5.A.1 Invariant subspaces

1. Formula
   Observation

   Appears in the handwritten definition as $U \subseteq V$ and in $T \in \mathcal{L}(V)$.

V

The ambient vector space containing the subspace U.

Vector space.

1. Formula
   Observation

   Appears in the handwritten definition as $U \subseteq V$ and in the set-builder notation \{Tu : $u \in U$\}.

U

A subset or subspace of V whose image under T is being considered.

Subset of V; later specified as a subspace of V.

1. Formula
   Observation

   Appears in the handwritten definition as $T \in \mathcal{L}(V)$ and in the expression $T(U)$.

T

A linear operator on the vector space V.

$\mathcal{L}(V)$, the set of linear operators from V to V.

1. Formula
   Observation

   Appears in the handwritten definition as $T \in \mathcal{L}(V)$.

$\mathcal{L}(V)$

The set of all linear operators on the vector space V.

Set of linear maps from V to V.

1. Formula
   Observation

   Appears in the handwritten set-builder notation as $u \in U$.

u

An arbitrary element of the subset or subspace U.

Element of U.

1. Formula
   Observation

   Appears in the handwritten definition as $T(U)$ = \{Tu : $u \in U$\}.

$T(U)$

The image of the set U under the mapping T.

Subset of V.

1. Formula
   Observation

   Appears in the handwritten annotation as range(T) = $T(V)$.

range(T)

The range of the operator T, defined as the image of the entire vector space V.

Subset of V.

1. Formula
   Observation

   The board repeatedly uses $V$ in $U \le V$, $T \in \mathcal{L}(V)$, and $T|_U : U \to V$.

V

The ambient vector space under discussion.

Vector space

1. Formula
   Observation

   The board writes $U \le V$, $T(U)$, $u \in U$, and $T|_U$ throughout the definitions and criterion.

U

A subspace of $V$ being tested for invariance or used as the domain of a restriction.

Subspace of $V$

1. Formula
   Observation

   The board states $T \in \mathcal{L}(V)$ and uses $T(u)$, $T(U)$, and $T|_U$.

2. Audio
   Observation

   The lecturer says to take $T$ to be our operator again.

T

A linear operator on $V$.

Linear operator on $V$

1. Formula
   Observation

   The notation $\mathcal{L}(V)$ appears in $T \in \mathcal{L}(V)$ and $T|_U \in \mathcal{L}(U)$. The symbol is written as a calligraphic L.

$\mathcal{L}(V)$

The set of linear operators on $V$.

Set of linear maps/operators

1. Formula
   Observation

   The first definition line reads $T(U)=\{Tu:u\in U\}$.

$T(U)$

The image of $U$ under $T$, i.e. the set of all vectors $Tu$ with $u\in U$.

Subset of $V$

1. Audio
   Observation

   First off, the image of a set U under a mapping T is denoted by T of U, and this is all points of the form T times u where little u is an element of big U.

2. Formula
   Observation

   Handwritten text: 'Def The image of $U \subseteq V$ under $T \in \mathcal{L}(V)$ is $T(U)$ = \{Tu : $u \in U$\}.'

The image of a subset U of a vector space V under a linear operator T is the set of all vectors obtained by applying T to each element of U.

1. U is a subset of $V (U \subseteq V)$.

2. T is a linear operator on $V (T \in \mathcal{L}(V))$.

1. Audio
   Observation

   So suppose we have an operator T, then we say a subspace U is invariant under T if it gets mapped back into itself. So in other words, if the image of U is contained back in U again.

2. Formula
   Observation

   Handwritten text: 'Def Let $T \in \mathcal{L}(V)$. A subspace $U \subseteq V$ is invariant under T iff $T(U) \subseteq U$.'

A subspace U of a vector space V is called invariant under a linear operator T if applying T to any vector in U results in a vector that is still within U.

1. T is a linear operator on $V (T \in \mathcal{L}(V))$.

2. U is a subspace of $V (U \subseteq V)$.

1. Definition of the image of a set under a linear map

1. Formula
   Observation

   The first handwritten line defines the image of $U\le V$ under $T\in\mathcal{L}(V)$ as $T(U)=\{Tu:u\in U\}$.

2. Audio
   Observation

   The spoken explanation matches the displayed definition of the image of $U$ under $T$.

For a subspace $U$ of $V$ and a linear operator $T$ on $V$, the image of $U$ under $T$ is the collection of all outputs obtained by applying $T$ to elements of $U$. This is the set-theoretic image of $U$ under the map $T$.

1. $U\le V$

2. $T\in\mathcal{L}(V)$

1. Formula
   Observation

   The second handwritten line states: $\text{Def Let } T\in\mathcal{L}(V).\ \text{A subspace } U\le V \text{ is invariant under } T \iff T(U)\subseteq U.$

2. Audio
   Observation

   The lecturer explains that to check invariance one picks an arbitrary $u\in U$ and checks that $T(u)$ is back in $U$ again.

A subspace $U$ of $V$ is called invariant under $T$ exactly when applying $T$ to any element of $U$ never leaves $U$. Equivalently, the whole image $T(U)$ is contained in $U$.

1. $T\in\mathcal{L}(V)$

2. $U\le V$

1. Image of a subspace under a linear operator

1. Formula
   Observation

   The third handwritten line reads: $\text{So to check if } U \text{ is invariant, pick an arbitrary } u\in U \text{ and check that } T(u)\in U.$

2. Audio
   Observation

   The lecturer says: pick an arbitrary element of $U$ and check that $T$ of it is back in $U$ again.

To verify that $U$ is invariant under $T$, take an arbitrary vector $u\in U$ and show directly that its image $T(u)$ also lies in $U$. This reduces the set-inclusion condition $T(U)\subseteq U$ to an elementwise test.

1. $U\le V$

2. $T\in\mathcal{L}(V)$

1. Invariant subspace under a linear operator

1. Formula
   Observation

   The board writes: $\text{Def Let } T\in\mathcal{L}(V).\ \text{If } U\le V \text{ is a subspace, then the restriction of } T \text{ to } U \text{ is } T|_U:U\to V \text{ by } T|_U(u)=T(u)\ \text{for } u\in U.$

2. Audio
   Observation

   The lecturer introduces another way to think about invariant subspaces using the idea of a restriction operator, then says the restriction is denoted with a bar and subscript $U$ and is a map from $U$ to $V$ defined by whatever $T$ did.

Given a linear operator $T$ on $V$ and a subspace $U\le V$, the restriction of $T$ to $U$ is the map obtained by keeping the same rule as $T$ but only allowing inputs from $U$. On the board this restricted map is denoted $T|_U$ and is initially viewed as a map from $U$ into $V$.

1. $T\in\mathcal{L}(V)$

2. $U\le V$

1. Image of a subspace under a linear operator

1. Formula
   Observation

   The final handwritten statement reads: $\text{Then } U \text{ is invariant under } T \iff T|_U\in\mathcal{L}(U),\ \text{i.e., } \operatorname{range} T|_U\subseteq U.$

2. Audio
   Observation

   The lecturer says that with this idea, $U$ is invariant under $T$ if and only if the restricted operator is an operator on $U$, and then adds that this means the range of $T$ restricted to $U$ is contained in $U$.

Using the restriction operator gives a second viewpoint on invariance: $U$ is invariant under $T$ exactly when the restricted map $T|_U$ is not merely a map from $U$ to $V$, but actually an operator on $U$ itself. That happens precisely when every output of $T|_U$ stays inside $U$.

1. $T\in\mathcal{L}(V)$

2. $U\le V$

1. Invariant subspace under a linear operator
2. Restriction of an operator to a subspace

1. Formula
   Observation

   Def The image of $\mathcal{U}\le V$ under $T\in\mathcal{L}(V)$ is $T(\mathcal{U})$ = \{ Tu : $u\in\mathcal{U}$ \}.

For a subspace $\mathcal{U}$ of V and a linear operator T on V, the image of $\mathcal{U}$ under T is the set of all vectors obtained by applying T to elements of $\mathcal{U}$.

1. $T \in \mathcal{L}(V)$

2. $\mathcal{U} \le V$

1. Formula
   Observation

   Def Let $T\in\mathcal{L}(V)$. A subspace $\mathcal{U}\le V$ is invariant under T iff $T(\mathcal{U})\subseteq \mathcal{U}$.

2. Audio
   Observation

   The lecturer explains that invariant subspaces let us break V into smaller chunks and study how T acts on each chunk.

A subspace $\mathcal{U}$ of V is invariant under T exactly when applying T to any vector in $\mathcal{U}$ produces another vector still lying in $\mathcal{U}$. Equivalently, the image of $\mathcal{U}$ under T is contained in $\mathcal{U}$.

1. $T \in \mathcal{L}(V)$

2. $\mathcal{U} \le V$

1. Definition of the image of a subspace under a linear map

1. Formula
   Observation

   So to check if $\mathcal{U}$ is invariant, pick an arbitrary $u\in\mathcal{U}$ and check that $T(u)\in\mathcal{U}$.

To verify that a subspace is invariant under T, take an arbitrary element u of the subspace and show that its image $T(u)$ also belongs to the subspace.

1. $\mathcal{U} \le V$

2. $T \in \mathcal{L}(V)$

1. Definition of an invariant subspace

1. Formula
   Observation

   Def Let $T\in\mathcal{L}(V)$. If $\mathcal{U}\le V$ is a subspace, then the restriction of T to $\mathcal{U}$ is T|_{$\mathcal{U}$}:\mathcal{U}$\to V$ by T|_{$\mathcal{U}$}(u)=$T(u)$ for $u\in\mathcal{U}$.

Given a linear operator T on V and a subspace $\mathcal{U}$, the restriction of T to $\mathcal{U}$ is the map that agrees with T on every vector of $\mathcal{U}$, but whose domain is only $\mathcal{U}$.

1. $T \in \mathcal{L}(V)$

2. $\mathcal{U} \le V$

1. Definition of the image of a subspace under a linear map

1. Formula
   Observation

   Then $\mathcal{U}$ is invariant under T iff T|_{$\mathcal{U}$} $\in \mathcal{L}(\mathcal{U})$, i.e., ran(T|_{$\mathcal{U}$}) $\subseteq \mathcal{U}$.

A subspace $\mathcal{U}$ is invariant under T precisely when the restricted map T|_{$\mathcal{U}$} can be regarded as a linear operator on $\mathcal{U}$; equivalently, the range of the restricted map lies inside $\mathcal{U}$.

1. $T \in \mathcal{L}(V)$

2. $\mathcal{U} \le V$

1. Definition of an invariant subspace
2. Definition of the restriction of a linear map to a subspace

1. Formula
   Observation

   A green annotation at the top right reads $\operatorname{range} T = T(V)$, followed by $\text{but } U \text{ can be smaller}$.

For the operator $T$ under discussion, $\operatorname{range} T = T(V)$.

1. $T\in\mathcal{L}(V)$

For the fixed operator $T$ on $V$.

1. Formula
   Observation

   The board states $U\le V$ is invariant under $T \iff T(U)\subseteq U$.

2. Audio
   Observation

   The lecturer explains the same condition elementwise by saying to pick arbitrary $u\in U$ and check $T(u)$ is back in $U$.

A subspace $U\le V$ is invariant under $T$ if and only if $T(U)\subseteq U$.

1. $T\in\mathcal{L}(V)$

2. $U\le V$

For all $u\in U$, the condition is equivalent to $T(u)\in U$.

1. Formula
   Observation

   The final line states $U$ is invariant under $T \iff T|_U\in\mathcal{L}(U)$, i.e. $\operatorname{range} T|_U\subseteq U$.

2. Audio
   Observation

   The lecturer says $U$ is invariant under $T$ if and only if the restricted operator is an operator on $U$, and then restates this as the range of $T$ restricted to $U$ being contained in $U$.

For $T\in\mathcal{L}(V)$ and $U\le V$, $U$ is invariant under $T$ if and only if $T|_U\in\mathcal{L}(U)$, equivalently $\operatorname{range} T|_U\subseteq U$.

1. $T\in\mathcal{L}(V)$

2. $U\le V$

For all $u\in U$, $T|_U(u)=T(u)$, and the equivalence concerns whether these values remain in $U$.

1. Formula
   Observation

   Prop For $T\in\mathcal{L}(V)$, \{0\}, V, null T, ran T are invariant.

2. Audio
   Observation

   The lecturer states that the trivial space V, the null space, and the range of T are all invariant.

For any linear operator T on V, the subspaces \{0\}, V, null T, and ran T are invariant under T.

1. $T \in \mathcal{L}(V)$

For all $T \in \mathcal{L}(V)$.

1. Formula
   Observation

   The board moves from the earlier definition $T(U)\subseteq U$ to the later statement $T|_U\in\mathcal{L}(U)$, i.e. $\operatorname{range} T|_U\subseteq U$.

2. Audio
   Observation

   The lecturer explicitly says there is another way to think about invariant subspaces using the restriction operator, then concludes that $U$ is invariant under $T$ iff the restricted operator is an operator on $U$, meaning its range is contained in $U$.

1. The video states the equivalence and explains it verbally, but does not write out a formal proof line by line.

1. Expression

   $$T(U)=\{Tu:u\in U\}$$
   Explanation

   Start from the definition of the image of $U$ under $T$.

   Justification

   Definition shown on the board.

   Shown in the video
2. Expression

   $$U\text{ invariant under }T\iff T(U)\subseteq U$$
   Explanation

   Invariance is defined by requiring the image of $U$ to lie inside $U$.

   Justification

   Definition shown on the board.

   Shown in the video
3. Expression

   $$T|_U:U\to V,\quad T|_U(u)=T(u)\text{ for }u\in U$$
   Explanation

   Introduce the restriction of $T$ to $U$ as the same rule as $T$, but with inputs restricted to $U$.

   Justification

   Definition shown on the board.

   Shown in the video
4. Expression

   $$\operatorname{range} T|_U = T(U)$$
   Explanation

   Because $T|_U(u)=T(u)$ for every $u\in U$, the set of outputs of the restricted map is exactly the image of $U$ under $T$.

   Justification

   Derived from the displayed definition of $T|_U$ and the definition of image.

   Derived from the video
5. Expression

   $$T|_U\in\mathcal{L}(U)\iff \operatorname{range} T|_U\subseteq U$$
   Explanation

   For the restricted map to be an operator on $U$, its outputs must land back in $U$.

   Justification

   Explanation given verbally and written as the final equivalence on the board.

   Shown in the video
6. Expression

   $$U\text{ invariant under }T\iff T|_U\in\mathcal{L}(U)$$
   Explanation

   Combining the previous observations yields the advertised equivalence between the two viewpoints on invariance.

   Justification

   Chain of the displayed definitions plus the derived identity of ranges.

   Derived from the video

The subspace $U$ is invariant under $T$ exactly when the restriction $T|_U$ is an operator on $U$, equivalently when $\operatorname{range} T|_U\subseteq U$.

1. Formula
   Observation

   Proof T(\{0\}) = \{0\}

2. Formula
   Observation

   $T(V) \subseteq V$

3. Formula
   Observation

   T(null T) = \{0\} $\subseteq$ null T

4. Formula
   Observation

   ran $T = T(V) \subseteq V \implies T$(ran T) $\subseteq T(V)$ = ran T.

5. Audio
   Observation

   The lecturer verbally justifies each line while writing it.

1. Expression

   $$T(\{0\}) = \{0\}$$
   Explanation

   The only element of the trivial subspace is 0, and a linear map sends 0 to 0, so the image of \{0\} is again \{0\}.

   Justification

   Linearity gives $T(0)=0$; definition of image of a subspace.

   Shown in the video
2. Expression

   $$T(V) \subseteq V$$
   Explanation

   Since T is an operator on V, its codomain is V, so the image of all of V is automatically contained in V.

   Justification

   Definition of $T \in \mathcal{L}(V)$.

   Shown in the video
3. Expression

   $$T(null T) = \{0\} \subseteq null T$$
   Explanation

   Every vector in null T is sent to 0 by T, so the image of null T is exactly \{0\}; because null T is a subspace, it contains 0, hence \{0\} $\subseteq$ null T.

   Justification

   Definition of null space and subspace property.

   Shown in the video
4. Expression

   $$ran T = T(V) \subseteq V$$
   Explanation

   The range of T is the image of the whole space V.

   Justification

   Definition of range/image.

   Shown in the video
5. Expression

   $$T(ran T) \subseteq T(V) = ran T$$
   Explanation

   Applying T to both sides of ran $T \subseteq V$ gives T(ran T) $\subseteq T(V)$, and $T(V)$ is exactly ran T.

   Justification

   Monotonicity of images under a map plus definition of range.

   Shown in the video

Therefore \{0\}, V, null T, and ran T are all invariant under T.

1. Audio
   Observation

   Just to distinguish here, the range of T is T of the entire vector space V, but U can be smaller.

2. Diagram
   Observation

   Green handwritten annotation appears above the definition: 'range(T) = $T(V)$, but U can be smaller' with an arrow pointing to $T(U)$.

1. Text 'range(T) = $T(V)$'

2. Text 'but U can be smaller'

3. Arrow pointing to $T(U)$

1. Green text and arrow are drawn on the screen.

1. The black handwritten definition of $T(U)$ remains visible.

The visual annotation clarifies that while the range of T is the image of the entire space V, the image $T(U)$ can be taken over a smaller subset U.

1. Audio
   Observation

   And so maybe it's worth emphasizing that this is typically not equal, just contained in. It might be equal, but it's often strictly smaller.

2. Diagram
   Observation

   Blue handwritten annotation appears below the definition: 'need not be equal' with an arrow pointing to the subset symbol $\subseteq$ in $T(U) \subseteq U$.

1. Text 'need not be equal'

2. Arrow pointing to $\subseteq$

1. Blue text and arrow are drawn on the screen.

1. The black handwritten definition of an invariant subspace remains visible.

The visual annotation highlights that the condition for an invariant subspace requires the image $T(U)$ to be a subset of U, which does not necessarily mean they are equal sets.

1. Diagram
   Observation

   The entire clip is a white digital board with the title '5.A.1 Invariant subspaces' at top left, black handwritten mathematical text in the main area, and a green annotation at top right.

1. Title '5.A.1 Invariant subspaces'

2. Black handwritten definitions and statements

3. Green annotation at top right

1. New lines are added progressively beneath the existing definitions as the lecture advances.

1. The title remains fixed at the top left.

2. The green note about $\operatorname{range} T=T(V)$ remains visible at the top right.

3. The earlier definitions stay on screen while later ones are added.

The visual organization presents a sequence of definitions culminating in an equivalent characterization of invariant subspaces.

1. Diagram
   Observation

   At the top right, green handwriting reads $\operatorname{range} T = T(V)$, but $U$ can be smaller, with a small arrow pointing toward the main definition area.

1. Green text $\operatorname{range} T = T(V)$

2. Green text 'but $U$ can be smaller'

3. Small arrow

1. The note stays visible throughout the clip.

The annotation warns that the full range of $T$ is computed on all of $V$, whereas the subspace $U$ under discussion may be a smaller domain whose image is $T(U)$.

1. Animation
   Observation

   A red dot appears under the $T(u)$ in the formula $T|_U(u)=T(u)$, then shifts to the $U$ in '$U\le V$'.

1. The exact intended emphasis is not verbally stated at that instant.

1. Red dot

2. Formula $T|_U(u)=T(u)$

3. Text '$U\le V$'

1. The red dot first marks the right-hand side $T(u)$ in the restriction formula.

2. It then moves to the $U$ in the hypothesis '$U\le V$'.

1. The underlying formulas remain unchanged while the pointer moves.

The motion visually links the rule defining $T|_U$ back to the original operator $T$ and to the subspace hypothesis on $U$.

1. Diagram
   Observation

   The screen shows a white digital board titled '5.A.1 Invariant subspaces' with handwritten definitions, a green annotation at top right, and later a proposition and proof written below.

1. Title '5.A.1 Invariant subspaces'

2. Handwritten definitions

3. Green annotation 'ran $T = T(V)$, but U can be smaller'

4. Proposition line

5. Proof lines

1. At about 20 seconds, the lecturer begins writing 'Prop'.

2. From about 20 to 57 seconds, the proposition listing \{0\}, V, null T, ran T is written.

3. From about 60 to 147 seconds, the proof lines are added one by one beneath the proposition.

1. The earlier definitions remain visible above while the new proposition and proof are written below.

2. The board remains a single static handwritten page without scene cuts.

The visual progression mirrors the logical structure: first definitions, then a proposition about standard invariant subspaces, then a direct verification using images and containment.

1. Audio
   Observation

   Just to distinguish here, the range of T is T of the entire vector space V, but U can be smaller. And so that's what distinguishes it from the range as we've seen before.

Assuming that $T(U)$ always refers to the range of the operator T.

The range of T is specifically $T(V)$, the image of the entire vector space. $T(U)$ is the image of a potentially smaller subset U.

1. Audio
   Observation

   And so maybe it's worth emphasizing that this is typically not equal, just contained in. It might be equal, but it's often strictly smaller.

Believing that for a subspace to be invariant, its image must be exactly equal to the subspace itself ($T(U) = U$).

The definition only requires that the image is contained within the subspace ($T(U) \subseteq U$). It can be strictly smaller.

1. Diagram
   Observation

   The green note explicitly says $\operatorname{range} T = T(V)$, but $U$ can be smaller.

One might think $T(U)$ is the same as $\operatorname{range} T$ just because both are images under $T$.

The board distinguishes them: $\operatorname{range} T=T(V)$ uses all inputs from $V$, while $T(U)$ uses only inputs from the smaller subspace $U$.

1. Formula
   Observation

   The restriction is first written as $T|_U:U\to V$ by $T|_U(u)=T(u)$ for $u\in U$.

2. Formula
   Observation

   Only later does the board state $T|_U\in\mathcal{L}(U)$, i.e. $\operatorname{range} T|_U\subseteq U$.

After defining $T|_U$, one might assume it is already an operator on $U$.

The video first defines $T|_U$ as a map from $U$ to $V$. It becomes an operator on $U$ exactly when $U$ is invariant, equivalently when $\operatorname{range} T|_U\subseteq U$.