Bored at a lecture, I began tinkering with 2-by-2 matrices and observed the following. Suppose we want to diagonalize the matrix $$\begin{pmatrix}a & b\\c & d\end{pmatrix}$$ in the most naive way possible. We seek a matrix $$\begin{pmatrix}\alpha & \beta \\ \gamma & \delta \end{pmatrix}$$ with $\alpha\delta - \beta\gamma \ne 0$ such that $$\frac{1}{\alpha\delta - \beta\gamma} \begin{pmatrix} \delta & -\beta \\ -\gamma & \alpha \end{pmatrix} \begin{pmatrix}a & b\\c & d\end{pmatrix} \begin{pmatrix}\alpha & \beta \\ \gamma & \delta \end{pmatrix} = \begin{pmatrix}* & 0 \\ 0 & *\end{pmatrix}.$$ Grinding through the algebra, it turns out that the off-diagonal terms are $Q(\alpha, \gamma)$ and $-Q(\beta, \delta)$, up to a scalar multiple, where $$Q(x, y) = cx^2 + (a-d)xy - by^2.$$ So to find a diagonalizing matrix, we seek solutions to $Q(x, y) = 0$. Because $Q$ is homogeneous, we are actually seeking solutions in $\mathbb{P}^1$. The condition $\alpha\delta - \beta\gamma \ne 0$ is equivalent to $$[\alpha:\gamma] \ne [\beta:\delta].$$ This viewpoint is totally new to me. For $n$-by-$n$ matrices, are there analogous homogeneous forms coming from trying to diagonalize in this way?

  • $\begingroup$ Yes there are... $\endgroup$ Aug 3, 2017 at 18:06
  • $\begingroup$ Well, what are they? I guess there is more than one form for $n > 2$ (there are three off-diagonal entries up to transposition, if $n=3$). Is there a general theorem about this? I'm really not well-versed in linear algebra. $\endgroup$
    – Unit
    Aug 4, 2017 at 1:38

1 Answer 1


If you want to do the same for an $n \times n$ matrix $A$, let $P=(p_{i,j})$ be a matrix such that $P^{-1} A P$ is diagonal. Then considering the adjugate matrix $\text{adj}(P) = ((-1)^{i+j}P_{j,i})$, one has for $i\ne j$, $$\sum_{k,l} (-1)^{i+k} P_{k,i} a_{k,l} p_{l,j} = 0$$ where $P_{i,j}$ is the $(i, j)$ minor of $P$. These $n(n-1)$ relations are homogeneous polynomials of degree $n$ in the coefficients $p_{i,j}$. I don't know where you want to go from here.


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