---
# System prepended metadata

title: "\U0001F31E Week 3"

---

# 🌞 Week 3 

## Notes on <u>Determinants</u>

> **Prepared by:** _Roshan Naidu_
> **Instructor:** _Professor Leon Johnson_
> **Course:** _INFO H611_

> [!NOTE]
> Inline math must be wrapped within `$ ... $` and display math within `$$ ... $$` in HackMD.
> 
> Switching to the **Preview Mode** (👁️) would render eveything here to be viewed properly.


## Conventions and Definitions used throughout the writeup
• Vectors are bold lowercase: $\mathbf{u},\mathbf{v}\in\mathbb{R}^n$
• Scalars (real numbers) are plain: $a,b,c\in\mathbb{R}$
• Matrices are uppercase:  $A,B\in\mathbb{R}^{m\times n}$
• Norm / Magnitude of a vector: $\|\mathbf{v}\|$
• Dot product: $\mathbf{u}\cdot\mathbf{v}$
• Cross product: $\mathbf{u}\times \mathbf{v}$
• Unit vector: $\hat{\mathbf{v}}=\frac{\mathbf{v}}{\|\mathbf{v}\|}$


---

## Q.1. If we define "term similarity" based on the cosine of the angle drawn between term-vectors, then term 1 is more similar to term 2 than it is to term 3.

I said: **True**.  
Correct: **False**. (Used cosine **distance** to see why)


## Matrix (from the notebook)

Rows are term-vectors:

$$
A = 
\begin{bmatrix}
2 & 3 & 0 & 1\\[4pt]
4 & 0 & 2 & 2\\[4pt]
8 & 3 & 2 & 4\\[4pt]
2 & 2 & 5 & 1
\end{bmatrix}
$$

Thus
$$
\mathbf{t}_1=(2,3,0,1),\qquad
\mathbf{t}_2=(4,0,2,2),\qquad
\mathbf{t}_3=(8,3,2,4)
$$


Cosine similarity between $\mathbf{u}$ and $\mathbf{v}$:

$$
\cos\theta(\mathbf{u},\mathbf{v}) \;=\; \frac{\mathbf{u}\cdot\mathbf{v}}{\|\mathbf{u}\|\,\|\mathbf{v}\|}.
$$

Cosine **distance** (a dissimilarity measure):

$$
d(\mathbf{u},\mathbf{v}) \;=\; 1 - \cos\theta(\mathbf{u},\mathbf{v}).
$$

Smaller \(d\) means more similar.

---

## Step–by–step numeric check

### Norms (computed exactly)


\begin{aligned}
\|\mathbf{t}_1\|^2 &= 2^2 + 3^2 + 0^2 + 1^2 = 4 + 9 + 0 + 1 = 14
&\Rightarrow&\;\|\mathbf{t}_1\|=\sqrt{14}.\\[6pt]
\|\mathbf{t}_2\|^2 &= 4^2 + 0^2 + 2^2 + 2^2 = 16 + 0 + 4 + 4 = 24
&\Rightarrow&\;\|\mathbf{t}_2\|=\sqrt{24}.\\[6pt]
\|\mathbf{t}_3\|^2 &= 8^2 + 3^2 + 2^2 + 4^2 = 64 + 9 + 4 + 16 = 93
&\Rightarrow&\;\|\mathbf{t}_3\|=\sqrt{93}.
\end{aligned}


### Dot products


\begin{aligned}
\mathbf{t}_1\cdot\mathbf{t}_2 &= 2\cdot4 + 3\cdot0 + 0\cdot2 + 1\cdot2 = 8 + 0 + 0 + 2 = 10,\\[6pt]
\mathbf{t}_1\cdot\mathbf{t}_3 &= 2\cdot8 + 3\cdot3 + 0\cdot2 + 1\cdot4 = 16 + 9 + 0 + 4 = 29.
\end{aligned}


### Cosine similarities (exact form and decimal)


\begin{aligned}
\cos(\theta_{1,2})
&= \frac{10}{\sqrt{14}\,\sqrt{24}}
= \frac{10}{\sqrt{336}}
\approx 0.5455447256,\\[6pt]
\cos(\theta_{1,3})
&= \frac{29}{\sqrt{14}\,\sqrt{93}}
= \frac{29}{\sqrt{1302}}
\approx 0.8036972934.
\end{aligned}


### Cosine distances


\begin{aligned}
d_{1,2} &= 1 - \cos(\theta_{1,2}) \approx 1 - 0.5455447256 = 0.4544552744,\\[6pt]
d_{1,3} &= 1 - \cos(\theta_{1,3}) \approx 1 - 0.8036972934 = 0.1963027066.
\end{aligned}


---

## Conclusion (why the statement is **False**)

- Cosine distance: \(d_{1,3}\approx 0.1963\) is **smaller** than \(d_{1,2}\approx 0.4545\).  
- Therefore **term 1 is more similar to term 3 than to term 2** (under cosine similarity).  
- The original statement (“term 1 more similar to term 2 than term 3”) is **False**.


## Real-world example

If term vectors are word occurrence profiles across the documents, then the two words can appear in many of the similar documents but with different relative frequency patterns. Here the Cosine similarity captures those frequency patterns (i.e. directions) and not just the raw overlap. Helpful in finding similarities.


---

## Q.2. There exists a "special collection" of documents which can only be created by linear combinations of all four columns of the matrix.


I said: **False**. 
Correct: **False**.

---

## Matrix from the notebook

The notebook's term–document matrix (columns = documents) is

$$
A =
\begin{bmatrix}
2 & 3 & 0 & 1\\[4pt]
4 & 0 & 2 & 2\\[4pt]
8 & 3 & 2 & 4\\[4pt]
2 & 2 & 5 & 1
\end{bmatrix},
$$

so the four column-vectors (documents) are
$$
\mathbf{c}_1 = \begin{bmatrix}2\\4\\8\\2\end{bmatrix},\quad
\mathbf{c}_2 = \begin{bmatrix}3\\0\\3\\2\end{bmatrix},\quad
\mathbf{c}_3 = \begin{bmatrix}0\\2\\2\\5\end{bmatrix},\quad
\mathbf{c}_4 = \begin{bmatrix}1\\2\\4\\1\end{bmatrix}.
$$

---

## If my answer was *True* / *False* (explaining why I was right/wrong)

- I answered **False** which is correct.
- **Why false:**  
  The statement shows a *necessity* that any "special collection" requires the combining of all four columns simultaneously. This is a much stronger claim than the usual linear-algebra fact that every vector in the column space can be written as a linear combination of the columns, but would **not necessarily require every column**. If the columns are linearly dependent, some columns are redundant and any vector in the span can be produced using a smaller subset (a basis). Thus there is no document that *must* be using all four columns.


### (a) Real-world example

Suppose the columns represent four documents of news articles and I was trying to synthesize a new article as a linear blend (i.e. weighted sum) of them. Then if one article is literally a shorter rewrite of another (e.g., News article 4 is just new article 1 with every word count halved), then new article 4 adds no new *direction* (i.e. topic) to the document space. Thus I can make any synthetic article using all four articles 1, 2 and 3. Therefore there is **no** synthetic article that *must* use all of the four documents. In this case, column 4 is redundant.

### (b) Theoretical (linear algebra) example
If a matrix $B\in\mathbb{R}^{n\times m}$ has $\operatorname{rank}(B)=r<m$, then the column space is an $r$ dimensional subspace. There exists a basis of $r$ columns (a subset of the columns) that spans the column space. Hence any vector in $\operatorname{Col}(B)$ can be written as a linear combination of only those $r$ basis columns. Thus no vector in the column-space *requires* coefficients on all $m$ columns. The statement "requires all $m$ columns" is false whenever $(r<m)$. In our concrete matrix $A$, $r=3<4$, so the theory matches the computed result.

>[!Note]
Because the four columns are linearly dependent (in fact $\mathbf{c}_4 = \tfrac12\mathbf{c}_1$ the column space is 3 Dimensional and there is **no** special document that can only be formed by using all four columns.Thus the statement is therefore **False**.


---

## Q.3. The columns of this matrix are all linearly independent.

I answered: **False**.  
Correct answer: **False**.

---

## Matrix (from the notebook)

The term–document matrix (columns = documents) is:

$$
A =
\begin{bmatrix}
2 & 3 & 0 & 1\\[4pt]
4 & 0 & 2 & 2\\[4pt]
8 & 3 & 2 & 4\\[4pt]
2 & 2 & 5 & 1
\end{bmatrix}
$$

with column-vectors

$$
\mathbf{c}_1 = \begin{bmatrix}2\\4\\8\\2\end{bmatrix},\quad
\mathbf{c}_2 = \begin{bmatrix}3\\0\\3\\2\end{bmatrix},\quad
\mathbf{c}_3 = \begin{bmatrix}0\\2\\2\\5\end{bmatrix},\quad
\mathbf{c}_4 = \begin{bmatrix}1\\2\\4\\1\end{bmatrix}.
$$

---

## If my answer was *True* / *False* (explaining why I was right/wrong)

- I answered **False** which is **correct**.
- Why False: the claim “all columns are linearly independent” would require the only solution
  $\alpha_1\mathbf{c}_1+\alpha_2\mathbf{c}_2+\alpha_3\mathbf{c}_3+\alpha_4\mathbf{c}_4=\mathbf{0}$
  to be $\alpha_1=\alpha_2=\alpha_3=\alpha_4=0$. 
  
  But for this matrix there exists a **nontrivial** solution, so the columns are linearly dependent. Thus my answer **False** is correct.

---

## Correct answer and formal explanation (why the statement is False)

- Computing an explicit relation between columns. Observe that

  $$
  \mathbf{c}_4 \;=\; \tfrac{1}{2}\mathbf{c}_1,
  $$

  because

  $$
  \tfrac{1}{2}\mathbf{c}_1 = \tfrac{1}{2}\begin{bmatrix}2\\4\\8\\2\end{bmatrix}
  = \begin{bmatrix}1\\2\\4\\1\end{bmatrix} = \mathbf{c}_4.
  $$

- Therefore the columns are **linearly dependent**
- Equivalently, the matrix rank is

  $$
  \operatorname{rank}(A)=3 < 4,
  $$

  so the four columns cannot be all independent. Hence the statement “the columns are all linearly independent” is **False**.

### (a) Real-world example
Imagine four pdfs where pdf 4 is literally a condensed version of pdf 1 with every term frequency halved (e.g. like an abstract obtained by removing half the word counts). In this case, pdf 4 brings no new topical direction. It is just a scalar multiple of pdf 1. So here the set of four pdfs is linearly dependent. Thus the claim “all pdfs are independent” is completely false.

### (b) Theoretical example
Let

B =
$$
\begin{bmatrix}1 & 0 & 0 & 2\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\end{bmatrix}
$$

Here column 4 equals 2x column 1. So the four columns of $B$ are dependent and $\operatorname{rank}(B)=3<4$. This mirrors the same situation in our matrix \(A\).

>[!Note]
Because $\mathbf{c}_4 = \tfrac12\mathbf{c}_1$ the columns are linearly dependent and my answer which was **False** is correct and the statement is therefore **False**.


---

## Q.4. This matrix also represents an invertible linear transformation. That is, it does **NOT** "squash" a 4-dimensional vector into a 3-dimensional space.

I answered: **False**.  
Correct answer: **False**.

---

## If my answer was *True* / *False* (explaining why I was right/wrong)

- I answered **False** which is **correct**.

- **Why the answer is False:**
  To be invertible a $4\times4$ matrix must have full rank $4$ (which equivalently a nonzero determinant). If any column is a linear combination of the others, then the columns are dependent, the rank is $(<4)$, and the linear map is not invertible in such a case. It maps $\mathbb{R}^4$ into a proper subspace (called a "squash" to lower dimension). Observing such a dependence in the notebook's matrix leads tto directly the conclusion that the transformation is not at all invertible.



## Correct answer and formal explanation (why the statement is **False**)

The notebook matrix is

$$
A =
\begin{bmatrix}
2 & 3 & 0 & 1\\[4pt]
4 & 0 & 2 & 2\\[4pt]
8 & 3 & 2 & 4\\[4pt]
2 & 2 & 5 & 1
\end{bmatrix}
$$

with columns

$$
\mathbf{c}_1=\begin{bmatrix}2\\4\\8\\2\end{bmatrix},\quad
\mathbf{c}_2=\begin{bmatrix}3\\0\\3\\2\end{bmatrix},\quad
\mathbf{c}_3=\begin{bmatrix}0\\2\\2\\5\end{bmatrix},\quad
\mathbf{c}_4=\begin{bmatrix}1\\2\\4\\1\end{bmatrix}
$$

**Observation:** $(\mathbf{c}_4 = \tfrac{1}{2}\mathbf{c}_1)$

$$
\tfrac12\mathbf{c}_1 = \tfrac12\begin{bmatrix}2\\4\\8\\2\end{bmatrix}
= \begin{bmatrix}1\\2\\4\\1\end{bmatrix} = \mathbf{c}_4
$$

Because one column is an exact scalar multiple of another, the columns are linearly dependent. From linear algebra:

- If the columns of a square matrix are linearly dependent then $\det(A)=0$
- $\det(A)=0\) \(\Longrightarrow\) \(A$ is **not invertible**
- Similarily, $\operatorname{rank}(A) < 4$
Here, the matrix $\operatorname{rank}(A)=3$

Therefore $(A)$ maps $\mathbb{R}^4$ onto a 3 Dimensional subspace where it "squashes" 4D to 3D, and thus it is **not invertible**.

>[!Tip]
a matrix with two proportional columns has zero determinant, so is never invertible

### (a) Real-world example
Lets imagine the Data compression pipeline where one of the four feature channels is exactly a scaled copy of another (like sensor B records exactly half the counts of sensor A). No matter how we weight all four channels, but  the information content lives in at most three independent directions from which we cannot recover a unique 4D original from the compressed output. This mapping is therefore non invertible. Many different 4D inputs collapse to the same 3D output spaces.

### (b) Theoretical (linear algebra) example
Let

B=\begin{bmatrix}
1 & 0 & 0 & 2\\[4pt]
0 & 1 & 0 & 0\\[4pt]
0 & 0 & 1 & 0\\[4pt]
0 & 0 & 0 & 0
\end{bmatrix}

Column 4 = \(2\)×column 1, so columns are dependent here $\operatorname{rank}(B)=3$, and $\det(B)=0$. $(B)$ is not invertible and maps $\mathbb{R}^4$ to a 3 Dimensional subspace.

---

>[!Note]
Because $\mathbf{c}_4=\tfrac12\mathbf{c}_1$ the matrix \(A\) has dependent columns and $\det(A)=0$. Therefore \(A\) is **not invertible** and the claim that it does **not** squash 4D to 3D is **False**.


## Q.5. The determinant of $(M_3)$ is negative because we subtracted a constant value (3) from all the values in matrix $(M)$. In other words, when you subtract a constant value from all elements in a matrix, the sign of the determinant will change."

I answered: **True**
Correct answer: **False**

---

## If my answer was *True* / *False* (explaining why I was right/wrong)

The transformation from $(M)$ to $(M_3)$ looked drastic, and the determinant became negative. It seemed to me that subtracting the same value from all elements would pull the matrix down via flipping the determinant’s sign.


## Correct answer and formal explanation (why the statement is **False**)

Subtracting a constant value from **all** entries of a matrix is **not** a linear transformation. The determinant is not simply a function of the individual values in the matrix. it is a function of the **linear relationships** between the rows or the columns of matrix.

### Example Matrices:-

Original matrix $(M)$:


M =
\begin{bmatrix}
2 & 5 & 1 & 2 & 4 \\
5 & 2 & 1 & 2 & 3 \\
1 & 2 & 1 & 3 & 1 \\
5 & 4 & 1 & 5 & 4 \\
1 & 4 & 3 & 2 & 1 \\
\end{bmatrix}


Transformed matrix $(M_3)$ (probably derived from subtracting 3 and then dividing):


$M_3$ =
\begin{bmatrix}
-0.5 & 1.0 & -1.0 & -0.5 & 0.5 \\
1.0 & -0.5 & -1.0 & -0.5 & 0.0 \\
-1.0 & -0.5 & -1.0 & 0.0 & -1.0 \\
1.0 & 0.5 & -1.0 & 1.0 & 0.5 \\
-1.0 & 0.5 & 0.0 & -0.5 & -1.0 \\
\end{bmatrix}

Although the determinant of $(M_3)$ is negative. It still does **not** prove that subtracting a constant would flip the sign. We can see this with a **counterexample** here:


### Counter example: 2×2 Matrix

Original matrix:


A = \begin{bmatrix} 4 & 2 \\ 3 & 1 \end{bmatrix}

$$
\quad \det(A) = 4 \cdot 1 - 2 \cdot 3 = -2
$$


Now here subtract 3 from **every entry**:


B = \begin{bmatrix} 1 & -1 \\ 0 & -2 \end{bmatrix},

$$
\quad \det(B) = 1 \cdot (-2) - (-1) \cdot 0 = -2
$$


>[!Note]
The determinant **did not change sign**. It stayed negative.
>- The determinant is affected by:
>  - Scaling rows or columns (scales the determinant)
> - Swapping rows (flips sign)
>  - Adding multiples of rows to each other (preserves determinant)
> But **subtracting a constant from every entry** is not a valid linear operation in this context.


>[!Tip]
> **The determinant is governed by linear algebraic structure, not by uniformly subtracting constants.**
So even though the determinant of $(M_3)$ is negative, it's **not because** we subtracted a constant from all entries of $(M)$. This operation does **not** predictably flip the sign of the determinant.


---

## Q.6. There exists a matrix \(B\) so that \(MB=I\) and \(BM=I\), where \(I\) is the identity matrix.

I answered: **True**
Correct answer: **True**

---

## If my answer was *True* / *False* (explaining why I was right/wrong)

I  recognized that the matrix \(M\) in the notebook is a **square** $5\times5$ matrix and from earlier work I knew that its determinant is a nonzero:
$$
\det(M)=36\neq 0
$$
A nonzero determinant for a square matrix fits the standard criterion for invertibility. From that fact one immediately concludes, there **exists** a unique matrix $B$ (the inverse $M^{-1}$) satisfying both

$$
MB=I\quad\text{and}\quad BM=I
$$

So my reasoning to check the invertibility via determinant and conclude existence of an inverse is correct.


## Correct answer and formal explanation (why the statement is **True**)

>[!Important]
>A square matrix $M\in\mathbb{R}^{n\times n}$ is invertible if $\det(M)\neq 0$. 

When invertible, its inverse $M^{-1}$ is the unique matrix $(B)$ such that

$$
MB=I_n \quad\text{and}\quad BM=I_n
$$

For the concrete matrix from the notebook,

M=
\begin{bmatrix}
2 & 5 & 1 & 2 & 4\\
5 & 2 & 1 & 2 & 3\\
1 & 2 & 1 & 3 & 1\\
5 & 4 & 1 & 5 & 4\\
1 & 4 & 3 & 2 & 1
\end{bmatrix}

we computed

$$
\det(M)=36\neq 0
$$
Therefore \(M\) is invertible and the matrix $B=M^{-1}$ exists and satisfies $MB=BM=I_5$


### (a) Real-world example
Let us consider a linear encoding/decoding pipeline for signals. The encoder applies an invertible linear transform $M$ to an input signal vector like a combination of sensors. If $M$ is invertible, then there is a decoder $B=M^{-1}$ that exactly recovers the original signal $B(Mx)=x$ and $M(Bx)=x$ 

This is exactly the property $MB=BM=I$ in practice of the lossless linear encoding.

### (b) Theoretical example
A simple $2\times2$ invertible matrix:

M=\begin{bmatrix}2&1\\1&1\end{bmatrix}

$$
\qquad \det(M)=2\cdot1-1\cdot1=1\neq0
$$

Its inverse is

$$
M^{-1}=\begin{bmatrix}1 & -1\\ -1 & 2\end{bmatrix}
$$
and one checks $MM^{-1}=M^{-1}M=I_2$


---

## Q.7. The columns of $(M_3)$ are all linearly independent.

I answered: **True**
Correct answer: **True**


$$
M_3 =
\begin{bmatrix}
-0.5 &  1   & -1   & -0.5 &  0.5\\[6pt]
 1   & -0.5 & -1   & -0.5 &  0   \\[6pt]
-1   & -0.5 & -1   &  0   & -1   \\[6pt]
 1   &  0.5 & -1   &  1   &  0.5 \\[6pt]
-1   &  0.5 &  0   & -0.5 & -1
\end{bmatrix}
$$


---

## If my answer was *True* / *False* (explaining why I was right/wrong)

I answered **True**. A correct and standard way to justify this is:

- Check whether the matrix is **invertible** (equivalently whether its determinant is nonzero)
- For a square $n\times n$ matrix, the following are equivalent:
  - the columns are linearly independent
  - the matrix has full rank \(n\)
  - the matrix is invertible
  - the determinant is nonzero

So if we observed (or computed) that $\det(M_3)\neq 0$ (or that $\operatorname{rank}(M_3)=5$), then concluding the columns are independent would be a correct reasoning.

Concretely (numerical computation from the notebook):

$$
\operatorname{rank}(M_3)=5,\qquad \det(M_3)=-5.4375\neq 0
$$

Because $\det(M_3)$ is nonzero and the rank equals $5$, the only solution of $M_3\mathbf{x}=\mathbf{0}$ is the trivial $\mathbf{x}=\mathbf{0}$. Hence the columns are linearly independent.

---

## Correct answer and formal explanation (why the statement is **True**)

**Correct answer: True.**

**Formal justification.** Suppose $\mathbf{c}_1,\dots,\mathbf{c}_5$ are the columns of $M_3$. If there were a nontrivial linear relation

$$
\alpha_1\mathbf{c}_1+\alpha_2\mathbf{c}_2+\cdots+\alpha_5\mathbf{c}_5=\mathbf{0}
$$

then the vector ${\alpha}=(\alpha_1,\dots,\alpha_5)^\top$ would be a nonzero element of the nullspace of $(M_3)$. But since $\det(M_3)\neq0$, the nullspace is ${\mathbf{0}}$ and no nontrivial relation exists here. Thus,   the columns are linearly independent.

### (a) Real-world example
Suppose each column of $(M_3)$ is the reading over time of a different independent sensor channel (like an accelerometer, temperature, pressure, etc.). If the columns are linearly independent, no sensor is a linear combination of the others an thus  monitoring and fusing these  channels provides  independent degrees of information. This is exactly the case when the data matrix (columns = channels) is invertible. Here we cannot reconstruct one sensor from the others.

### (b) Theoretical example
A random $5\times5$ matrix with entries drawn from a continuous distribution   is almost surely invertible  with the probability of 1 where its columns are linearly independent. The given $(M_3)$ is an explicit numeric instance where that generic condition holds rank $(5)$ with the determinant $\neq0$


---

## Q.8. $(M_3)$ can be created from $(M)$ using basic matrix operations (matrix subtraction, matrix/scalar multiplication, etc.).
*Hint:* consider the matrix \(J\), the matrix of all ones.

I answered: **False**
Correct answer: **True**

## Matrix (from .ipynb notebook shared)

M =
\begin{bmatrix}
2 & 5 & 1 & 2 & 4 \\
5 & 2 & 1 & 2 & 3 \\
1 & 2 & 1 & 3 & 1 \\
5 & 4 & 1 & 5 & 4 \\
1 & 4 & 3 & 2 & 1
\end{bmatrix}



$M_3$ =
\begin{bmatrix}
-0.5 & 1 & -1 & -0.5 & 0.5 \\
1 & -0.5 & -1 & -0.5 & 0 \\
-1 & -0.5 & -1 & 0 & -1 \\
1 & 0.5 & -1 & 1 & 0.5 \\
-1 & 0.5 & 0 & -0.5 & -1
\end{bmatrix}


---

## If my answer was *True* / *False* (explaining why I was right/wrong)

- At first glance, $(M_3)$ looks completely different from $(M)$ (because of negative values and fractions).
- It might feel like no “simple” operations on $(M)$ could produce such a transformed matrix.
- The trick is realizing the role of the **all-ones matrix i.e $(J)$**, which shifts every entry of $(M)$ by the same amount before rescaling.

---

>[!Tip]
>- Even if a resulting matrix looks very different, we should always check if it can be expressed as  
>  $$
  M_3 = \alpha M + \beta J
  $$
>  for some scalars $\alpha, \beta$  
>- Here that is what exactly  happens, so thus the statement is **True**.


## Correct answer and formal explanation (why the statement is **True**)

- The key is the **matrix of ones** $(J)$, where every entry is 1.  
  For a $5 \times 5$ matrix: 
 
  J = 
  \begin{bmatrix}
  1 & 1 & 1 & 1 & 1 \\
  1 & 1 & 1 & 1 & 1 \\
  1 & 1 & 1 & 1 & 1 \\
  1 & 1 & 1 & 1 & 1 \\
  1 & 1 & 1 & 1 & 1
  \end{bmatrix}

- By combining \(M\), scalar multiplication and the subtraction with \(J\), we can easily construct $(M_3)$  
  In the notebook:
  
  $$
  M_3 = \frac{1}{2}(M - 3J)
  $$
  
  equivalently $(M_2 = M - 3J)$, 
  then $(M_3 = \tfrac{1}{2}M_2$).

- This only uses the **basic matrix operations**: subtraction and scalar multiplication stated as exactly as the problem statement which is allowed.


---

## 9. Optional LaTeX Practice

Some useful math symbols:
$\alpha,\beta,\gamma,\pi,\infty$
$\nabla, \int_0^1 f(x)\,dx, \sum_{i=1}^n i^2$
$\lim_{x\to 0}\frac{\sin x}{x} = 1$

Actions:
Fractions: $\tfrac{1}{2}$
Summation: $\sum_{i=1}^n i^2\) → \(\sum_{i=1}^n i^2$
Integral: $\int_0^1 x^2\,dx = \tfrac{1}{3}$
Determinant: $\det(A)$ 
Inverse: $A^{-1}$

---

## 10. 🌕 Final Reflection


This week’s assignment deepened my understanding of **matrix operations**, **linear independence**, and the **matrix invertibility** through concrete examples with matrices $(M)$ and $(M_3)$. At first, many of the True/False statements felt uncomfortable and difficult  but then working through the problems with definitions and real life scenarios, computations, and counterexamples clarified all of my concepts important for this course topic.

### Key Lessons
1. **Cosine similarity and orthogonality**  
   - Cosine distance can reveal whether vectors (or columns of a matrix) are orthogonal, which directly connects to linear independence.  
   - Even when values “look” independent, cosine distance helps confirm or deny this formally.

2. **Linear independence**  
   - Checking whether columns are linearly independent requires examining the rank of the matrix.  
   - $(M_3)$ provided a clear case where columns *were* independent, which confirmed the importance of systematic verification.

3. **Invertibility**  
   - For a square matrix, invertibility is equivalent to having linearly independent columns and a nonzero determinant.  
   - Statements about whether a matrix transformation “squashes” vectors into lower dimensions became much clearer with this perspective.

4. **Determinant misconceptions**  
   - Simply subtracting a constant from all entries of a matrix does *not* flip the determinant’s sign.  
   - Counterexamples with \(2 \times 2\) matrices highlighted the importance of relying on formal properties rather than assumptions.

5. **Matrix transformations using \(J\) (the all-ones matrix)**  
   - A big insight was that complicated-looking matrices like $(M_3)$ can be derived from \(M\) using only subtraction and scalar multiplication with \(J\).  
   - This shows the power of expressing matrices in terms of simple building blocks.

---