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My attempt

Proof. Suppose $A$ is a triangular $3\times3$ matrix. Then so is $A-\lambda_j\operatorname{Id}$ for $1\le j\le3,$ which means that $\det(A-\lambda_j\operatorname{Id})$ is the product of its diagonal entries. Thus $p(\lambda)=\prod(A_{ii}-\lambda),$ and hence the roots of the characteristic polynomial are the diagonal entries $A_{11}$, $A_{22}$, $A_{33}$, as claimed.

I'm not sure whether my approach here is correct. My understanding is that the problem asks us to show that every triangular $3\times3$ matrix $A$ satisfies the determinant formula whereby the lambdas are the roots of the characteristic polynomial and hence eigenvalues of $A$?

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    $\begingroup$ I think the question should have been to show $(A-\lambda_1I)(A-\lambda_2I)(A-\lambda_3I)=0$ where $\lambda_i=A_{ii}$ $\endgroup$
    – Shubham Johri
    May 17, 2021 at 18:51
  • $\begingroup$ @Shubnam Johri. I couldn't agree more. $\endgroup$
    – Karam
    May 17, 2021 at 18:53
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    $\begingroup$ The claim is not that $(A_{11}-\lambda_1 I)(A_{22}-\lambda_2 I)(A_{33}-\lambda_3 I)=0.$ Rather $A-\lambda_i I$ is a $3\times3$ matrix, and the multiplication is matrix multiplication: $(A-\lambda_1 I)(A-\lambda_2 I)(A-\lambda_3 I)=0. \qquad$ $\endgroup$
    – Michael Hardy
    May 17, 2021 at 18:56
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    $\begingroup$ It is not particularly tedious to note that the $[A]_{kk}$ are the eigenvalues and just perform the multiplication (you don't need to track every entry). $\endgroup$
    – copper.hat
    May 17, 2021 at 19:27
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    $\begingroup$ @Karam : The problem is still to show that if you multiply three specified matrices, you get the zero matrix, not to show that if you multiply three specified numbers you get the number $0. \qquad$ $\endgroup$
    – Michael Hardy
    May 18, 2021 at 0:03

1 Answer 1

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$A$ is an upper triangular $n\times n$ matrix, $\lambda_1$, $\ldots$, $\lambda_n$ its diagonal elements.

Let's show that the matrix $$\prod_{i=1}^n ( A- \lambda_i I)$$ applied to any vector produces the zero vector. That is enough to see that the matrix is $0$.

The main observation is that $A- \lambda_k I$ applied to a vector with the last $n-k$ components $0$ gets us a vector with the last $n-k+1$ components $0$.

Now, start with any vector $v$. Note that the vector $$(A- \lambda_n I) v$$ has the last component $0$. Now apply $(A- \lambda_{n-1}I)$ and get $$(A- \lambda_{n-1}I) ((A- \lambda_{n}I)v$$ with last $2$ components $0$. Continue $n$ steps and get $$(A- \lambda_1 I) \cdots (A- \lambda_n I) v$$ the vector with all components $0$.

$\bf{Added:}$

In fact we showed that a certain product of matrices with prescribed $0$ positions is the $0$ matrix. Below is the case $n=3$, $\times$ denotes an arbitrary element: $$\left( \begin{matrix} 0 & \times& \times \\0 & \times & \times \\ 0 &0 &\times \end{matrix}\right)\cdot\left( \begin{matrix} \times & \times& \times \\0 & 0 & \times \\ 0 &0 &\times \end{matrix}\right)\cdot \left( \begin{matrix} \times & \times& \times \\0 & \times & \times \\ 0 &0 &0 \end{matrix}\right) = 0_3$$

as this calculation with WA shows.

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  • $\begingroup$ Explanation: $A$ upper triangular means $A$ invariates a (standard) flag. On the tsuccesive quotients $A$ acts as the multiplications by $\lambda_i$. Hence on the $i$-th quotient $A- \lambda_i I$ is $0$, that is $A-\lambda_i I$ takes $V_i$ to $V_{i-1}$. So the product take $V_n = V$ to $V_0 = 0$. $\endgroup$
    – orangeskid
    May 18, 2021 at 4:21

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