How many involutary matrices of order 2013 and trace = 2013 exist having a57 = -1?

My Attempt: An involutary matrix is the one which satisfies : $A^2$ = I . Now, if k is an eigen value of A, then it satisfies : $k^2$ - 1 =0 which gives the possible eigen values of the involutary matrix A as 1 or -1 .

Now, for a matrix of order 2013, trace = 2013 clearly means that all it's eigen values are 1.

Now. i understand that had the condition a57 = -1 NOT been given , the IDENTITY matrix is one of the matrices satisfying the above conditions.

BUT how do i prove that the identity matrix is the only involutary matrix of order 2013 which has it's trace= 2013? Thank you.

(this question came up before diagonalization came up in the course which i am reading which i think might be early for the question to crop up :) )


You need to show, in addition to the fact that the characteristic polynomial is $(X-1)^{2013}$ which you already did, that every involution is diagonalisable. Depending on what you already know, you have two options. You can either reason that $X^2-1=(X+1)(X-1)$ splits into distinct (the question seems to imply this is not over a field of characteristic$~2$) linear factors, and any matrix that satisfies such a polynomial (equation) is diagonalisable. Or, if you dont know about that fact, you can explicitly show that for any involution$~A$, every vector can be written as a sum of eigenvectors for$~A$, so that the sum of the eigenspaces fills the whole space. For the latter, just split any vector in its parts fixed and anti-fixed under the involution : $v=\frac12(v+A\cdot v)+\frac12(v-A\cdot v)$. Of course given that there is no eigenvalue$~-1$, the term $\frac12(v-A\cdot v)$ must actually be zero here.

  • $\begingroup$ Thank you Marc . Although, i have read that A would be diagonalizable in such cases, this question came up before diagonalization came up in the course. $\endgroup$ – MathMan Oct 3 '13 at 6:52
  • $\begingroup$ Well, the second argument shows that $v=\frac12(v+A\cdot v)$ for all $v$, in other words $v=A\cdot v$, once you know that $-1$ is not an eigenvalue of$~A$. $\endgroup$ – Marc van Leeuwen Oct 3 '13 at 7:05

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