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I am told the following.

"It is important to note that the reduced row echelon form of $(A -\lambda I)$ will always have at least one row of zeroes. The number of zero rows equals the dimension of the eigenspace."

This is clear to me, since $(A - \lambda I)$ must not be invertible, by the definition of eigenvalues and eigenvectors.

However, I am then given the example of finding the eigenspaces for the upper triangular matrix $$ \begin{matrix} 1 & 2 & 6 \\ 0 & 3 & 5 \\ 0 & 0 & 4 \\ \end{matrix} $$

which has the reduced row echelon form $$ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{matrix} $$

The accompanying explanation is as follows.

Since $A$ is upper triangular, so is $(A - \lambda I)$ and hence its determinant is just the product of the diagonal. Thus, the characteristic equation is ...

However, since this matrix is full rank (it is the identity matrix), it must be invertible. This is in contradiction of the original statement, which noted that the reduced row echelon form of $(A -\lambda I)$ will always have at least one row of zeroes, by the definitional requirement of eigenvectors to be nonzero.

I would appreciate it if someone could please explain if my reasoning is incorrect. If so, why? What is the correct reasoning?

Thank you.

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    $\begingroup$ There is no contradiction. You have written down $A$, not $A-\lambda I$. $A$ has the eigenvalues $\lambda = 1,3,4$. If for example $\lambda = 1$ then $A - \lambda I = \pmatrix{0 & 2 & 6\\ 0 & 2 & 5\\ 0 & 0 & 3}$. $\endgroup$ – Winther Nov 18 '16 at 11:56
  • $\begingroup$ @Winther So the $-\lambda I$ terms of the expression $A-\lambda I$ would have to make one of the rows of $A$ all-zero, in order to ensure non-invertibility of $A-\lambda I$? $\endgroup$ – The Pointer Nov 18 '16 at 11:58
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    $\begingroup$ It might be instructive to try to explicitly compute the reduced form of $A-\lambda I$ for the three eigenvalues of this matrix here and see that you always get a row of all zeros. $\endgroup$ – Winther Nov 18 '16 at 12:04

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