Let $A$ be a $4\times4$ matrix with real entries and eigenvalues $1$, $-1$, $2$ and $-2$, then which of the following statements are true? If $B$ is a matrix defined as $B=A^4-5A^2+5I$, where $I$ is a $4\times4$ identity matrix, then 


*

*$\det(A+B)=0$

*$\det(B)=0$

*$\operatorname{tr}(A+B)=4$


From the given conditions, I could only conclude that trace of matrix $A$ is $0$, and the value of determinant of matrix $B$ is $-11$. Now how to calculate the trace of matrix $B$?
 A: Hint: The characteristic polynomial of $A$ is $$P(x) = (x-1)(x+1)(x-2)(x+2) = x^4-5x^2+4,$$ therefore $B=P(A)+I$. For the third equation, you can use linearity of the trace and the fact that if the eigenvalues of $M$ are $\lambda_i$, then $\operatorname{tr}(M^k)=\sum_i\lambda_i^k$, but it’s much simpler if you take advantage of the above expression for $B$.

Update: Knowing this, the determinants can be found by inspection. By the Cayley-Hamilton Theorem, $P(A)=0$, therefore $B=I_4$ and $\det B=1$. Similarly, $\det(A+B)=\det(A+I)=0$ because $-1$ is an eigenvalue of $A$. The trace of $B$ can be computed directly using properties of the trace, i.e., $$\begin{align}
\operatorname{tr}B &= \operatorname{tr}A^4-5\operatorname{tr}A^2+5\operatorname{tr}I_4 \\
&=(1^4+(-1)^4+2^4+(-2)^4)+5(1^2+(-1)^2+2^2+(-2)^2)+5\cdot4,
\end{align}$$ but it, too can be found by inspection once the expression for $B$ has been simplified.
A: After suitable base change (take eigenvectors of the distinct eigenvalues of $A$), we have
$$A=\begin{pmatrix}1&0&0&0\\0&-1&0&0\\0&0&2&0\\0&0&0&-2\end{pmatrix}, B=\begin{pmatrix}1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix}.$$
Not that hard to answer these questions now, right?
A: Here's a way you might like better. If $v$ is an eigenvector of $A$ with eigenvalue $\lambda$, then from
$$
B = A^4 - 5A^2 + 5I,
$$
we compute
$$
Bv = (\lambda^4 - 5 \lambda^2 + 5)v.
$$
Hence $v$ is an eigenvector of $B$ with eigenvalue 
\begin{equation}
\mu = \lambda^4 - 5 \lambda^2 + 5.
\ref{1}
\end{equation}
Since we know the eigenvalues of $A$,
\begin{align*}
\lambda_1 &= 1 \\
\lambda_2 &= -1 \\
\lambda_3 &= 2 \\
\lambda_4 &= -2,
\end{align*}
this tells us the eigenvalues of $B$. They are 
\begin{align*}
\mu_1 &= 1 \\
\mu_2 &= 1 \\
\mu_3 &= 1 \\
\mu_4 &= 1
\end{align*}
if you do the computation using \eqref{1}. Using the facts that

  
*
  
*The trace is the sum of the eigenvalues,
  
*The determinant is the product of the eigenvalues,
  

We can compute the quantities in 2. and 3. this way. (In computing 3., we also use the fact that trace is a linear function.)
However, the determinant is not a linear function! Nonetheless, we can compute the quantity in 1. because $B$ and $A$ have common eigenvectors; hence its eigenvalues are found by adding corresponding eigenvalues of $A$ and $B$. That is, $(A+B)v = Av + Bv = \lambda v + \mu v = (\lambda + \mu)v$. Then multiplying $(\lambda_1 + \mu_1) \cdots (\lambda_4 + \mu_4)$ to find $\det(A+B)$ gives $0$.
