Linear-Algebra-Done-Right · Unit 12 · Video 1 · Interactive Practice

Upper-Triangular Matrices and Invariant Spans

IKey Formulas

Formula	Name	Hypotheses
$Tv_k = A_{1,k}v_1 + \cdots + A_{n,k}v_n$	Column $k$ of $\mathcal{M}(T)$ holds the coordinates of $Tv_k$	$V$ finite-dimensional, $T \in \mathcal{L}(V)$ , $v_1, \ldots, v_n$ a basis of $V$
$A_{j,k} = 0$ whenever $j > k$	Upper triangular: at least $\frac{n(n-1)}{2}$ of the $n^2$ entries are $0$	$\mathcal{M}(T)$ is $n$ -by- $n$
$\mathcal{M}(T)$ upper triangular $\iff$ $\text{span}(v_1, \ldots, v_k)$ invariant under $T$ for each $k$ $\iff$ $Tv_k \in \text{span}(v_1, \ldots, v_k)$ for each $k$	Conditions for an upper-triangular matrix	$T \in \mathcal{L}(V)$ , $v_1, \ldots, v_n$ a basis of $V$ , $k = 1, \ldots, n$
$Tv = \lambda v$ , basis $v, v_2, \ldots, v_n$ $\Rightarrow$ column $1$ is $(\lambda, 0, \ldots, 0)$	An eigenvector first clears column $1$ below the diagonal	$V$ finite-dimensional, $v \neq 0$ ; such $v$ exists when $\mathbf{F} = \mathbf{C}$ and $V \neq \{0\}$

Key Insight: Being upper triangular is a property of $T$ together with an ordered basis. The theorem turns it into $n$ vector checks: $Tv_k$ may use only $v_1, \ldots, v_k$ .

IIAn Eigenvector First Clears Column 1

On $\mathbf{R}^2$ , $\mathcal{M}(T, (v_1, v_2))$ is upper triangular exactly when $Tv_1 \in \text{span}(v_1)$ , that is, when $v_1$ is an eigenvector.

operator

💡 Challenge: find the two lines through the origin on which $v_1$ makes $A_{2,1} = 0$ for $T(x, y) = (x + y, 2y)$ . For the rotation there is none: over $\mathbf{R}$ it has no eigenvalue.

IIIHow Many Entries Must Be Zero

Row $j$ of an $n$ -by- $n$ upper-triangular matrix has $j - 1$ forced zeros, so $\frac{n(n-1)}{2}$ in all.

size n 4

IVSame Vectors, Different Order

For $T(x, y, z) = (2x + y,\ 5y + 3z,\ 8z)$ , only one of the six orders of $e_1, e_2, e_3$ gives an upper-triangular matrix.

VQuiz Questions

Problem 1 · Differentiation, Basis Reversed

Given: $D \in \mathcal{L}(\mathcal{P}_2(\mathbf{R}))$ defined by $Dp = p'$ , and the basis $x^2, x, 1$ of $\mathcal{P}_2(\mathbf{R})$ — find $\mathcal{M}(D, (x^2, x, 1))$ and decide whether it is upper triangular.

What is the matrix?

Is it upper triangular?

✅ Correct! Reversing the basis turns the upper-triangular

\mathcal{M}(D, (1, x, x^2))

into a matrix with

2

and

1

below the diagonal: the check

Dv_1 \in \text{span}(v_1)

already fails.

❌ That is the matrix for the basis $1, x, x^2$ . With

v_1 = x^2

, column

1

holds the coordinates of

Dx^2 = 2x

in the list

x^2, x, 1

, namely

(0, 2, 0)

❌ The coordinates went into rows. The coordinates of

Dv_k

fill column

k

: column

1

(0, 2, 0)

because

Dx^2 = 2x

❌ Check the coefficients.

Dx^2 = 2x

gives a

2

in the

x

-slot of column

1

, and

Dx = 1

gives a

1

in the

1

-slot of column

2

❌ Not quite. Write

Dx^2

Dx

D1

in terms of

x^2, x, 1

and stand the coordinate lists side by side as columns.

❌ The zeros must be below the diagonal. Upper triangular means

A_{j,k} = 0

for

j > k

; here

A_{2,1} = 2

and

A_{3,2} = 1

❌ The basis order matters. Upper triangularity belongs to

D

together with an ordered basis; in the order

x^2, x, 1

the first check

Dx^2 \in \text{span}(x^2)

fails.

❌ The diagonal may be anything. Diagonal entries, including

0

, are unrestricted; the matrix fails because

A_{2,1} = 2 \neq 0

❌ Not quite. Apply condition (3) at

k = 1

: is

Dx^2

a multiple of

x^2

Show solution

Write each image in the basis $v_1 = x^2$ , $v_2 = x$ , $v_3 = 1$ :

Dx^2 = 2x \ \to\ (0, 2, 0), \qquad Dx = 1 \ \to\ (0, 0, 1), \qquad D1 = 0 \ \to\ (0, 0, 0).

These are the columns:

\mathcal{M}\big(D, (x^2, x, 1)\big) = \begin{pmatrix} 0 & 0 & 0 \\ 2 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix}.

The entries $A_{2,1} = 2$ and $A_{3,2} = 1$ lie below the diagonal, so the matrix is not upper triangular. Condition (3) fails at $k = 1$ : $Dx^2 = 2x \notin \text{span}(x^2)$ , so $\text{span}(x^2)$ is not invariant.

In the order $1, x, x^2$ every check passes ( $D1 = 0$ , $Dx = 1$ , $Dx^2 = 2x$ ), and the invariant spans are $\text{span}(1)$ , $\text{span}(1, x)$ , $\mathcal{P}_2(\mathbf{R})$ : differentiation never raises degree.

Problem 2 · Choosing the Order

Given: $T \in \mathcal{L}(\mathbf{F}^3)$ , $T(x, y, z) = (3x,\ x + 2y,\ y + z)$ — find the order of $e_1, e_2, e_3$ in which $\mathcal{M}(T)$ is upper triangular.

✅ Correct!

Te_3 = e_3

Te_2 = 2e_2 + e_3

Te_1 = 3e_1 + e_2

: each image uses only the current and earlier basis vectors.

❌ The standard order fails at $k = 1$ .

Te_1 = (3, 1, 0) = 3e_1 + e_2 \notin \text{span}(e_1)

; the standard matrix has the entries

1

and

1

below its diagonal.

❌ This fails at $k = 1$ .

Te_2 = (0, 2, 1) = 2e_2 + e_3 \notin \text{span}(e_2)

❌ This passes $k = 1$ but fails at $k = 2$ .

Te_1 = 3e_1 + e_2 \notin \text{span}(e_3, e_1)

❌ Not quite. Check

Tv_k \in \text{span}(v_1, \ldots, v_k)

for

k = 1, 2, 3

Show solution

Compute the images of the standard basis vectors:

Te_1 = (3, 1, 0) = 3e_1 + e_2, \qquad Te_2 = (0, 2, 1) = 2e_2 + e_3, \qquad Te_3 = (0, 0, 1) = e_3.

By condition (3), $Te_1$ uses $e_2$ , so $e_2$ must come before $e_1$ ; $Te_2$ uses $e_3$ , so $e_3$ must come before $e_2$ . The only order satisfying both is $e_3, e_2, e_1$ , and then

\mathcal{M}\big(T, (e_3, e_2, e_1)\big) = \begin{pmatrix} 1 & 1 & 0 \\ 0 & 2 & 1 \\ 0 & 0 & 3 \end{pmatrix}.

The invariant spans are $\text{span}(e_3)$ (the $z$ -axis) and $\text{span}(e_3, e_2)$ (the $y$ - $z$ plane).

Problem 3 · A Basis Off the Axes

Given: $T \in \mathcal{L}(\mathbf{R}^3)$ , $T(x, y, z) = (y,\ x,\ x + y + 2z)$ , and the basis $v_1 = (1, 1, 0)$ , $v_2 = (1, -1, 0)$ , $v_3 = (0, 0, 1)$ — find $\mathcal{M}(T, (v_1, v_2, v_3))$ and which spans are invariant.

What is the matrix?

Which statement is true?

✅ Correct!

Tv_1 = v_1 + 2v_3

leaves the plane

\text{span}(v_1, v_2)

, so neither span is invariant, even though

Tv_2 \in \text{span}(v_1, v_2)

❌ The coordinates went into rows.

Tv_1 = v_1 + 2v_3

fills column

1

with

(1, 0, 2)

; this transposed matrix only looks upper triangular.

❌ That is the matrix in the standard basis. In the basis

v_1, v_2, v_3

, column

k

holds the coordinates of

Tv_k

in terms of

v_1, v_2, v_3

❌ Check $Tv_2$ .

T(1, -1, 0) = (-1, 1, 0) = -v_2

, so

A_{2,2} = -1

❌ Not quite. Compute

Tv_1

Tv_2

Tv_3

and write each as a combination of

v_1, v_2, v_3

❌ Check $k = 1$ .

Tv_1 = (1, 1, 2)

is not a multiple of

v_1 = (1, 1, 0)

❌ Invariance needs every vector of the span.

Tv_2 \in \text{span}(v_1, v_2)

is not enough:

v_1

is in the span too, and

Tv_1 = (1, 1, 2)

has third coordinate

2

, so it is not.

❌ Check $k = 1$ .

Tv_1 = (1, 1, 2) = v_1 + 2v_3

is not in

\text{span}(v_1)

❌ Not quite. A span is invariant only if

T

maps every vector in it, in particular each

v_j

with

j \le k

, back into it.

Show solution

Images of the basis vectors, written in the basis:

Tv_1 = (1, 1, 2) = v_1 + 2v_3, \qquad Tv_2 = (-1, 1, 0) = -v_2, \qquad Tv_3 = (0, 0, 2) = 2v_3.

So the columns are $(1, 0, 2)$ , $(0, -1, 0)$ , $(0, 0, 2)$ :

\mathcal{M}\big(T, (v_1, v_2, v_3)\big) = \begin{pmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 2 & 0 & 2 \end{pmatrix}.

The entry $A_{3,1} = 2$ lies below the diagonal, so the matrix is not upper triangular.

$\text{span}(v_1)$ : $Tv_1 \notin \text{span}(v_1)$ , so it is not invariant. $\text{span}(v_1, v_2)$ is the plane $z = 0$ , and $v_1$ lies in it while $Tv_1 = (1, 1, 2)$ does not, so it is not invariant either. Condition (3) holds at $k = 2$ (since $Tv_2 = -v_2$ ), yet condition (2) fails at $k = 2$ : the two conditions agree as statements about all $k$ at once, not $k$ by $k$ .

Problem 4 · Reorder to Repair

Given: the operator $T(x, y, z) = (y,\ x,\ x + y + 2z)$ of Problem 3 and the same three vectors $(1, 1, 0)$ , $(1, -1, 0)$ , $(0, 0, 1)$ — find an order of them that makes $\mathcal{M}(T)$ upper triangular.

✅ Correct!

(1, -1, 0)

and

(0, 0, 1)

are each sent to multiples of themselves, and

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

uses only itself and an earlier vector.

❌ This fails at $k = 1$ .

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

is not a multiple of

(1, 1, 0)

❌ This passes $k = 1$ but fails at $k = 2$ .

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

needs

(0, 0, 1)

, which comes later: an eigenvector first is not enough by itself.

❌ The order is a free choice. Problem 3's order fails, but upper triangularity depends on the order: place

(0, 0, 1)

before

(1, 1, 0)

❌ Not quite. Check

Tv_k \in \text{span}(v_1, \ldots, v_k)

for each

k

Show solution

Write $a = (1, 1, 0)$ , $b = (1, -1, 0)$ , $c = (0, 0, 1)$ . From Problem 3:

Ta = a + 2c, \qquad Tb = -b, \qquad Tc = 2c.

Condition (3) holds automatically for $b$ and $c$ wherever they sit, since each image is a multiple of the vector itself. For $a$ , the image uses $c$ , so $c$ must come before $a$ . The order $b, c, a$ does this:

\mathcal{M}\big(T, (b, c, a)\big) = \begin{pmatrix} -1 & 0 & 0 \\ 0 & 2 & 2 \\ 0 & 0 & 1 \end{pmatrix}.

The orders $c, a, b$ and $c, b, a$ work as well; $a, b, c$ and $a, c, b$ fail at $k = 1$ , and $b, a, c$ fails at $k = 2$ .

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