Linear-Algebra-Done-Right · Unit 12 · Video 2 · Interactive Practice

Eigenvalues Sit on the Upper-Triangular Diagonal

IKey Formulas

Formula	Statement	Hypotheses
$(T - \lambda_k I)v_k \in \operatorname{span}(v_1, \ldots, v_{k-1})$	Key fact, read from column $k$ ; for $k = 1$ it says $Tv_1 = \lambda_1 v_1$	$V$ finite-dimensional, $T \in \mathcal{L}(V)$ , $\mathcal{M}(T, (v_1, \ldots, v_n))$ upper triangular with diagonal $\lambda_1, \ldots, \lambda_n$
$(T - \lambda_1 I)\cdots(T - \lambda_n I) = 0$	Equation satisfied by an operator with an upper-triangular matrix; the factors may be taken in any order	Same basis and diagonal
eigenvalues of $T$ $= \{\lambda_1, \ldots, \lambda_n\}$	Determination of eigenvalues from an upper-triangular matrix; a repeated diagonal entry is one eigenvalue	$V$ finite-dimensional, $T \in \mathcal{L}(V)$ , $\mathcal{M}(T)$ upper triangular with respect to some basis of $V$
$q(z) = (z - \lambda_1)\cdots(z - \lambda_n)$	$q(T) = 0$ , so $q$ is a polynomial multiple of the minimal polynomial of $T$	Same basis and diagonal

Key Insight: One inclusion drives both theorems: $T - \lambda_k I$ maps $\operatorname{span}(v_1, \ldots, v_k)$ into $\operatorname{span}(v_1, \ldots, v_{k-1})$ . Applied factor by factor it makes the product $0$ ; with a dimension count it makes each $\lambda_k$ an eigenvalue. The hypothesis matters: $\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$ has diagonal $0, 0$ but eigenvalues $1, -1$ , and it is not upper triangular.

IIThe Staircase of Zeros

For $T(x, y, z) = (2x + y,\, 5y + 3z,\, 8z)$ , the first $m$ factors of $(T - 2I)(T - 5I)(T - 8I)$ already send $e_1, \ldots, e_m$ to $0$ .

factors m 1 basis vector k 1

IIIDown the Chain, One Factor at a Time

In written order each factor drops a vector one step down $V \supset \operatorname{span}(e_1, e_2) \supset \operatorname{span}(e_1) \supset \{0\}$ .

x 1 y 2 z −1 order factors applied 0

IVEach Diagonal Entry Has an Eigenvector

$T - \lambda_k I$ squeezes $\operatorname{span}(e_1, \ldots, e_k)$ into dimension $k - 1$ , forcing a nonzero $v$ with $(T - \lambda_k I)v = 0$ .

diagonal entry a 1 b 1 c 0

💡 The dimension count gives the eigenvalues $2$ , $5$ , $8$ . There is no fourth: $q(z) = (z - 2)(z - 5)(z - 8)$ satisfies $q(T) = 0$ , so it is a polynomial multiple of the minimal polynomial, and every eigenvalue is a zero of $q$ .

VQuiz Questions

Problem 1 · Read the Diagonal

Given: $T \in \mathcal{L}(\mathbf{F}^4)$ has, with respect to the standard basis, $\mathcal{M}(T) = \begin{pmatrix} 3 & 1 & 0 & 2 \\ 0 & -1 & 4 & 1 \\ 0 & 0 & 3 & 5 \\ 0 & 0 & 0 & 0 \end{pmatrix}.$ Find the set of eigenvalues of $T$ .

✅ Correct! The matrix is upper triangular, so the eigenvalues are exactly the diagonal entries

3, -1, 3, 0

; the repeated

3

is one eigenvalue.

❌ $0$ is on the diagonal. The entry

\lambda_4 = 0

is an eigenvalue like any other:

T - 0I = T

maps

\mathbf{F}^4

into

\operatorname{span}(e_1, e_2, e_3)

, of dimension

3

, so

T

is not injective.

❌ That is the first row. The first row

3, 1, 0, 2

holds the

e_1

-coefficients of

Te_1, \ldots, Te_4

. The theorem reads the diagonal:

3, -1, 3, 0

❌ Those entries sit just above the diagonal. The entries above the diagonal (the

*

region) may be anything and play no part; the eigenvalues are the diagonal entries

3, -1, 3, 0

❌ Not quite. Check that the matrix is upper triangular, then read its diagonal.

Show solution

Every entry below the diagonal is $0$ , so $\mathcal{M}(T)$ is upper triangular with diagonal $\lambda_1, \lambda_2, \lambda_3, \lambda_4 = 3, -1, 3, 0$ . By the determination of eigenvalues from an upper-triangular matrix, the eigenvalues of $T$ are precisely these entries. The value $3$ appears twice on the diagonal but is a single eigenvalue, so the set is $\{-1, 0, 3\}$ .

In particular $0$ is an eigenvalue, so $T$ is not injective and not invertible.

Problem 2 · Keep the Repeated Factor

Given: $T \in \mathcal{L}(\mathbf{F}^3)$ has, with respect to the standard basis, $\mathcal{M}(T) = \begin{pmatrix} 2 & 1 & 3 \\ 0 & 2 & 4 \\ 0 & 0 & 5 \end{pmatrix}.$ Find the operator that equals $0$ .

✅ Correct! The diagonal is

2, 2, 5

, so the theorem gives

(T - 2I)(T - 2I)(T - 5I) = 0

: one factor for each diagonal entry, repeats included.

❌ A repeated entry still contributes a factor. Here

(T - 5I)e_2 = (1, -3, 0)

and

(T - 2I)(1, -3, 0) = (-3, 0, 0) \neq 0

, so

(T - 2I)(T - 5I) \neq 0

❌ The wrong factor is squared.

(T - 5I)^2 e_2 = (-6, 9, 0)

and

(T - 2I)(-6, 9, 0) = (9, 0, 0) \neq 0

. The entry that repeats on the diagonal is

2

, not

5

❌ The factor for $5$ is missing.

(T - 2I)e_3 = (3, 4, 3)

and

(T - 2I)(3, 4, 3) = (13, 12, 9) \neq 0

. Since

5

is an eigenvalue, an eigenvector

v

for

5

gives

(T - 2I)^2 v = 9v \neq 0

❌ Not quite. List the diagonal with repeats and take one factor

T - \lambda_k I

for each entry.

Show solution

The matrix is upper triangular with diagonal $\lambda_1, \lambda_2, \lambda_3 = 2, 2, 5$ , so the equation satisfied by an operator with an upper-triangular matrix gives

(T - 2I)(T - 2I)(T - 5I) = (T - 2I)^2(T - 5I) = 0.

The other three are not $0$ , as one basis vector each shows:

$(T - 2I)(T - 5I)e_2 = (T - 2I)(1, -3, 0) = (-3, 0, 0)$ ;
$(T - 2I)(T - 5I)^2 e_2 = (T - 2I)(-6, 9, 0) = (9, 0, 0)$ ;
$(T - 2I)^2 e_3 = (T - 2I)(3, 4, 3) = (13, 12, 9)$ .

Dropping the repeat can work for some matrices, but nothing guarantees it; the theorem guarantees only the product with every diagonal entry.

Problem 3 · Pin Down the Minimal Polynomial

Given: $T \in \mathcal{L}(\mathbf{F}^4)$ has, with respect to the standard basis, $\mathcal{M}(T) = \begin{pmatrix} 1 & 2 & 0 & 1 \\ 0 & 3 & 1 & 0 \\ 0 & 0 & 1 & 2 \\ 0 & 0 & 0 & 3 \end{pmatrix}.$ Find a polynomial multiple of the minimal polynomial $p$ of $T$ , then $p$ itself.

Which polynomial is a multiple of p?

What is p?

✅ Correct!

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

sends

T

0

, and none of its smaller monic divisors with zeros

1

and

3

does, so here

p = q

❌ The theorem keeps the repeats. The diagonal is

1, 3, 1, 3

, so the guaranteed polynomial has four factors. And

(T - I)(T - 3I)e_3 = (2, 0, 0, 0) \neq 0

, so

(z - 1)(z - 3)

is not a multiple of

p

❌ $2$ is not on the diagonal. It is an entry above the diagonal. The theorem builds its polynomial from the diagonal

1, 3, 1, 3

only, and this one does not send

T

0

(T - I)(T - 2I)(T - 3I)e_3 = (-2, 0, 0, 0) \neq 0

, so it is not a multiple of

p

❌ That uses the first row $1, 2, 0, 1$ . The theorem uses the diagonal

1, 3, 1, 3

. A multiple of

p

has every eigenvalue as a zero, and

3

is not a zero of

z(z - 1)^2(z - 2)

❌ Not quite. Build

q(z) = (z - \lambda_1)\cdots(z - \lambda_4)

from the diagonal;

q(T) = 0

makes it a multiple of

p

❌ This does not send $T$ to $0$ .

(T - I)(T - 3I)e_3 = (T - I)(0, 1, -2, 0) = (2, 0, 0, 0) \neq 0

❌ This does not send $T$ to $0$ .

(T - 3I)e_4 = (1, 0, 2, 0)

, then

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

, then

(T - I)(0, 2, 0, 0) = (4, 4, 0, 0) \neq 0

❌ This does not send $T$ to $0$ .

(T - 3I)^2 e_3 = (T - 3I)(0, 1, -2, 0) = (2, -2, 4, 0)

and

(T - I)(2, -2, 4, 0) = (-4, 0, 0, 0) \neq 0

❌ Not quite.

p

divides

q

and has zeros exactly

1

and

3

; test which candidate sends

T

0

Show solution

A multiple of $p$ . The matrix is upper triangular with diagonal $1, 3, 1, 3$ . With $q(z) = (z - 1)(z - 3)(z - 1)(z - 3) = (z - 1)^2(z - 3)^2$ , the first theorem gives $q(T) = 0$ , and every polynomial that sends $T$ to $0$ is a polynomial multiple of $p$ .

Narrowing down $p$ . The zeros of $p$ are exactly the eigenvalues of $T$ , which are the diagonal entries $1$ and $3$ . Since $p$ is monic and divides $q$ , it is one of $(z - 1)(z - 3)$ , $(z - 1)^2(z - 3)$ , $(z - 1)(z - 3)^2$ , $(z - 1)^2(z - 3)^2$ . Testing the first three on a basis vector, with $T - I$ and $T - 3I$ read from the matrix:

(T - I)(T - 3I)e_3 = (2, 0, 0, 0),\qquad (T - I)^2(T - 3I)e_4 = (4, 4, 0, 0),\qquad (T - I)(T - 3I)^2 e_3 = (-4, 0, 0, 0).

None is $0$ , so $p(z) = (z - 1)^2(z - 3)^2$ , of degree $4 = \dim \mathbf{F}^4$ .

Problem 4 · Invertibility From the Diagonal

Given: $T \in \mathcal{L}(\mathbf{F}^4)$ has, with respect to the standard basis, $\mathcal{M}(T) = \begin{pmatrix} 2 & 5 & 0 & 1 \\ 0 & -3 & 1 & 4 \\ 0 & 0 & 2 & 5 \\ 0 & 0 & 0 & 6 \end{pmatrix}.$ Decide whether $T$ is invertible, then find every $c \in \mathbf{F}$ for which $T - cI$ is not invertible.

Is T invertible?

For which c is T − cI not invertible?

✅ Correct!

T - cI

fails to be invertible exactly when

c

is an eigenvalue, and the eigenvalues are the diagonal entries

2, -3, 6

; since

0

is not among them,

T

itself is invertible.

❌ A repeat does not matter.

T

is invertible exactly when

0

is not an eigenvalue. The repeated

2

is simply the eigenvalue

2

, listed twice.

❌ Only the diagonal decides. The zeros below the diagonal are what make the matrix upper triangular, and the

0

above it plays no part; no diagonal entry is

0

❌ The diagonal already decides. The matrix is upper triangular, so its eigenvalues are

2, -3, 6

;

0

is not one, so

T

is injective, hence invertible.

❌ Not quite.

T

is invertible exactly when

0

is not an eigenvalue of

T

❌ That answers the first question about $T = T - 0I$ . And even there

0

fails: it is not an eigenvalue, so

T - 0I

is invertible.

❌ Check the sign.

T - cI

is not invertible when

c

itself is an eigenvalue: for

c = 2

the diagonal of

\mathcal{M}(T) - 2I

0, -5, 0, 4

. For

c = -2

it is

4, -1, 4, 8

, with no zero.

❌ Only diagonal entries count. For

c = 5

, the matrix

\mathcal{M}(T) - 5I

is upper triangular with diagonal

-3, -8, -3, 1

, so

0

is not an eigenvalue of

T - 5I

and

T - 5I

is invertible.

❌ Not quite.

T - cI

is not injective exactly when

c

is an eigenvalue of

T

Show solution

The matrix is upper triangular with diagonal $2, -3, 2, 6$ , so the eigenvalues of $T$ are $2$ , $-3$ and $6$ .

For an operator on a finite-dimensional space, invertible is equivalent to injective, and $T - cI$ fails to be injective exactly when $c$ is an eigenvalue of $T$ . Hence $T - cI$ is not invertible precisely for $c \in \{-3, 2, 6\}$ .

Taking $c = 0$ : $0$ is not an eigenvalue, so $T = T - 0I$ is invertible.

The same conclusion comes from the matrix of $T - cI$ , which is $\mathcal{M}(T) - cI$ : still upper triangular, with diagonal $2 - c$ , $-3 - c$ , $2 - c$ , $6 - c$ . Its eigenvalues are these entries, and $0$ is among them exactly when $c \in \{-3, 2, 6\}$ .

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