Question

If A is similar to B, then there exists one and only one invertible matrix S such that $S^{-1}AS=B$.

Short answer

The above statement is false.

If A is similar to B, then there doesn’t exist one and only one invertible matrix S such that $S^{-1}AS=B$.

Step-by-step solution

Definition of similar matrix

Let us suppose two square matrices A and B. Then the matrix B is said to be similar to matrix A if there exists an invertible matrix S such that

$S^{-1}AS=B........\left(1\right)$

To prove the invertible matrix S is not unique 

Let us consider a matrix P such that

$P=2S$

$\Rightarrow P^{-1}=\frac{1}{2}{S}^{-1}$

$\Rightarrow 2P^{-1}=S^{-1}$

Now,

$$\begin{gathered} P^{-1}AP=\left(\frac{1}{2}{S}^{-1}A\left(2S\right)\right)=S^{-1}AS=B \\ \Rightarrow P^{-1}AP=B \end{gathered}$$

This implies that there exists an invertible matrix P other than S such that

$B=P^{-1}AP$

Final Answer

If we take a matrix P such that P = 2S, then we have$B=P^{-1}AP,$ which shows that there exists another invertible matrix P such that the matrix A is similar to the matrix B.

Hence, the statement “If A is similar to B, then there exists one and only one invertible matrix S such that$S^{-1}AS=B$ is false.