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Orthogonal Complements

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This lecture segment introduces orthogonal complements using a concrete example in R4\mathbb{R}^4. The instructor defines a 2D subspace WW and its orthogonal complement W⊥W^\perp, showing that their direct sum equals R4\mathbb{R}^4. A specific vector is then decomposed into components belonging to WW and W⊥W^\perp, illustrating the unique representation property of direct sums. This 23-second excerpt is the closing portion of a linear-algebra whiteboard lesson on orthogonal complements. The board already displays a concrete example in R4\mathbb{R}^4: WW has basis {(1,0,0,0)T,(0,1,0,0)T}\{(1,0,0,0)^T,(0,1,0,0)^T\}, W⊥W^{\perp} has basis {(0,0,1,0)T,(0,0,0,1)T}\{(0,0,1,0)^T,(0,0,0,1)^T\}, and the lecturer states W⊕W⊥=R4W \oplus W^{\perp} = \mathbb{R}^4. A general vector (a,b,c,d)T(a,b,c,d)^T is decomposed as (a,b,0,0)T+(0,0,c,d)T(a,b,0,0)^T + (0,0,c,d)^T, with a handwritten note emphasizing that the representation is unique. In the audio, the speaker explicitly says there is no other way of writing that decomposition, then concludes that this gives a basic idea of orthogonal complements and ends the session.

Use the learning inspector for key ideas and moments, or open the reading tabs for the complete notes.

Chapters

0:00Defining Subspaces W and W_perp1:00Direct Sum and Vector Decomposition Example2:20Uniqueness of Direct Sum Representation2:40Uniqueness of the displayed decomposition2:47Closing summary of the orthogonal-complement example

Learning script

Generated from the video's visuals and explanation; not verbatim speech.

We begin by defining a specific subspace WW within the 4-dimensional real vector space R4\mathbb{R}^4. We assign WW a basis consisting of the standard unit vectors for the first two dimensions: (1000)\begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix} and (0100)\begin{pmatrix} 0 \\ 1 \\ 0 \\ 0 \end{pmatrix}. This makes WW a 2-dimensional subspace.

Next, we define the orthogonal complement of WW, denoted W⊥W^\perp. We give it a basis of the remaining standard unit vectors: (0010)\begin{pmatrix} 0 \\ 0 \\ 1 \\ 0 \end{pmatrix} and (0001)\begin{pmatrix} 0 \\ 0 \\ 0 \\ 1 \end{pmatrix}. By construction, every vector in WW is orthogonal to every vector in W⊥W^\perp.

Because WW and W⊥W^\perp share no non-zero vectors and together span all of R4\mathbb{R}^4, their direct sum is the entire space: W⊕W⊥=R4W \oplus W^\perp = \mathbb{R}^4. This means any vector in R4\mathbb{R}^4 can be expressed as the sum of a vector in WW and a vector in W⊥W^\perp.

To illustrate, consider the arbitrary vector (43−26)\begin{pmatrix} 4 \\ 3 \\ -2 \\ 6 \end{pmatrix}. We can decompose it by separating the components corresponding to our chosen bases. The first two components form a vector in WW: (4300)\begin{pmatrix} 4 \\ 3 \\ 0 \\ 0 \end{pmatrix}. The last two components form a vector in W⊥W^\perp: (00−26)\begin{pmatrix} 0 \\ 0 \\ -2 \\ 6 \end{pmatrix}.

The key property highlighted here is that this direct sum representation is unique. There is only one way to split a given vector in R4\mathbb{R}^4 into a part belonging to WW and a part belonging to W⊥W^\perp based on these specific subspaces.

The clip opens on a finished whiteboard example titled "Orthogonal Complements." The board has already fixed the ambient space as R4\mathbb{R}^4, assigned WW the basis {(1,0,0,0)T,(0,1,0,0)T}\left\{(1,0,0,0)^T,(0,1,0,0)^T\right\}, and assigned W⊥W^{\perp} the basis {(0,0,1,0)T,(0,0,0,1)T}\left\{(0,0,1,0)^T,(0,0,0,1)^T\right\}. Underneath, it states W⊕W⊥=R4W \oplus W^{\perp} = \mathbb{R}^4 and rewrites a general vector (a,b,c,d)T(a,b,c,d)^T as (a,b,0,0)T+(0,0,c,d)T(a,b,0,0)^T + (0,0,c,d)^T.

At this point the lecturer's key verbal point is uniqueness: when a vector is written as a sum of one element from WW and one element from W⊥W^{\perp}, that representation is not arbitrary. He says there is no other way of writing it, matching the handwritten note on the board that reads "This representation is UNIQUE."

After that emphasis, the remainder of the excerpt functions as a wrap-up rather than a new derivation. The speaker says the example gives a basic idea of orthogonal complements and indicates that the topic will continue in a later lesson.

The final seconds contain course-closing remarks only. No new formula, proof step, or example is introduced; the same static board remains on screen as the lecturer signs off.

Knowledge cards

01

Subspace W in R4R^4

A 2-dimensional subspace of R4\mathbb{R}^4 defined by the basis vectors (1000)\begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix} and (0100)\begin{pmatrix} 0 \\ 1 \\ 0 \\ 0 \end{pmatrix}. Any vector in WW has zeros in its third and fourth components.

W=span{(1000),(0100)}W = \text{span}\left\{ \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 1 \\ 0 \\ 0 \end{pmatrix} \right\}
02

Orthogonal Complement W^perp

The subspace of all vectors in R4\mathbb{R}^4 that are orthogonal to every vector in WW. In this example, it is spanned by (0010)\begin{pmatrix} 0 \\ 0 \\ 1 \\ 0 \end{pmatrix} and (0001)\begin{pmatrix} 0 \\ 0 \\ 0 \\ 1 \end{pmatrix}. Any vector in W⊥W^\perp has zeros in its first and second components.

W⊥=span{(0010),(0001)}W^\perp = \text{span}\left\{ \begin{pmatrix} 0 \\ 0 \\ 1 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 0 \\ 0 \\ 1 \end{pmatrix} \right\}
03

Direct Sum Decomposition

The equation W⊕W⊥=R4W \oplus W^\perp = \mathbb{R}^4 signifies that every vector v∈R4v \in \mathbb{R}^4 can be written uniquely as v=w+uv = w + u, where w∈Ww \in W and u∈W⊥u \in W^\perp. For the vector (43−26)\begin{pmatrix} 4 \\ 3 \\ -2 \\ 6 \end{pmatrix}, the decomposition is (4300)+(00−26)\begin{pmatrix} 4 \\ 3 \\ 0 \\ 0 \end{pmatrix} + \begin{pmatrix} 0 \\ 0 \\ -2 \\ 6 \end{pmatrix}.

(43−26)=(4300)+(00−26)\begin{pmatrix} 4 \\ 3 \\ -2 \\ 6 \end{pmatrix} = \begin{pmatrix} 4 \\ 3 \\ 0 \\ 0 \end{pmatrix} + \begin{pmatrix} 0 \\ 0 \\ -2 \\ 6 \end{pmatrix}
04

Orthogonal complements in a concrete R4\mathbb{R}^4 example

The board defines a subspace WW of R4\mathbb{R}^4 using the first two standard basis vectors and defines W⊥W^{\perp} using the last two. This is the concrete setup used to illustrate orthogonal complements.

W=span⁡{(1000),(0100)},W⊥=span⁡{(0010),(0001)}W = \operatorname{span}\left\{\begin{pmatrix}1\\0\\0\\0\end{pmatrix},\begin{pmatrix}0\\1\\0\\0\end{pmatrix}\right\},\quad W^{\perp} = \operatorname{span}\left\{\begin{pmatrix}0\\0\\1\\0\end{pmatrix},\begin{pmatrix}0\\0\\0\\1\end{pmatrix}\right\}
05

Direct-sum identity shown on the board

With the above choices, the lecture states that the whole space splits as the direct sum of WW and its orthogonal complement.

W⊕W⊥=R4W \oplus W^{\perp} = \mathbb{R}^4
06

Coordinate decomposition of a general vector

A vector in R4\mathbb{R}^4 is split into a part supported in the first two coordinates and a part supported in the last two coordinates. The first summand lies in WW, and the second lies in W⊥W^{\perp}.

(abcd)=(ab00)+(00cd)\begin{pmatrix}a\\b\\c\\d\end{pmatrix} = \begin{pmatrix}a\\b\\0\\0\end{pmatrix} + \begin{pmatrix}0\\0\\c\\d\end{pmatrix}
07

Uniqueness of the W⊕W⊥W \oplus W^{\perp} representation

The lecturer explicitly stresses that this way of writing the vector as a sum of one element of WW and one element of W⊥W^{\perp} is unique. The board reinforces the point with the note "This representation is UNIQUE."

08

Closing role of the excerpt

After the uniqueness remark, the clip serves as a conclusion to the lesson: the speaker says the example gives a basic idea of orthogonal complements and that the topic will continue next time. No additional mathematics is introduced in the final seconds.

Detailed learning notes

Explore conditions, steps and evidence. Supplementary explanations are labeled separately from content shown in the video.

Symbols · 10

R4\mathbb{R}^4

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    Written as R4\mathbb{R}^4 on the whiteboard.

  2. Audio
    Observation

    Spoken as 'R four'.

Symbol

R4\mathbb{R}^4

Meaning

The vector space of all 4-dimensional real vectors.

Domain

Vector space

W

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    Written as WW on the whiteboard.

  2. Audio
    Observation

    Spoken as 'W'.

Symbol

W

Meaning

A subspace of R4\mathbb{R}^4 with basis (1000),(0100)\begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 1 \\ 0 \\ 0 \end{pmatrix}.

Domain

Subspace of R4\mathbb{R}^4

W⊥W^\perp

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    Written as W⊥W^\perp on the whiteboard.

  2. Audio
    Observation

    Spoken as 'W perp'.

Symbol

W⊥W^\perp

Meaning

The orthogonal complement of WW, a subspace of R4\mathbb{R}^4 with basis (0010),(0001)\begin{pmatrix} 0 \\ 0 \\ 1 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 0 \\ 0 \\ 1 \end{pmatrix}.

Domain

Subspace of R4\mathbb{R}^4

R4\mathbb{R}^4

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    The whiteboard shows the ambient space as R4\mathbb{R}^4 in the lines "Let's take R4\mathbb{R}^4" and "Now W⊕W⊥=R4W \oplus W^{\perp} = \mathbb{R}^4".

Symbol

R4\mathbb{R}^4

Meaning

The four-dimensional real vector space used as the ambient space for the example.

Domain

Vector space over R\mathbb{R} with dimension 4.

W

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    The board states "Let WW have basis ..." and later uses WW in W⊕W⊥=R4W \oplus W^{\perp} = \mathbb{R}^4.

Symbol

W

Meaning

A subspace of R4\mathbb{R}^4 whose displayed basis is {(1000),(0100)}\left\{\begin{pmatrix}1\\0\\0\\0\end{pmatrix},\begin{pmatrix}0\\1\\0\\0\end{pmatrix}\right\}.

Domain

Subspace of R4\mathbb{R}^4; in this example it is the span of the first two standard coordinate vectors.

W⊥W^{\perp}

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    The board states "Let W⊥W^{\perp} have basis ..." and later uses W⊥W^{\perp} in W⊕W⊥=R4W \oplus W^{\perp} = \mathbb{R}^4.

Symbol

W⊥W^{\perp}

Meaning

The orthogonal complement of WW in R4\mathbb{R}^4, shown with basis {(0010),(0001)}\left\{\begin{pmatrix}0\\0\\1\\0\end{pmatrix},\begin{pmatrix}0\\0\\0\\1\end{pmatrix}\right\}.

Domain

Subspace of R4\mathbb{R}^4; here it is the span of the last two standard coordinate vectors.

⊕\oplus

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    The equation on the board reads W⊕W⊥=R4W \oplus W^{\perp} = \mathbb{R}^4.

Symbol

⊕\oplus

Meaning

Direct sum of subspaces, indicating that every vector in the ambient space is written as a sum of one vector from each subspace in a unique way.

Domain

Operation on subspaces of a vector space.

(abcd)\begin{pmatrix}a\\b\\c\\d\end{pmatrix}

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    The board writes "a vector in R4=(abcd)\mathbb{R}^4 = \begin{pmatrix}a\\b\\c\\d\end{pmatrix} can be written as ...".

Symbol

(abcd)\begin{pmatrix}a\\b\\c\\d\end{pmatrix}

Meaning

A general vector in R4\mathbb{R}^4 with scalar coordinates a,b,c,da,b,c,d.

Domain

a,b,c,d∈Ra,b,c,d \in \mathbb{R}.

(ab00)\begin{pmatrix}a\\b\\0\\0\end{pmatrix}

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    The decomposition line shows the first summand as (ab00)\begin{pmatrix}a\\b\\0\\0\end{pmatrix}.

Symbol

(ab00)\begin{pmatrix}a\\b\\0\\0\end{pmatrix}

Meaning

The component of (a,b,c,d)T(a,b,c,d)^T lying in WW, obtained by keeping the first two coordinates and zeroing the last two.

Domain

Element of W=span⁡{(1,0,0,0)T,(0,1,0,0)T}W = \operatorname{span}\{(1,0,0,0)^T,(0,1,0,0)^T\}.

(00cd)\begin{pmatrix}0\\0\\c\\d\end{pmatrix}

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    The decomposition line shows the second summand as (00cd)\begin{pmatrix}0\\0\\c\\d\end{pmatrix}.

Symbol

(00cd)\begin{pmatrix}0\\0\\c\\d\end{pmatrix}

Meaning

The component of (a,b,c,d)T(a,b,c,d)^T lying in W⊥W^{\perp}, obtained by zeroing the first two coordinates and keeping the last two.

Domain

Element of W⊥=span⁡{(0,0,1,0)T,(0,0,0,1)T}W^{\perp} = \operatorname{span}\{(0,0,1,0)^T,(0,0,0,1)^T\}.

Knowledge points · 6

Direct Sum of Subspaces

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    Written as W⊕W⊥=R4W \oplus W^\perp = \mathbb{R}^4 on the whiteboard.

  2. Audio
    Observation

    Spoken as 'W direct sum W perp, well, that's equal to R four'.

Definition
Explanation

The direct sum of two subspaces UU and VV, denoted U⊕VU \oplus V, is the set of all vectors that can be written uniquely as the sum of a vector in UU and a vector in VV. In this example, W⊕W⊥=R4W \oplus W^\perp = \mathbb{R}^4.

Formula
W⊕W⊥=R4W \oplus W^\perp = \mathbb{R}^4
Conditions
  1. WW and W⊥W^\perp are subspaces of R4\mathbb{R}^4

  2. W∩W⊥={0}W \cap W^\perp = \{\mathbf{0}\}

Prerequisites
  1. Subspace
  2. Orthogonal Complement

Subspace

Clear evidence
Shown in the video
Evidence
  1. Audio
    Observation

    Spoken as 'let's say that W is the subspace that has these two vectors as a basis'.

Definition
Explanation

A subset of a vector space that is itself a vector space under the same operations. Here, WW is a 2-dimensional subspace of R4\mathbb{R}^4.

Conditions
  1. Closed under vector addition

  2. Closed under scalar multiplication

  3. Contains the zero vector

Orthogonal Complement

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    Written as W⊥W^\perp on the whiteboard.

  2. Audio
    Observation

    Spoken as 'we will let W perp have basis...'.

Definition
Explanation

For a subspace WW of an inner product space VV, the orthogonal complement W⊥W^\perp is the set of all vectors in VV that are orthogonal to every vector in WW.

Conditions
  1. Defined within an inner product space

Prerequisites
  1. Subspace

Concrete orthogonal-complement pair in R4\mathbb{R}^4

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    The whiteboard heading is "Orthogonal Complements" and the worked setup gives bases for WW and W⊥W^{\perp} inside R4\mathbb{R}^4.

  2. Audio
    Observation

    The lecturer says this gives "a nice basic idea of orthogonal complements to work with."

Definition
Explanation

The clip presents a specific example in R4\mathbb{R}^4: WW is spanned by the first two standard basis vectors, and W⊥W^{\perp} is spanned by the last two. This illustrates how a subspace and its orthogonal complement split the ambient space into coordinate blocks.

Formula
W=span⁡{(1000),(0100)},W⊥=span⁡{(0010),(0001)}W = \operatorname{span}\left\{\begin{pmatrix}1\\0\\0\\0\end{pmatrix},\begin{pmatrix}0\\1\\0\\0\end{pmatrix}\right\},\quad W^{\perp} = \operatorname{span}\left\{\begin{pmatrix}0\\0\\1\\0\end{pmatrix},\begin{pmatrix}0\\0\\0\\1\end{pmatrix}\right\}
Conditions
  1. The ambient space is R4\mathbb{R}^4.

  2. The displayed bases are taken as given on the board.

Direct-sum decomposition of a vector in R4\mathbb{R}^4

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    The board explicitly writes W⊕W⊥=R4W \oplus W^{\perp} = \mathbb{R}^4.

  2. Formula
    Observation

    Below it, a general vector (a,b,c,d)T(a,b,c,d)^T is rewritten as (a,b,0,0)T+(0,0,c,d)T(a,b,0,0)^T + (0,0,c,d)^T.

Formula
Explanation

Using the chosen subspaces, every vector in R4\mathbb{R}^4 is decomposed into one part in WW and one part in W⊥W^{\perp}. The first summand keeps the first two coordinates, and the second keeps the last two.

Formula
(abcd)=(ab00)+(00cd)\begin{pmatrix}a\\b\\c\\d\end{pmatrix} = \begin{pmatrix}a\\b\\0\\0\end{pmatrix} + \begin{pmatrix}0\\0\\c\\d\end{pmatrix}
Conditions
  1. a,b,c,d∈Ra,b,c,d \in \mathbb{R}.

  2. WW and W⊥W^{\perp} are the subspaces listed on the board.

Prerequisites
  1. Concrete orthogonal-complement pair in R4\mathbb{R}^4

Uniqueness of the W⊕W⊥W \oplus W^{\perp} representation

Clear evidence
Shown in the video
Evidence
  1. Audio
    Observation

    The lecturer says, "of two individual subspaces, the way that I write it is unique. There's no other way of writing that."

  2. Formula
    Observation

    The board has the handwritten note "This representation is UNIQUE." next to the decomposition.

Method
Explanation

For the displayed decomposition, the lecturer emphasizes that writing a vector as a sum of one element of WW and one element of W⊥W^{\perp} is unique: there is no alternative pair of summands producing the same vector under this direct-sum setup.

Formula
Conditions
  1. Applies to the decomposition associated with W⊕W⊥=R4W \oplus W^{\perp} = \mathbb{R}^4 shown on the board.

Prerequisites
  1. Direct-sum decomposition of a vector in R4\mathbb{R}^4
Claims and conditions · 2

The example satisfies W⊕W⊥=R4W \oplus W^{\perp} = \mathbb{R}^4

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    The board states "Now W⊕W⊥=R4W \oplus W^{\perp} = \mathbb{R}^4".

Proposition
Statement

For the subspaces displayed on the board, W⊕W⊥=R4W \oplus W^{\perp} = \mathbb{R}^4.

Hypotheses
  1. WW has basis {(1,0,0,0)T,(0,1,0,0)T}\left\{(1,0,0,0)^T,(0,1,0,0)^T\right\}.

  2. W⊥W^{\perp} has basis {(0,0,1,0)T,(0,0,0,1)T}\left\{(0,0,1,0)^T,(0,0,0,1)^T\right\}.

Quantifiers

Universal over all vectors of R4\mathbb{R}^4 once the two subspaces are fixed.

The displayed decomposition is unique

Clear evidence
Shown in the video
Evidence
  1. Audio
    Observation

    The lecturer says the way of writing the vector using the two subspaces is unique and that there is no other way of writing it.

  2. Formula
    Observation

    The board note reads "This representation is UNIQUE."

Proposition
Statement

In the shown decomposition of a vector in R4\mathbb{R}^4 into a WW-part plus a W⊥W^{\perp}-part, the representation is unique.

Hypotheses
  1. The vector is written using the specific subspaces WW and W⊥W^{\perp} listed on the board.

Quantifiers

For each vector in R4\mathbb{R}^4, there is exactly one such decomposition relative to the displayed WW and W⊥W^{\perp}.

Derivations and proofs · 2

Decomposition into Direct Sum

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    Written as (43−26)=(4300)+(00−26)\begin{pmatrix} 4 \\ 3 \\ -2 \\ 6 \end{pmatrix} = \begin{pmatrix} 4 \\ 3 \\ 0 \\ 0 \end{pmatrix} + \begin{pmatrix} 0 \\ 0 \\ -2 \\ 6 \end{pmatrix} on the whiteboard.

  2. Audio
    Observation

    Spoken as 'it can be written as a vector from this subspace plus a vector from this subspace'.

Proof
Steps
  1. Expression
    (43−26)\begin{pmatrix} 4 \\ 3 \\ -2 \\ 6 \end{pmatrix}
    Explanation

    Start with an arbitrary vector in R4\mathbb{R}^4.

    Justification

    Given in the problem statement.

    Shown in the video
  2. Expression
    =(4300)+(00−26)= \begin{pmatrix} 4 \\ 3 \\ 0 \\ 0 \end{pmatrix} + \begin{pmatrix} 0 \\ 0 \\ -2 \\ 6 \end{pmatrix}
    Explanation

    Split the vector into a part in WW (first two components) and a part in W⊥W^\perp (last two components).

    Justification

    Definition of direct sum and the specific bases chosen for WW and W⊥W^\perp.

    Shown in the video
  3. Expression
    (4300)∈W\begin{pmatrix} 4 \\ 3 \\ 0 \\ 0 \end{pmatrix} \in W
    Explanation

    The first vector is a linear combination of the basis vectors of WW: 4(1000)+3(0100)4\begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix} + 3\begin{pmatrix} 0 \\ 1 \\ 0 \\ 0 \end{pmatrix}.

    Justification

    Definition of span/basis.

    Derived from the video
  4. Expression
    (00−26)∈W⊥\begin{pmatrix} 0 \\ 0 \\ -2 \\ 6 \end{pmatrix} \in W^\perp
    Explanation

    The second vector is a linear combination of the basis vectors of W⊥W^\perp: −2(0010)+6(0001)-2\begin{pmatrix} 0 \\ 0 \\ 1 \\ 0 \end{pmatrix} + 6\begin{pmatrix} 0 \\ 0 \\ 0 \\ 1 \end{pmatrix}.

    Justification

    Definition of span/basis.

    Derived from the video
Conclusion

Any vector in R4\mathbb{R}^4 can be uniquely decomposed into a sum of a vector in WW and a vector in W⊥W^\perp.

Deriving the displayed split into WW and W⊥W^{\perp} components

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    The board writes a general vector (a,b,c,d)T(a,b,c,d)^T and then rewrites it as (a,b,0,0)T+(0,0,c,d)T(a,b,0,0)^T + (0,0,c,d)^T.

Intuitive argument
Steps
  1. Expression
    (abcd)\begin{pmatrix}a\\b\\c\\d\end{pmatrix}
    Explanation

    Start with an arbitrary vector in R4\mathbb{R}^4.

    Justification

    Given directly on the board as the object to decompose.

    Shown in the video
  2. Expression
    (ab00)+(00cd)\begin{pmatrix}a\\b\\0\\0\end{pmatrix} + \begin{pmatrix}0\\0\\c\\d\end{pmatrix}
    Explanation

    Separate the first two coordinates from the last two coordinates.

    Justification

    Coordinate-wise addition reproduces the original vector exactly.

    Shown in the video
  3. Expression
    (ab00)∈W,(00cd)∈W⊥\begin{pmatrix}a\\b\\0\\0\end{pmatrix} \in W,\quad \begin{pmatrix}0\\0\\c\\d\end{pmatrix} \in W^{\perp}
    Explanation

    Identify the first summand as lying in WW and the second as lying in W⊥W^{\perp} using the bases shown above.

    Justification

    Each summand is a linear combination of the corresponding displayed basis vectors.

    Derived from the video
Conclusion

The example visually demonstrates the decomposition R4=W⊕W⊥\mathbb{R}^4 = W \oplus W^{\perp} by splitting coordinates into two orthogonal blocks.

Worked examples · 2

Decomposing a Vector in R4R^4

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    Written as (43−26)=(4300)+(00−26)\begin{pmatrix} 4 \\ 3 \\ -2 \\ 6 \end{pmatrix} = \begin{pmatrix} 4 \\ 3 \\ 0 \\ 0 \end{pmatrix} + \begin{pmatrix} 0 \\ 0 \\ -2 \\ 6 \end{pmatrix} on the whiteboard.

  2. Audio
    Observation

    Spoken as 'let's say for example... some random vector 4, 3, minus 2, 6'.

Problem

Write the vector (43−26)\begin{pmatrix} 4 \\ 3 \\ -2 \\ 6 \end{pmatrix} as a sum of a vector in WW and a vector in W⊥W^\perp.

Given
  1. W=span{(1000),(0100)}W = \text{span}\left\{ \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 1 \\ 0 \\ 0 \end{pmatrix} \right\}

  2. W⊥=span{(0010),(0001)}W^\perp = \text{span}\left\{ \begin{pmatrix} 0 \\ 0 \\ 1 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 0 \\ 0 \\ 1 \end{pmatrix} \right\}

  3. Target vector: (43−26)\begin{pmatrix} 4 \\ 3 \\ -2 \\ 6 \end{pmatrix}

Goal

Find w∈Ww \in W and v∈W⊥v \in W^\perp such that w+v=(43−26)w + v = \begin{pmatrix} 4 \\ 3 \\ -2 \\ 6 \end{pmatrix}.

Steps
  1. Expression
    (43−26)=(4300)+(00−26)\begin{pmatrix} 4 \\ 3 \\ -2 \\ 6 \end{pmatrix} = \begin{pmatrix} 4 \\ 3 \\ 0 \\ 0 \end{pmatrix} + \begin{pmatrix} 0 \\ 0 \\ -2 \\ 6 \end{pmatrix}
    Explanation

    Separate the components corresponding to the bases of WW and W⊥W^\perp.

    Justification

    Direct sum definition and chosen bases.

    Shown in the video
Answer

w=(4300)w = \begin{pmatrix} 4 \\ 3 \\ 0 \\ 0 \end{pmatrix}, v=(00−26)v = \begin{pmatrix} 0 \\ 0 \\ -2 \\ 6 \end{pmatrix}

Verification

Check that w∈Ww \in W, v∈W⊥v \in W^\perp, and w+vw+v equals the original vector.

Worked example: decomposing a vector in R4\mathbb{R}^4 using WW and W⊥W^{\perp}

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    The full worked board content gives bases for WW and W⊥W^{\perp}, states W⊕W⊥=R4W \oplus W^{\perp} = \mathbb{R}^4, and decomposes (a,b,c,d)T(a,b,c,d)^T.

  2. Audio
    Observation

    The lecturer closes by saying this gives a basic idea of orthogonal complements.

Problem

Show how a general vector in R4\mathbb{R}^4 is written as a sum of one vector from WW and one vector from W⊥W^{\perp} for the subspaces listed on the board.

Given
  1. Ambient space: R4\mathbb{R}^4.

  2. Basis of WW: {(1,0,0,0)T,(0,1,0,0)T}\left\{(1,0,0,0)^T,(0,1,0,0)^T\right\}.

  3. Basis of W⊥W^{\perp}: {(0,0,1,0)T,(0,0,0,1)T}\left\{(0,0,1,0)^T,(0,0,0,1)^T\right\}.

  4. General vector: (a,b,c,d)T(a,b,c,d)^T.

Goal

Produce the explicit W⊕W⊥W \oplus W^{\perp} decomposition and state whether it is unique.

Steps
  1. Expression
    (abcd)=(ab00)+(00cd)\begin{pmatrix}a\\b\\c\\d\end{pmatrix} = \begin{pmatrix}a\\b\\0\\0\end{pmatrix} + \begin{pmatrix}0\\0\\c\\d\end{pmatrix}
    Explanation

    Split the vector into first-two-coordinate and last-two-coordinate parts.

    Justification

    Direct coordinate addition.

    Shown in the video
  2. Expression
    (ab00)∈W,(00cd)∈W⊥\begin{pmatrix}a\\b\\0\\0\end{pmatrix} \in W,\quad \begin{pmatrix}0\\0\\c\\d\end{pmatrix} \in W^{\perp}
    Explanation

    Recognize the two summands as belonging to the displayed subspaces.

    Justification

    They are linear combinations of the respective bases written on the board.

    Derived from the video
  3. Expression
    Explanation

    Conclude that the representation is unique.

    Justification

    Stated explicitly by the lecturer and written on the board as "This representation is UNIQUE."

    Shown in the video
Answer

(a,b,c,d)T=(a,b,0,0)T+(0,0,c,d)T(a,b,c,d)^T = (a,b,0,0)^T + (0,0,c,d)^T, with the first term in WW, the second in W⊥W^{\perp}, and the decomposition unique.

Verification

Adding the two displayed summands coordinatewise recovers (a,b,c,d)T(a,b,c,d)^T; membership follows from the bases shown on the board.

Visual events · 3

Writing Basis Vectors

Clear evidence
Shown in the video
Evidence
  1. Animation
    Observation

    Instructor writes the basis vectors for WW and W⊥W^\perp on the whiteboard.

Objects
  1. Whiteboard

  2. Marker

  3. Basis vectors for WW

  4. Basis vectors for W⊥W^\perp

Changes
  1. Vectors are written sequentially on the board.

Invariants
  1. The mathematical meaning of the vectors remains constant once written.

Interpretation

Establishes the specific subspaces WW and W⊥W^\perp being used in the example.

Writing Direct Sum Equation and Decomposition

Clear evidence
Shown in the video
Evidence
  1. Animation
    Observation

    Instructor writes the equation W⊕W⊥=R4W \oplus W^\perp = \mathbb{R}^4 and then the vector decomposition.

Objects
  1. Whiteboard

  2. Marker

  3. Equation W⊕W⊥=R4W \oplus W^\perp = \mathbb{R}^4

  4. Vector decomposition equation

Changes
  1. Equations are written to show the relationship between the spaces and the specific vector split.

Invariants
  1. The underlying mathematical truth of the direct sum property.

Interpretation

Visually demonstrates how an arbitrary vector in R4\mathbb{R}^4 is broken down into components belonging to WW and W⊥W^\perp.

Static summary board for the orthogonal-complement example

Clear evidence
Shown in the video
Evidence
  1. Diagram
    Observation

    The screen remains on a static whiteboard layout with the title "Orthogonal Complements", the subspace bases, the equation W⊕W⊥=R4W \oplus W^{\perp} = \mathbb{R}^4, the decomposition line, and the note "This representation is UNIQUE."

  2. Animation
    Observation

    No new writing or diagrammatic transformation occurs during the clip; only the lecturer gestures in the webcam window.

Objects
  1. Title "Orthogonal Complements".

  2. Lines defining bases for WW and W⊥W^{\perp}.

  3. Equation W⊕W⊥=R4W \oplus W^{\perp} = \mathbb{R}^4.

  4. Decomposition of (a,b,c,d)T(a,b,c,d)^T.

  5. Handwritten uniqueness note.

Changes
  1. No mathematical content is added or erased during the 23-second excerpt.

Invariants
  1. The displayed subspaces, equation, and decomposition remain fixed throughout.

  2. The uniqueness statement remains visible beside the decomposition.

Interpretation

The visual functions as a summary board: it fixes the example, states the direct-sum identity, and highlights the uniqueness conclusion while the lecturer verbally wraps up.

Misconceptions · 1

Mistaking the direct-sum decomposition for a non-unique rewriting

Clear evidence
Shown in the video
Evidence
  1. Audio
    Observation

    The lecturer explicitly counters the idea of alternate decompositions by saying, "There's no other way of writing that."

  2. Formula
    Observation

    The board note says "This representation is UNIQUE."

Misconception

One might think a vector could be split into a WW-part and a W⊥W^{\perp}-part in multiple different ways.

Clarification

In the displayed W⊕W⊥W \oplus W^{\perp} setting, the decomposition is unique; the lecturer stresses that there is no other way to write the same vector using these two subspaces.

Concept relations · 4

Direct Sum of Subspaces → Subspace

Clear evidence
Shown in the video
Evidence
  1. Audio
    Observation

    Spoken as 'that's what the direct sum means'.

Application
Explanation

The concept of direct sum is applied to the specific subspaces WW and W⊥W^\perp to show they span R4\mathbb{R}^4.

Orthogonal Complement → Subspace

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    W⊥W^\perp is defined relative to WW.

Special case
Explanation

The orthogonal complement is a specific type of subspace related to another subspace via orthogonality.

Concrete orthogonal-complement pair in R4\mathbb{R}^4 → Direct-sum decomposition of a vector in R4\mathbb{R}^4

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    The bases for WW and W⊥W^{\perp} are written immediately above the equation W⊕W⊥=R4W \oplus W^{\perp} = \mathbb{R}^4 and the coordinate split.

Application
Explanation

The concrete choice of bases for WW and W⊥W^{\perp} is what makes the displayed coordinate decomposition possible.

Direct-sum decomposition of a vector in R4\mathbb{R}^4 → Uniqueness of the W⊕W⊥W \oplus W^{\perp} representation

Clear evidence
Shown in the video
Evidence
  1. Audio
    Observation

    The uniqueness remark is spoken directly about the just-written decomposition.

  2. Formula
    Observation

    The note "This representation is UNIQUE." is placed next to the decomposition line.

Proof dependency
Explanation

The uniqueness claim is about the specific representation produced by the direct-sum decomposition shown on the board.

Find an answer · 5

Why is the representation of a vector as a direct sum of subspaces unique?

Clear evidence
Shown in the video
Evidence
  1. Audio
    Observation

    Spoken as 'what's nice about this representation... is unique'.

Knowledge points
  1. Direct Sum of Subspaces

How do you decompose a vector in R4R^4 into parts belonging to W and W perp?

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    Vector decomposition is shown on the board.

Knowledge points
  1. Direct Sum of Subspaces
  2. Decomposing a Vector in R4R^4

What is the concrete orthogonal-complement example shown on the board in R4\mathbb{R}^4?

Clear evidence
Supplementary explanation
Evidence
  1. Formula
    Observation

    The board contains a fully written example with WW, W⊥W^{\perp}, and a vector decomposition in R4\mathbb{R}^4.

Knowledge points
  1. Concrete orthogonal-complement pair in R4\mathbb{R}^4
  2. Direct-sum decomposition of a vector in R4\mathbb{R}^4

Why does the lecturer say this representation is unique?

Clear evidence
Supplementary explanation
Evidence
  1. Audio
    Observation

    The lecturer says the way of writing the vector using the two subspaces is unique.

  2. Formula
    Observation

    The board note reads "This representation is UNIQUE."

Knowledge points
  1. Uniqueness of the W⊕W⊥W \oplus W^{\perp} representation

How is a general vector (a,b,c,d)T(a,b,c,d)^T split into parts lying in WW and W⊥W^{\perp}?

Clear evidence
Supplementary explanation
Evidence
  1. Formula
    Observation

    The decomposition line explicitly rewrites (a,b,c,d)T(a,b,c,d)^T as (a,b,0,0)T+(0,0,c,d)T(a,b,0,0)^T + (0,0,c,d)^T.

Knowledge points
  1. Direct-sum decomposition of a vector in R4\mathbb{R}^4
  2. Worked example: decomposing a vector in R4\mathbb{R}^4 using WW and W⊥W^{\perp}
Coverage and review notes

Covered · Full segment covers the definition of subspaces W and W_perp, their direct sum, and an example of vector decomposition.

Covered · Audio states that the decomposition using the two subspaces is unique; the board simultaneously shows the full example and the uniqueness note.

Covered · The lecturer summarizes that this gives a basic idea of orthogonal complements and closes the lesson while the static board remains visible.

Covered · Final trailing second contains no additional mathematical content beyond the closing of the same static summary segment.

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