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I am using Axler's text, where he proves the following theorem by induction. However, I would like to prove in a simpler way that if the matrix of a linear operator $T$ on $V$ is upper triangular with respect to some basis, then any eigenvalue $\lambda$ appears on the diagonal $\ker (T-\lambda I)^{\dim V}$ times.

My idea, roughly is that to observe that as matrix, $(T-\lambda I)^{\dim V}$ has exactly $n$ rows of $0$, where $n$ is the number of times $\lambda$ appears on the diagonal. The dimension of the kernel of that map is the number of free variable when considering the RREF of the matrix. Then the dimension of the image is the number of non-zero rows, $\dim V - n$. Which means that $n$ corresponds to the dimension of the kernel.

Is this a valid proof?

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    $\begingroup$ Now try asking a question about Euclid's Geometry text, and see whether you can get Euclid to answer. $\endgroup$ Commented Sep 28, 2016 at 7:21
  • $\begingroup$ @GerryMyerson It was indeed a pleasant surprise :) $\endgroup$
    – lEm
    Commented Sep 28, 2016 at 8:01

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The statement that "as a matrix, $(T - \lambda I)^{\dim V}$ has exactly $n$ rows of $0$, where $n$ is the number of times $\lambda$ appears on the diagonal" is not correct. For example, take $$ T = \left[\begin{array}{ccc}1 & 1 & 1\\ 0 & 0 & 1 \\ 0& 0& 1 \end{array}\right] $$ and take $\lambda = 0$. Because $0$ appears once on the diagonal, we have $n = 1$. However, $$ T^3 = \left[\begin{array}{ccc}1 & 1 & 5\\ 0 & 0 & 1 \\ 0& 0& 1 \end{array}\right], $$ which has no rows of $0$.

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    $\begingroup$ Thanks for the answer, I wasn't thinking clearly enough. Also I would like to thank you for writing such an excellent book, it is as clear as it can be and helps me a lot. $\endgroup$
    – lEm
    Commented Sep 28, 2016 at 7:56

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