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How can you see straight away, without any calculations, that 2x2 matrices like $A_1$ has eigenvalues 0 and 0 while $A_2$ has eigenvalues 1 and 2? Is it because there is a zero in the left bottom corner and the number in the top right corner does not matter?

$$ A_1 = \begin{bmatrix}0 & 1 \\ 0 & 0\end{bmatrix} \\ A_2 = \begin{bmatrix}1 & 3 \\ 0 & 2\end{bmatrix} $$

Please don't use math-heavy reasoning, I have a very basic understanding of linear algebra.

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  • Check this: https://math.stackexchange.com/q/264969/42969, or this: https://math.stackexchange.com/q/803313/42969 – Martin R May 30 '21 at 09:48
  • If you write out the characteristic polynomial , you will notice why in fact the lower left $0$ is the reason and in fact the upper right entry is irrelevant. – Peter May 30 '21 at 09:49
  • An easier argument : The first matrix has determinant $0$ because it has a column containing of only zeros. We get such a matrix in the case of $A_2$ after subtracting $1$ or $2$ from the diagonal elements (in the latter case, we have a row containing only zeros). But this argument does not show that $0$ is the only eigenvalue of $A_1$ – Peter May 30 '21 at 09:51
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    This is because the determinant of a triangular matrix is the product of the elements in the diagonal and therefor the characteristic polynomials are, respectively, $\lambda^2$ and $(\lambda-1)(\lambda-2)$. – Bernard May 30 '21 at 09:54
  • I tried my best without any calculations and failed. –  May 30 '21 at 11:01

2 Answers2

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I think 3b1b made this recent video just for you!

DatBoi
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In the particular case of a triangular matrix, the eigen values of the matrix are just the diagonal elements.

(Just calculate the characteristic polynomial in this case.)

math
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