1
$\begingroup$

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 } $?

$\endgroup$
  • $\begingroup$ 1. 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. $\endgroup$ – Tunococ May 20 '14 at 20:16
  • $\begingroup$ 2. 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. $\endgroup$ – Tunococ May 20 '14 at 20:19
  • $\begingroup$ Tell me if the following sentences are logically equivalent: (1) $x \in S$ if and only if $x \in \{1, 2, 3\}$ (2) $S = \{1, 2, 3\}$. $\endgroup$ – Tunococ May 28 '14 at 1:20
  • $\begingroup$ @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? $\endgroup$ – Greek - Area 51 Proposal May 28 '14 at 9:37
  • $\begingroup$ 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 $\{a_{11}, a_{22}, a_{33}\}$". $\endgroup$ – Tunococ May 28 '14 at 9:48
2
$\begingroup$

It's equivalent to showing that an upper triangular matrix is injective iff none of its diagonal entries are $0$.

If one of the diagonal entries, say the $i$th, is $0$, then restricting the operator to the span of the first $i$ basis vectors gives a map into the span of the first $i-1$ basis vectors, which by Rank-Nullity cannot be injective.

Now, if all of the diagonal entries are nonzero, then the columns must be independent: the last column is the only one with a nonzero $n$th entry, so its coefficient has to be $0$. Continuing this way, you can kill all the coefficients.

$\endgroup$
  • $\begingroup$ Would you please elucidate how this answers my questions? I'm not asking for other proofs? $\endgroup$ – Greek - Area 51 Proposal May 28 '14 at 9:36
1
$\begingroup$

Not sure I get your question perfectly but :

$\lambda$ is an eigenvalue of A $\Leftrightarrow \lambda $ is a root of the characteristic polynom of A $\Leftrightarrow |A -\lambda Id| =0 \Leftrightarrow \Pi_{i}^n (a_{ii}-\lambda)=0 \Leftrightarrow \exists i\ \lambda =a_{ii}$.

Most equivalence are quite obvious ( you need to know that the determinant of a tringular matrix is the product of its diagonal elements and that the kernel of a matrix which determinant is 0 has a non trivial kernel).

$\endgroup$
  • $\begingroup$ Thank you; I forgot to elucidate earlier my desire to omit determinants from this proof, so would you please recast your answer? $\endgroup$ – Greek - Area 51 Proposal May 22 '14 at 15:02
0
$\begingroup$

Expand $\det(A - \lambda I)$ using minors of the first column.

The only term that survives is $(a_{11}-\lambda)\det(B)$ where:

$B = \begin{bmatrix}a_{22}-\lambda&a_{23}\\0&a_{33}-\lambda \end{bmatrix}$, which clearly has determinant:

$(a_{22} - \lambda)(a_{33} - \lambda)$.

A similar strategy works for any $n \times n$ upper triangular matrix.

This shows that every eigenvalue (root of $\det(A - \lambda I)$) is a diagonal entry of $A$ and vice-versa.

Surely you can see that (in the $3\times3$ case) if $a_{33} - \lambda = 0$ that the last ROW is $0$, recall column rank = row rank.

If $a_{22} - \lambda = 0$, then the 2nd row is a scalar multiple of the 3rd row, so after row-reduction, we'll have at LEAST one zero row at the bottom.

The logic is a bit more involved for an $n \times n$ upper triangular matrix, but if one of the diagonal elements of $A - \lambda I$ is $0$, it should be clear that THAT row is a linear combination of the rows below it.

$\endgroup$
  • $\begingroup$ Thank you; I forgot to elucidate earlier my desire to omit determinants from this proof, so would you please recast your answer? $\endgroup$ – Greek - Area 51 Proposal May 22 '14 at 15:02
0
$\begingroup$

Let $\lambda$ be a diagonal entry of $A.$ In $A-\lambda I,$ the corresponding diagonal entry is $0.$ Look at the first $0$ entry on the diagonal of $A-\lambda I.$ (first from $(1,1)$th entry through $(n,n)$th.) If this occurs on the $k$th column, then the $k$th column is a linear combination of earlier columns. Thus ${\rm rank}(A-\lambda I)<n.$ And, ${\rm null}(A-\lambda I)\geq 1.$ Thus $Av=\lambda v$ for some $v\neq 0.$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.