3

Source: Linear Algebra by David Lay (4 edn 2011). p. 269 Theorem 5.1.1.

For simplicity, consider the $3\times 3$ case. If $A$ is upper triangular, then $ A-\lambda I= \begin{bmatrix} a_{11} & a_{12} & a_{13}\\ 0 & a_{22} & a_{23}\\ 0 & 0 & a_{33} \end{bmatrix} - \begin{bmatrix} \lambda & 0 & 0\\ 0 & \lambda & 0\\ 0 & 0 & \lambda \end{bmatrix} =\begin{bmatrix} a_{11}-\lambda & a_{12} & a_{13}\\ 0 & a_{22}-\lambda & a_{23}\\ 0 & 0 & a_{33}-\lambda \\ \end{bmatrix}$

$\lambda$ is an eigenvalue of $A \iff$ The equation $(A-\lambda I)x=0$ has a nontrivial solution. $\iff$ $(A-\lambda I)x=0 $ has a free variable $ \iff $
Because of the zero entries in $A-\lambda I$, at least one of the entries on the diagonal of $A-\lambda I$ is zero.
$\color{red}{\iff} \lambda$ equals $\color{red}{one \, of } $ the entries $a_{11},\ a_{22},\ a_{33}$ in $A$. For the lower triangular case, see Question 5.1.28.

$1.$ I don't understand the red $\color{red}{\iff}$. $a_{11} - \lambda = 0 \iff$ The first column is $\mathbf{0}. \iff$ $x_1$ is a free variable. But what about the other entries?

$2.$ I linked an analogous question. How does the proof overhead proves that all of the eigenvalues = all its diagonal entries, when it states $\color{red}{one \, of } $?

  • If $a_{22} - \lambda = 0$, then the subproblem $\begin{pmatrix}a_{22} - \lambda & a_{23} \ 0 & a_{33} - \lambda\end{pmatrix} \begin{pmatrix}x_2\x_3\end{pmatrix} = 0$ has $x_2$ as a free variable. More generally, you look at the submatrix whose upper-left corner is the first zero on the diagonal of the original matrix and deduce that the corresponding $x_i$ is a free variable.
    • – Tunococ May 20 '14 at 20:16
  • If you put all the equivalences together, you'd get "$\lambda$ is an eigenvalue if and only if $\lambda \in {a_{11}, a_{22}, a_{33}}$". If that doesn't mean all eigenvalues are all the diagonal entries, I guess you must be considering multiplicities also. In that case, you need the characteristic polynomial to define multiplicities, which will involve $\det$. However, if you accept the concept of $\det$ from the beginning, the whole problem here becomes quite trivial.
    • – Tunococ May 20 '14 at 20:19
  • Tell me if the following sentences are logically equivalent: (1) $x \in S$ if and only if $x \in {1, 2, 3}$ (2) $S = {1, 2, 3}$. – Tunococ May 28 '14 at 01:20
  • @Tunococ Yes, I think they are. In (1), you are just working with an element of S? But I still don't see how this relates to my questions? –  May 28 '14 at 09:37
  • 1
    I am not very sure what you're confused about now. I believe my first comment should have answered your first question, and my last comment above should have answered your second question. Or if you want me to expand more, here is my attempt. "$\lambda$ is an eigenvalue of $A$ if and only if $\lambda$ is one of the entries $a_{11}, a_{22}, a_{33}$" is equivalent to "the set of eigenvalues of $A$ is ${a_{11}, a_{22}, a_{33}}$". – Tunococ May 28 '14 at 09:48
  • If this still doesn't answer your question, I believe you might be looking at the multiset of eigenvalues instead of just the set of eigenvalues. In that case, I am not quite sure how to determine the multiplicities without the notion of determinant. – Tunococ May 28 '14 at 09:51
  • @Tunococ For $\lambda$ to be an eigenvalue of A, don't we need $\lambda$ = ALL of the entries ... ? You wrote $\lambda$ is ONE of the entries ..., so this means that $\lambda$ doesn't have to equal ALL of them? –  May 28 '14 at 10:26
  • $\lambda$ is a variable that can take on only one value at a time. The first statement implicitly implies the quantification "for all $\lambda$". I can expand it further as "$\forall \lambda [(\lambda \text{ is an eigenvalue of } A) \Leftrightarrow ((\lambda = a_{11}) \text{ or } (\lambda = a_{22}) \text{ or } (\lambda = a_{33}))]$". This should define ${a_{11}, a_{22}, a_{33}}$ as the set of eigenvalues of $A$. – Tunococ May 28 '14 at 22:32