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I want to show that if $A$ is a $2\times2$ matrix with repeated eigenvalue $c$, then $(A-c)^2=0$.

I'm aware that this is a simple case of the Cayley-Hamilton theorem, but this requires more background to follow than I would like to require. Is there an easier direct check we can do in the $2\times2$ case?

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  • $\begingroup$ Not $A-c$ but $A-cI$ $\endgroup$
    – Jean Marie
    Mar 27, 2017 at 19:05
  • $\begingroup$ @JeanMarie i think it's not a great crime to identify your scalars and scalar matrices. But sure if it bothers you. $\endgroup$
    – ziggurism
    Mar 27, 2017 at 19:06
  • $\begingroup$ I agree, there are surely most important crimes :) $\endgroup$
    – Jean Marie
    Mar 27, 2017 at 19:16

3 Answers 3

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Well, every such matrix is similar to an upper triangular matrix with the same eigenvalues*. And similarity transformations don't change the polynomials that a matrix satisfies.**

So can you show that $$ A = \begin{bmatrix} a & b \\ 0 & a \end{bmatrix} $$ satisfies a quadratic of the form $(A - c)^2$ for some constant $c$? What should $c$ be?

  • Hint: apply a row operation to a matrix by multiplying on the left by an elementary matrix, $E$, and then multiply by the inverse of $E$ on the right, performing a closely related column operation. Show that it's possible to choose $E$ to kill off the lower left entry of your original matrix.

** Why? Hint: write out $p(QAQ^{-1})$ for your example above, but write your example in the form $$(A - cI)^2.$$ Then try to notice a pattern that might apply to polynomials in general.

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  • $\begingroup$ How do we know every matrix is similar to upper triangular? I consider invoking Jordan normal form to be as big a barrier or bigger than Cayley-Hamilton. $\endgroup$
    – ziggurism
    Mar 27, 2017 at 18:53
  • $\begingroup$ I guess it's sufficient to observe it has an invariant subspace and choose a basis with an element in that subspace... $\endgroup$
    – ziggurism
    Mar 27, 2017 at 18:56
  • $\begingroup$ Did you read the first hint? It gives a recipe for doing so (except in the case where the upper left entry is $0$, which requires an extra step. $\endgroup$ Mar 27, 2017 at 18:57
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    $\begingroup$ No invariant subspaces needed...just do gaussian elimination to eliminate the $(2,1)$ entry via left-multiplication by an elementary matrix $E$; follow this with right-multiplication by $E^{-1}$ so that you've done conjugation. Sometimes things really ARE easy. $\endgroup$ Mar 27, 2017 at 18:58
  • $\begingroup$ @ziggurism The current answer is a bit too complicated, but the triangularisation is actually very easy. Just take any eigenvector $u$ and complete it to an ordered basis $\{u,v\}$. Let $P$ be the augmented matrix $[u|v]$. Then $P^{-1}AP$ is triangular. $\endgroup$
    – user1551
    Mar 27, 2017 at 19:06
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For any square matrix, the product of its eigenvalues (taking multiplicity into account) is equal to its determinant, and the sum of its eigenvalues is equal to its trace. This can be verified by using Vieta’s formulas or by a simple direct calculation for the $2\times2$ case.

Let $A=\pmatrix{\alpha&\beta\\\gamma&\delta}$. $A$ has a repeated eigenvalue, $c$, so $\operatorname{tr}A=\alpha+\delta=2c$ and $\det A=\alpha\delta-\beta\gamma=c^2$. We have $(A-cI)^2=A^2-2cA+c^2I$. Taking the first and third terms, $$A^2+c^2I=A^2+\det(A)I=\pmatrix{a^2+\beta\gamma&\beta(\alpha+\delta)\\\gamma(\alpha+\delta)&\beta\gamma+\delta^2}+\pmatrix{\alpha\delta-\beta\gamma&0\\0&\alpha\delta-\beta\gamma}=\pmatrix{\alpha(\alpha+\delta)&\beta(\alpha+\delta)\\\gamma(\alpha+\delta)&\delta(\alpha+\delta)}=\operatorname{tr}(A)A=2cA$$ therefore $(A-cI)^2=2cA-2cA=0$.

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  • $\begingroup$ That's exactly what I was looking for $\endgroup$
    – ziggurism
    Mar 28, 2017 at 1:26
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Let $v$ be an eigenvector of $A$ with eigenvalue $c$, so $Av=cv$, and choose an independent vector $w$, so that we have a basis. Then $Aw=av+bw$. Expressed relative to this basis, the linear operator has a matrix $\begin{pmatrix}c & a\\0 &b\end{pmatrix}$, an upper triangular matrix with eigenvalues $b,c$. By hypothesis, $A$ has repeated eigenvalue $c$, hence $b=c$. Therefore $Aw=av+cw$ or $(A-c)w=av$.

So $(A-c)^2v=0$ because a fortiori $(A-c)v=0$. And $(A-c)^2w=(A-c)av=0.$ Since $(A-c)^2$ vanishes on the basis, it vanishes on its whole domain, that is it is the zero operator.

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