I would like you to help me with a problem. If we consider a matrix
$$A = \begin{bmatrix}
4 & 2 &-1 & 6\\
3 & 0 & 2 & 4\\
-1 & 0 & 3 & 0\\
0 & 0 & 5 & 0
\end{bmatrix}$$
the problem says that the characteristic polynomial is $P_A(\lambda) = (2-\lambda)(4-\lambda)(-1-\lambda)(5-\lambda)$. However, when we calculate directly
$$P_A(\lambda) = \det(A – \lambda I),$$
where $I$ is the identity metrix, we arrive at
$$P_A(\lambda) = \lambda^4-7\lambda^3+5\lambda^2+52\lambda+40.$$
Hence, I have the following question. Is it possible to find another equivalent $A$ matrix using some property that preserves the characteristic polynomial and arrives at the proposed result? What am I doing wrong?
Best Answer
You could just take the diagonal matrix $$ B={\rm diag}(2,4,-1,5). $$ It has the above characteristic polynomial $$ P_B(t)=(t-2)(t-4)(t+1)(t-5). $$
The matrix $A$ has indeed the characteristic polynomial you have computed, namely $$ P_A(t)=t^4 - 7t^3 + 5t^2 + 52t + 40 $$ This polynomial is irreducible over $\Bbb Q$. In particular it has no rational root.
Why should $A$ and $B$ be "equivalent"? The characteristic polynomials are definitely different, so the matrices cannot be equivalent, or similar.