Let matrix A be real, square and nonsingular. If $\det (A^2 - 2A + I) = 0$, then 2 is an eigenvalue of matrix $(A + A^{-1})$.

Solution: Cayley-Hamilton theorem claims that $p(A) = 0$, where $p(\cdot)$ is characteristic polynomial. Hence, we can replace A with $\lambda$ and after solving quadratic equation find that $\lambda = 1$.

Next, I'm aware of the property that $A^{-1}$ has $\lambda^{-1}$, i.e. 1 is also an eigenvalue of $A^{-1}$.

My question is: what to do now? Simply, to sum $1+1=2$, but I don't see why eigenvalue of sum of matrices should be the sum of their eigenvalues. Could you help me figure it out?

  • 2
    $\begingroup$ The sum of the eigenvalues of a matrix is equal to its trace. The trace is a linear transformation. $\endgroup$ – André 3000 Jun 7 '14 at 4:56
  • $\begingroup$ Nice, thanks. $tr(A+A^{-1}) = trA + trA^{-1} = (1+1) + (1+1) = 4$. But there different ways of getting 4 = 3+1 = 5-1, not just 2+2. $\endgroup$ – Yal dc Jun 7 '14 at 5:05
  • $\begingroup$ Step 1: Prove that every eigenvector of $A$ is an eigenvector of $A^{-1}$ [or use the result if you know it]; Step 2: Prove that if $A$ and $B$ share an eigenvector for respective eigenvalues $\lambda_A$ and $\lambda_B$, then $\lambda_A + \lambda_B$ is an eigenvalue of $A+B$ (with the same eigenvector $x$) [or use this result if you know it]. $\endgroup$ – M. Vinay Jun 7 '14 at 5:17
  • $\begingroup$ By the way, $f(A) = \det(A^2 - 2A + I) = 0$ does not necessarily mean that $f(x)$ is the characteristic polynomial $p(x)$. All it tells you is that $f(x)$ and $p(x)$ have common factors, so at least one of the roots of $f(x)$ will be an eigenvalue. You should rephrase that argument. In this case, $f(x)$ has two equal roots, so you're lucky (because it's guaranteed to be an eigenvalue). $\endgroup$ – M. Vinay Jun 7 '14 at 5:28
  • $\begingroup$ step 1: yes, I know this result. step 2: $(A+B)x=(\lambda_A + \lambda_B) x \Leftrightarrow Ax + Bx=\lambda_A x + \lambda_B x$, so 2 is an eigenvalue. $\endgroup$ – Yal dc Jun 7 '14 at 5:48

Alternative Solution

$\det(A^2 - 2A + I) = 0 \Rightarrow\\ \det(A^{-1})\det(A^2 - 2A + I) = 0 \Rightarrow\\ \det(A - 2I + A^{-1}) = \det((A + A^{-1}) - 2I) = 0$

$\therefore$ $2$ is an eigenvalue of $A + A^{-1}$.

  • $\begingroup$ $|(A^{2}-2A+I) \cdot A^{-1}| \rightarrow |A+A^{-1}-2I| = 0$, but I don't see that it somehow tells me about 2. It doesn't need be characteristic equation. $\endgroup$ – Yal dc Jun 7 '14 at 5:37
  • $\begingroup$ Given a matrix $M$, $|M - \lambda I| = 0 \Rightarrow$? $\endgroup$ – M. Vinay Jun 7 '14 at 5:41
  • $\begingroup$ I don't know. The determinant isn't linear transformation. $\endgroup$ – Yal dc Jun 7 '14 at 6:06
  • $\begingroup$ $|M - \lambda I| = 0 \Rightarrow$ $\lambda$ is an eigenvalue of $M$. $\endgroup$ – M. Vinay Jun 7 '14 at 6:08
  • $\begingroup$ Oh, right! How could I not recognize it?! Thank you for your help. $\endgroup$ – Yal dc Jun 7 '14 at 6:11

Your application of the Cayley-Hamilton theorem is mistaken for many reasons: you don't know the characteristic polynomial here, and the conclusion of the theorem does not give you what you want (namely information about eigenvalues); and if you would know the characteristic polynomial of$~A$, then you could find the eigenvalues of$~A$ (as its roots) without using the C-H theorem.

However it happens that you can argue that $\lambda=1$ is an eigenvalue of $A$ here (and that without using that $A$ is invertible), but only due to a rather exceptional circumstance, namely that $X^2-2X+1=(X-1)^2$ has $1$ as unique root. Here's how: it is given that $A^2-2A+I$ is singular, so it has some nonzero vector $v$ in its kernel. Now the fact that $0=(A^2-2A+I)v=(A-I)(A-I)v$ means that either $(A-I)v=Av-v$ is an eigenvector of$~A$ for $\lambda=1$, namely if it is nonzero, or else (if $Av-v$) $v$ itself is an eignevector of$~A$ for $\lambda=1$. Either way $\lambda=1$ is an eigenvalue of$~A$. (Note that if your polynomial had factored into distinct linear factors, you would only get that one of their roots is an eigenvalue.)

Now with an eignevector $w$ of $A$ for $\lambda=1$ things are easy: $(A+A^{-1})w=\lambda w+\lambda^{-1}w=2w$ (like you wanted to argue; this part is correct because you are reasoning for a specific eigenvector) so $w$ is also an eigenvector of $A+A^{-1}$, for the eigenvalue$~2$.

Note that you can equally well conclud for any pair of scalars $a,b$e that $aA+bA^{-1}$ has an eigenvalue $a+b$ (or more generally that any combination of powers of $A$ has as eignevalue the corresponding combination of powers of $\lambda=1$), something that a clever manipulation of determinants (as in the asnwer by M. Vinay) would not give you.


If $A$ is an invertible matrix and $Av = \lambda v$ for some scalar $\lambda$ and non-zero vector $v,$ then $\lambda \neq 0$ (or $A$ could not be invertible), and $A^{-1}v = \frac{1}{\lambda} v$ (applying $A^{-1}$ to each side above, then dividing by $\lambda ).$ Hence $(A+A^{-1})v = \left( \lambda + \frac{1}{\lambda} \right) v,$ so $\lambda + \frac{1}{\lambda}$ is an eigenvalue of $(A+ A^{-1}).$ Apply this with $\lambda = 1$ in your case.


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