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Let $A$, be a real $4 \times 4$ matrix such that $-1,1,2,-2$ are its eigenvalues. If $B=A^4-5A^2+5I$, then which of the following are true?

  1. $\det(A+B)=0$
  2. $\det (B)=1$
  3. $\operatorname{trace}(A-B)=0 $
  4. $\operatorname{trace}(A+B)=4$

Using Cayley-Hamilton I get $B=I$, and I know that $\operatorname{trace}(A+B)=\operatorname{trace}(A)+\operatorname{trace}(B)$. From these facts we can obtain easily about 2,3,4 but I am confused in 1. How can I verify (1)? Thanks for your help.

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+1 for showing your efforts. – user1551 Dec 13 '12 at 14:35
The matrix $A$ is diagonalizable, since it has $4$ pairwise distinct eigenvalues. The four conditions you consider are invariant under similarity. So work with the diagonalized form of $A$. – 1015 Jan 27 '13 at 20:54


$$\det(A+I)=0 \Leftrightarrow \lambda=-1 \mbox{is an eigenvalue}$$

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The characteristic equation of $A$ is given by $(t-1)(t+1)(t+2)(t-2)=0 $ which implies $t^{4}-5t^{2}+4=0$. Now $A$ must satisfy its characteristic equation which gives that $A^{4}-5A^{2}+4I=0$ and so we see that $B=A^{4}-5A^{2}+4I+I=0+I=I$. Hence, the eigenvalues of $(A+B)$ is given by $(-1+1),(1+1),(2+1),(-2+1)$ that is $0,2,3,-1.$[Without the loss of generality, one can take $A$ to be diagonal matrix which would not change trace or determinant of the matrix. ]So we see that $det(A+B)$ is the product of its eigenvalues which is $0$. . Also we see that trace of $(A+B)$ is the sum of its eigenvalues which is $(0+2+3-1)=4.$ Also, B being the identity matrix, $det(B)=1.$ So the options $(1),(2) and (4)$ are true .

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You need to justify why it is so easy to calculate the eigenvalues of $A + B$. – Zhen Lin Dec 13 '12 at 14:47

Well, perhaps we can pass on to the Jordan Canonical form of $\,A\,$:

$$J_A=\begin{pmatrix}1&0&0&0\\0&\!\!\!-1&0&0\\0&0&2&0\\0&0&0&\!\!\!-2\end{pmatrix}\Longrightarrow B=J_A^4-5J_A^2+5I=\begin{pmatrix}1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix}$$

We can do the above since determinant and trace are invariant under similarity.

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I had just been answering a question about "without loss of generality". This is a nice example. Since $A$ is diagonalisable (it's characteristic polynomial is $(X+1)(X-1)(X-2)(X+2)$ which is split without multiple roots), we may assume without loss of generality that $A$ is diagonal (do a base change to a basis of eigenvectors, which changes neither determinants nor traces). You can do that case explicitly by computation.

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Just to note the following points:

Let $A_{n\times n}$ be a matrix with eigenvalues $\lambda_{1},...,\lambda_{n}$ and $f(x)$ a polynomial. Then $$\det(f(A))=f(\lambda_{1})...f(\lambda_{n})$$ and $$\text{trace}(f(A))=f(\lambda_{1})+...+f(\lambda_{n})$$

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