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.
)
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.
)
This is AI generated Response.
Asked by: Test User 100161
Created at: 2025-05-07 01:04:44
Final Answer:
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Let's solve the problem in **three steps** as requested, with explanation at each step:
---
### **Step 1: Understand the Structure of Matrix \( A \)**
**Explanation:**
Given \( A \) is a \( 20 \times 20 \) matrix where every diagonal entry is \( 20 \), and every off-diagonal entry is \( 1 \). That is,
\[
A_{ij} =
\begin{cases}
20 & \text{if } i = j \\
1 & \text{if } i \neq j
\end{cases}
\]
We can write \( A \) as:
\[
A = (20-1)I + J
\]
where \( I \) is the identity matrix and \( J \) is the \( 20 \times 20 \) matrix of all ones.
So,
\[
A = 19I + J
\]
---
### **Step 2: Find the Eigenvalues of \( A \)**
**Explanation:**
Because \( A \) is symmetric, its singular values are the absolute values of its eigenvalues.
Recall that \( J \) has rank 1, with one eigenvalue \( 20 \) (since all entries are 1, and the vector of all ones is an eigenvector) and the rest \( 0 \).
For \( A = 19I + J \):
- The eigenvector \( \mathbf{1} \) (vector of all ones) gets:
\[
A\mathbf{1} = 19\mathbf{1} + J\mathbf{1} = 19\mathbf{1} + 20\mathbf{1} = 39\mathbf{1}
\]
So, one eigenvalue is **39**.
- For any vector orthogonal to \( \mathbf{1} \) (sum of entries is zero), \( J\mathbf{x} = 0 \), so:
\[
A\mathbf{x} = 19\mathbf{x}
\]
Thus, the other 19 eigenvalues are **19**.
---
### **Step 3: Find the Largest Singular Value \( \sigma_1 \)**
**Explanation:**
The singular values of a symmetric matrix are the absolute values of its eigenvalues.
So singular values of \( A \) are: \( 39, 19, 19, \dots, 19 \) (19 times).
Therefore, the largest singular value \( \sigma_1 = 39 \).
---
## **Final Answer**
\[
\boxed{39}
\]
is the first singular value (\( \sigma_1 \)) of \( A \).
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