Show that every triangular $3\times3$ matrix $A$ satisfies $(A-\lambda_1 I)(A-\lambda_2 I)(A-\lambda_3 I)=0$, where $\lambda_i$ are eigenvalues of $A$ 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 $A$?
 A: $A$ 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.
