Show that $(A-c)^2=0$ for a $2\times2$ matrix with repeated eigenvalues without using Cayley-Hamilton

Question

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?

Answer 1

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.

Answer 2

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 $(A-cI)^2=2cA-2cA=0$.

Answer 3

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.