$\begin{pmatrix} 7 & -2\\2 & 2 \end{pmatrix}$

The eigenvalues for this matrix are $\lambda=6$ and $\lambda=3$

It also happens that $(A-6I)(A-3I)=0$

I've checked for various $2$ x $2$ matrices. If a matrix has eigenvalues $\lambda_1$ and $\lambda_2$ then $(A-\lambda_1 I)(A-\lambda_2 I)=0$. Even when they're not distinct this works. I cant show this however. I have tried for a $3$ x $3$ as well- pick any two distinct and it will work. However I have found a counterexample for which it doesnt work when they're not distinct. Is this result true? Why? Simple algebra for $2$ x $2$ case does not seem to yield it. What about for $n$ x $n$?

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    $\begingroup$ You may want to study the notion of $characteristic$ $polynomial$ of a matrix. $\endgroup$ – uniquesolution Sep 1 '15 at 21:21

The Cayley-Hamilton theorem says that if $p$ is the characteristic polynomial of a matrix $A$, then $p(A) = 0$. Because $A$ commutes with itself and $I$, you can factor this matrix polynomial just like its scalar equivalent. In 2 dimensions, you are just reproducing this fact, since you are including both roots. In 3 dimensions, however, you have been trying to get by with only two of the three roots, so it is hardly surprising you have found exceptions. I t is just luck, though, that you haven't yet found an counter-example with two distinct eigenvalues. You will find one in the matrix $$\begin{pmatrix} 1 && 0 && 0\\0 && 2 && 0\\0 && 0 && 3\end{pmatrix}$$ for example.


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