of $z^2$ , not a zero. At

z = -1

p(-1) = ((-1)^2 + 1)(-1 - 2) = -6 \neq 0

, so

-1

is not an eigenvalue and cannot appear on the diagonal.

❌ An eigenvalue is missing.

-i

is a zero of

p

, so it is an eigenvalue, and the diagonal entries of an upper-triangular matrix are all of the eigenvalues.

❌ That is the test over $\mathbf{R}$ . Over

\mathbf{C}

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

, so

p

is a product of factors

z - \lambda

and the theorem gives an upper-triangular matrix.

Show solution

Over $\mathbf{C}$ ,

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

a product of factors $z - \lambda$ with $\lambda \in \mathbf{C}$ , so some basis of $\mathbf{C}^3$ gives $T$ an upper-triangular matrix.

The diagonal entries of any upper-triangular matrix of $T$ are exactly the eigenvalues of $T$ , and the eigenvalues are the zeros of $p$ : $i$ , $-i$ , $2$ . Three diagonal slots must show three distinct eigenvalues, so the diagonal is $i, -i, 2$ in some order.

Problem 3 · A Rotation Block on $\mathbf{R}^4$

Given: $T \in \mathcal{L}(\mathbf{R}^4)$ , $T(x_1, x_2, x_3, x_4) = (-x_2,\ x_1,\ 2x_3 + x_4,\ 2x_4)$ — find its minimal polynomial, then decide whether it has an upper-triangular matrix.

What is the minimal polynomial of T?

Is there a basis giving an upper-triangular matrix?

✅ Correct!

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

has the non-real zeros

\pm i

, so

\mathbf{R}^4

fails; over

\mathbf{C}

it is

(z - i)(z + i)(z - 2)(z - 2)

❌ One factor $z - 2$ is too few.

(T - 2I)e_4 = e_3

and

(T^2 + I)e_3 = 5e_3

, so

(T^2 + I)(T - 2I)e_4 = 5e_3 \neq 0

❌ The rotation block survives.

(T - 2I)^2 e_1 = T^2 e_1 - 4Te_1 + 4e_1 = -e_1 - 4e_2 + 4e_1 = 3e_1 - 4e_2 \neq 0

❌ That only kills $e_1$ and $e_2$ .

Te_3 = 2e_3

, so

(T^2 + I)e_3 = 5e_3 \neq 0

❌ Not quite. Test each candidate on

e_1, \dots, e_4

❌ A real eigenvalue is not enough. Every factor of

p

must have the form

z - \lambda

with

\lambda \in \mathbf{R}

, and

z^2 + 1

does not.

❌ Over $\mathbf{C}$ the obstruction disappears.

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

, so the same formula on

\mathbf{C}^4

passes the test.

❌ The whole operator must pass. The

x_3, x_4

part is upper triangular, but

z^2 + 1

divides

p

and has no real zeros, so

p

is not a product of real factors

z - \lambda

❌ Not quite. Apply the test to

p

over each field.

Show solution

On the standard basis: $Te_1 = e_2$ , $Te_2 = -e_1$ , $Te_3 = 2e_3$ , $Te_4 = e_3 + 2e_4$ .

Minimal polynomial. $T^2 e_1 = -e_1$ and $T^2 e_2 = -e_2$ , so $T^2 + I$ kills $e_1, e_2$ . Also $(T - 2I)e_3 = 0$ and $(T - 2I)e_4 = e_3$ , so $(T - 2I)^2$ kills $e_3, e_4$ . Polynomials in $T$ commute, so $q(T)e_j = 0$ for each $j$ , and

q(z) = (z^2 + 1)(z - 2)^2 \quad\text{satisfies}\quad q(T) = 0.

The minimal polynomial divides $q$ . The feedback computations show that $(z^2 + 1)(z - 2)$ , $(z - 2)^2$ and $z^2 + 1$ do not send $T$ to $0$ . Every other proper monic divisor of $q$ divides one of these three, so $p = q$ .

Upper-triangular? Over $\mathbf{R}$ , $p(i) = 0$ with $i \notin \mathbf{R}$ , so $p$ is not a product of real factors $z - \lambda$ : no basis of $\mathbf{R}^4$ works. For the same formula on $\mathbf{C}^4$ , the basis $(1, -i, 0, 0)$ , $(1, i, 0, 0)$ , $e_3$ , $e_4$ gives

\begin{pmatrix} i & 0 & 0 & 0 \\ 0 & -i & 0 & 0 \\ 0 & 0 & 2 & 1 \\ 0 & 0 & 0 & 2 \end{pmatrix},

since $T(1, -i, 0, 0) = (i, 1, 0, 0) = i(1, -i, 0, 0)$ and $T(1, i, 0, 0) = (-i, 1, 0, 0) = -i(1, i, 0, 0)$ .

Problem 4 · Which Basis Vectors Are Eigenvectors?

Given: $T \in \mathcal{L}(\mathbf{C}^3)$ has minimal polynomial $(z - 1)^2(z + 3)$ , and $\mathcal{M}(T)$ is upper triangular with respect to the basis $v_1, v_2, v_3$ — find the statement that must be true.

✅ Correct! Column 1 of an upper-triangular matrix is

0

below its first entry, so

Tv_1 = \lambda_1 v_1

with

v_1 \neq 0

❌ The first diagonal entry could be $-3$ . If

Te_1 = -3e_1

Te_2 = e_2

Te_3 = e_2 + e_3

, then

T

has this minimal polynomial and an upper-triangular matrix with respect to

e_1, e_2, e_3

, with

Te_1 = -3e_1

❌ Later basis vectors need not be eigenvectors. In the example

Te_1 = -3e_1

Te_2 = e_2

Te_3 = e_2 + e_3

, the vector

e_3

is the last basis vector and

Te_3 \notin \operatorname{span}(e_3)

❌ That can never happen here. If every

v_k

were an eigenvector, with eigenvalue

1

-3

, then

(T - I)(T + 3I)

would kill each

v_k

, so the degree-2 polynomial

(z - 1)(z + 3)

would be a multiple of the degree-3 minimal polynomial.

Show solution

(a) is always true. Column 1 of $\mathcal{M}(T)$ holds the coefficients of $Tv_1$ , and in an upper-triangular matrix only its first entry $\lambda_1$ can be nonzero. So $Tv_1 = \lambda_1 v_1$ , and $v_1 \neq 0$ because it belongs to a basis.

(b) and (c) can fail. Define $T$ by $Te_1 = -3e_1$ , $Te_2 = e_2$ , $Te_3 = e_2 + e_3$ . Then $(T + 3I)(T - I)^2 = 0$ , while $(T - I)(T + 3I)e_3 = (T - I)(e_2 + 4e_3) = 4e_2 \neq 0$ and $(T - I)^2 e_1 = 16e_1 \neq 0$ , so the minimal polynomial is $(z - 1)^2(z + 3)$ . With respect to $e_1, e_2, e_3$ ,

\mathcal{M}(T) = \begin{pmatrix} -3 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix}

is upper triangular, $v_1 = e_1$ has eigenvalue $-3$ , and column 3 has the entry $1$ above the diagonal, so $v_3 = e_3$ is not an eigenvector.

(d) is always false. The eigenvalues are the zeros of the minimal polynomial, $1$ and $-3$ . If each $v_k$ were an eigenvector, then $(T - I)(T + 3I)v_k = 0$ for every $k$ (the two factors commute), so $(T - I)(T + 3I) = 0$ . A polynomial of degree $2$ would then be a multiple of the minimal polynomial, which has degree $3$ : impossible.

Solved: 0 / 4

The Minimal Polynomial Decides Upper-Triangularity

IKey Formulas

IIThe Zeros of ppp Must Lie in F\mathbf{F}F

IIIColumn by Column in the Video's Basis

IVThe Induction Step as a Block Matrix

VQuiz Questions

IIThe Zeros of $p$ Must Lie in $\mathbf{F}$