SOLUTION: Let Jn be the n × n matrix all of whose entries are 1. Let n > 1 and show that: (In − Jn)^−1 = In − 1/(n − 1)Jn

Question 1164527: Let Jn be the n × n matrix all of whose entries are 1. Let n > 1 and show that:
(In − Jn)^−1 = In − 1/(n − 1)Jn
Answer by CPhill(2261) (Show Source): You can put this solution on YOUR website!
To prove that $(I_n - J_n)^{-1} = I_n - \frac{1}{n-1}J_n$, we must show that their product results in the Identity matrix $I_n$.
### **Key Property of $J_n$**
Before starting, recall the definition of $J_n$: it is an $n \times n$ matrix where every entry is $1$.
The product $J_n \cdot J_n$ results in a matrix where each entry is the dot product of a row of $1$s and a column of $1$s. Since there are $n$ entries:
$$J_n^2 = n J_n$$
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### **The Proof**
Let $A = (I_n - J_n)$ and $B = (I_n - \frac{1}{n-1}J_n)$. We will compute $A \cdot B$:
**Step 1: Expand the product**
$$(I_n - J_n) \left( I_n - \frac{1}{n-1}J_n \right)$$
Using the distributive property (FOIL):
$$= I_n \cdot I_n - I_n \cdot \frac{1}{n-1}J_n - J_n \cdot I_n + J_n \cdot \frac{1}{n-1}J_n$$
**Step 2: Simplify the terms**
* $I_n \cdot I_n = I_n$
* $I_n \cdot \frac{1}{n-1}J_n = \frac{1}{n-1}J_n$
* $J_n \cdot I_n = J_n$
* $J_n \cdot \frac{1}{n-1}J_n = \frac{1}{n-1}J_n^2$
Substituting these back:
$$= I_n - \frac{1}{n-1}J_n - J_n + \frac{1}{n-1}J_n^2$$
**Step 3: Substitute $J_n^2 = n J_n$**
$$= I_n - \frac{1}{n-1}J_n - J_n + \frac{n}{n-1}J_n$$
**Step 4: Combine the $J_n$ terms**
Factor out $J_n$:
$$= I_n + \left( -\frac{1}{n-1} - 1 + \frac{n}{n-1} \right) J_n$$
Find a common denominator for the terms inside the parentheses:
$$= I_n + \left( \frac{-1 - (n-1) + n}{n-1} \right) J_n$$
$$= I_n + \left( \frac{-1 - n + 1 + n}{n-1} \right) J_n$$
$$= I_n + \left( \frac{0}{n-1} \right) J_n$$
**Step 5: Final Result**
$$= I_n + 0 = I_n$$
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### **Conclusion**
Since $(I_n - J_n)(I_n - \frac{1}{n-1}J_n) = I_n$, the two matrices are inverses of each other. Therefore:
$$(I_n - J_n)^{-1} = I_n - \frac{1}{n-1}J_n$$
$\blacksquare$
> **Note:** This inverse exists only if $n-1 \neq 0$, which is why the condition $n > 1$ is specified. If $n=1$, the matrix $I_1 - J_1$ becomes $[1] - [1] = [0]$, which is not invertible.
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