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I am solving a few exercise I found online.

My trouble is with solution 5.b) the second direction. Where it is stated that it follows that $(T_m - \lambda_j)$ is invertible on $U_m$ for $j = 1,...,m-1$. I understand all the previous steps but not this one. Why does it follow?

Here I post the solution I am having trouble with:

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The fact you need is that a non-zero multiple of the identity matrix plus a nilpotent matrix is invertible. (Proof: If $X^n = 0$ and $\lambda \neq 0$, then the inverse of $\lambda I + X$ is $\lambda^{-1}I - \lambda^{-2}X + \lambda^{-3} X^2 + \cdots + (-1)^{n-1} \lambda^{-n} X^{n-1}$.)

Then $T_m - \lambda_j = (\lambda_m - \lambda_j) I + (T_m - \lambda_m)$, where $I$ denotes the identity matrix. Since $j \neq m$, $\lambda_j \neq \lambda_m$ and $T_m - \lambda_m$ is nilpotent by the given argument.

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Let's think about $T_m- \lambda_j$. We can write $N=T_m-\lambda_m$ and then $T_m-\lambda_j=N+(\lambda_m-\lambda_j)I=N+ \lambda$ where $N$ is nilpotent and $\lambda=\lambda+m- \lambda_j$.

$N$ is nilpotent so we can suppose that $N^k=0$. Then $N=(N+\lambda)- \lambda$ and so expanding $N^k=0$ in the above way gives a polynomial in $N+ \lambda$. The constant term is just $(-\lambda)^k$ which we call $c$. Then if we write $p(N+ \lambda)-c=-c$, we see that we can factor out an $N+\lambda$ from the left and divide both sides by $-c$ to get the expression $(N+\lambda)q=1$ where $q=\frac{p(N+ \lambda)-c}{-c(N+\lambda)}$. Thus, $q$ is the inverse of $N+ \lambda$!

I've written out this argument as this technique of using polynomials to find inverses is one that often comes in handy.

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