give answer in 3 step with explanation at the end of each step and final answer at the end:

For
i
=
1
,
2
,
dots,
2
0
,
let vec
(
v
)
i
i
n
R
2
0
denote the vector whose
i
th entry is
2
0

Question

give answer in 3 step with explanation at the end of each step and final answer at the end:

For 
i
=
1
,
2
,
dots,
2
0
,
 ﻿let vec
(
v
)
i
i
n
R
2
0
 ﻿denote the vector whose 
i
 ﻿th entry is 
2
0
 ﻿and whose other entries are all 
1
 ﻿s 
.
 ﻿Let 
A
 ﻿be the 
2
0
×
2
0
 ﻿matrix 
[
v
e
c
(
v
)
1
v
e
c
(
v
)
2
c
d
o
t
s
v
e
c
(
v
)
2
0
]
 ﻿whose 
i
 ﻿th column is the vector vec
(
v
)
i
.
 ﻿That is
,
 
A
 ﻿is a 
2
0
×
2
0
 ﻿matrix whose diagonal entries are all 
2
0
 ﻿s and whose off
-
diagonal entries are all 
1
s
.
 ﻿You may assume that the reduced echelon form 
(
RREF
)
 ﻿of 
A
 ﻿is I, the 
2
0
×
2
0
 ﻿identity matrix.
Problem 
7
.
 
(
1
2
 ﻿points
)
 ﻿Find 
σ
1
,
 ﻿the first singular value of 
A
.
 
(
Hints: Note that 
A
 ﻿is symmetric. Again, you may find problem 
5
 ﻿on the Weeks 
9
-
1
0
 ﻿extra problem set to be helpful; and again, no credit will be given if all you do is enter the matrix into WolframAlpha or something similar and ask it to find the singular values.
)

VIPSolutions · Accepted Answer

Let's solve the problem in **3 steps as requested, with explanations after each step**:

---

### **Step 1: Understand the Structure of Matrix $ A $ and its Eigenvalues**

**Matrix $ A $:**
- $ A $ is a $ 20 \times 20 $ matrix.
- Diagonal entries: all **20**.
- Off-diagonal entries: all **1**.

This is a classic "diagonal-plus-constant" or "rank-one update" matrix.

**We can write:**
$$
A = 20I + (J - I)
$$
where $ I $ is the identity and $ J $ is the all-ones matrix.

So:
$$
A = 20I + (J - I) = 19I + J
$$

**Explanation:**  
This representation helps us find the eigenvalues, as the eigenvalues of $ J $ are well-known: $ n $ (for the all-ones vector) and $ 0 $ (for any vector orthogonal to the all-ones vector).

---

### **Step 2: Find the Eigenvalues of $ A $ (Since $ A $ is Symmetric, Singular Values are Absolute Values of Eigenvalues)**

Let $ n = 20 $.

- The all-ones vector $ \mathbf{1} $ is an eigenvector of $ J $ with eigenvalue $ n $.
- Any vector orthogonal to $ \mathbf{1} $ is an eigenvector of $ J $ with eigenvalue $ 0 $.

From $ A = 19I + J $:

- For $ \mathbf{1} $:
  $$
  A\mathbf{1} = 19\mathbf{1} + J\mathbf{1} = 19\mathbf{1} + 20\mathbf{1} = 39\mathbf{1}
  $$
  So, eigenvalue 39.

- For $ x $ with $ \mathbf{1}^T x = 0 $:
  $$
  A x = 19x + Jx = 19x + 0 = 19x
  $$
  So, eigenvalue 19 (with multiplicity 19).

**Explanation:**  
The largest eigenvalue is 39, corresponding to the all-ones vector, and the rest are 19.

---

### **Step 3: State the Largest Singular Value $\sigma_1$ of $A$**

- **Singular values of a symmetric matrix** are the absolute values of its eigenvalues.
- Here, all eigenvalues are positive: $39, 19, 19, ..., 19$.

So, **the largest singular value is $ \boxed{39} $.**

**Explanation:**  
The first singular value $\sigma_1$ is the largest eigenvalue in magnitude, which is 39.

---

## **Final Answer**

$$
\boxed{39}
$$

This is the first singular value ($\sigma_1$) of the matrix $A$.

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