If A and B are square matrices of order `n ,` then prove that `Aa n dB` will commute iff `A-lambdaIa n dB-lambdaI` commute for every scalar `lambdadot`

Options

A. `AB=BA`
B. `AB+BA=O`
C. `A=-B`
D. none of these

Explanation

To prove that two square matrices \( A \) and \( B \) of order \( n \) commute if and only if \( A - \lambda I \) and \( B - \lambda I \) commute for every scalar \( \lambda \), we will follow these steps:

### Step 1: Understand the Commutativity Condition
We start with the definition of commutativity. Two matrices \( A \) and \( B \) commute if:
\[ AB = BA \]

### Step 2: Assume \( A \) and \( B \) Commute
Assume \( A \) and \( B \) commute, i.e., \( AB = BA \). We need to show that \( A - \lambda I \) and \( B - \lambda I \) also commute.

### Step 3: Compute \( (A - \lambda I)(B - \lambda I) \)
Using the distributive property of matrix multiplication:
\[ (A - \lambda I)(B - \lambda I) = AB - A\lambda I - \lambda IB + \lambda^2 I \]
Since \( I \) is the identity matrix, we have:
\[ = AB - \lambda A - \lambda B + \lambda^2 I \]

### Step 4: Compute \( (B - \lambda I)(A - \lambda I) \)
Now compute the other product:
\[ (B - \lambda I)(A - \lambda I) = BA - B\lambda I - \lambda IA + \lambda^2 I \]
This simplifies to:
\[ = BA - \lambda B - \lambda A + \lambda^2 I \]

### Step 5: Use the Commutativity of \( A \) and \( B \)
Since \( AB = BA \), we can replace \( BA \) with \( AB \):
\[ = AB - \lambda B - \lambda A + \lambda^2 I \]

### Step 6: Compare Both Results
Now we have:
1. \( (A - \lambda I)(B - \lambda I) = AB - \lambda A - \lambda B + \lambda^2 I \)
2. \( (B - \lambda I)(A - \lambda I) = AB - \lambda B - \lambda A + \lambda^2 I \)

Both expressions are equal, thus:
\[ (A - \lambda I)(B - \lambda I) = (B - \lambda I)(A - \lambda I) \]

### Step 7: Assume \( A - \lambda I \) and \( B - \lambda I \) Commute
Now we need to show the converse: if \( A - \lambda I \) and \( B - \lambda I \) commute for every scalar \( \lambda \), then \( A \) and \( B \) commute.

### Step 8: Set Up the Commutativity Condition
Assume:
\[ (A - \lambda I)(B - \lambda I) = (B - \lambda I)(A - \lambda I) \]

### Step 9: Expand Both Sides
Expanding both sides gives:
\[ AB - A\lambda - B\lambda + \lambda^2 I = BA - B\lambda - A\lambda + \lambda^2 I \]

### Step 10: Cancel Out Common Terms
Cancel \( \lambda^2 I \) and the terms \( -A\lambda \) and \( -B\lambda \) from both sides:
\[ AB = BA \]

### Conclusion
Thus, we have shown that \( A \) and \( B \) commute if and only if \( A - \lambda I \) and \( B - \lambda I \) commute for every scalar \( \lambda \).