Question

A and B are two nonsingular matrices. If the characteristic equation of A is $ a_0\lambda^3 + a_1\lambda^2 + a_2\lambda + a_3 = 0 $ and characteristic equation of $ B^{-1}AB $ is $ b_0\lambda^3 + b_1\lambda^2 + b_2\lambda + b_3 = 0 $, then

• A. \( \frac{a_0}{b_0} = \frac{a_1}{b_1} = \frac{a_2}{b_2} = \frac{a_3}{b_3} \)
• B. \( \frac{a_0}{b_0} = \frac{a_1}{b_1} = \frac{a_2}{b_2} = \frac{a_3}{b_3} \)
• C. \( a_0+a_1+a_2+a_3 = b_0+b_1+b_2+b_3 \)
• D. \( a_0a_1a_2a_3 = b_0b_1b_2b_3 \)

Hint:

• Similar matrices \(A\) and \(P^{-1}AP\) have the same characteristic polynomial, eigenvalues, determinant, and trace.
• If \(p(\lambda) = c_n\lambda^n + \dots + c_0\) and \(q(\lambda) = d_n\lambda^n + \dots + d_0\) are two characteristic polynomials for similar matrices, they must be proportional, i.e., \(p(\lambda) = k \cdot q(\lambda)\) for some non-zero constant \(k\). This means \(c_i/d_i = k\) for all \(i\).

Answer 1

The Correct Option is A

Similar matrices have the same characteristic polynomial (and thus the same eigenvalues, determinant, trace, etc.).
The matrix \(B^{-1}AB\) is similar to matrix A. 
The characteristic equation of A is \(p_A(\lambda) = a_0\lambda^3 + a_1\lambda^2 + a_2\lambda + a_3 = 0\). 
The characteristic equation of \(B^{-1}AB\) is \(p_{B^{-1}AB}(\lambda) = b_0\lambda^3 + b_1\lambda^2 + b_2\lambda + b_3 = 0\). 
Since A and \(B^{-1}AB\) are similar, their characteristic polynomials must be the same, up to a non-zero constant multiplier. 
That is, \(p_A(\lambda) = k \cdot p_{B^{-1}AB}(\lambda)\) for some constant \(k \neq 0\). So, \(a_0\lambda^3 + a_1\lambda^2 + a_2\lambda + a_3 = k(b_0\lambda^3 + b_1\lambda^2 + b_2\lambda + b_3)\). 
Comparing coefficients of powers of \(\lambda\): \(a_0 = kb_0\) \(a_1 = kb_1\) \(a_2 = kb_2\) \(a_3 = kb_3\) From these, we can write the ratios (assuming \(b_i \neq 0\)): \(k = \frac{a_0}{b_0} = \frac{a_1}{b_1} = \frac{a_2}{b_2} = \frac{a_3}{b_3}\). 
This means the coefficients are proportional. Options (a) and (b) are identical in the image and state this proportionality. If the characteristic polynomial is defined as \(\det(A-\lambda I)\), then the leading coefficient \(a_0\) (coefficient of \(\lambda^3\)) would be \((-1)^3 = -1\) or \(1\) depending on convention of \(\det(\lambda I - A)\). 
If \(a_0\) and \(b_0\) are chosen to be 1 (monic polynomial), then \(k=1\) and \(a_i = b_i\) for all i. The given options imply that the characteristic polynomials are proportional. \[ \boxed{\frac{a_0}{b_0} = \frac{a_1}{b_1} = \frac{a_2}{b_2} = \frac{a_3}{b_3}} \]