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This is 5.18 from Axler's Linear Algebra Done Right:

Theorem: Suppose $T \in L(V)$ has an upper-triangular matrix with respect to some basis of $V$. Then the eigenvalues of $T$ consist precisely of the entries on the diagonal of that upper-triangular matrix.

Proof:

Suppose $(v_1, \ldots , v_n)$ is a basis of $V$ with respect to which $T$ has an upper-triangular matrix where the diagonal entries are $\lambda_1, \ldots, \lambda_n$.

Let $\lambda \in F$

Then for matrix $M(T - \lambda I$) where the diagonal entries are $\lambda_1 - \lambda, \ldots \lambda_n - \lambda.$ We can suppose we are dealing with complex vector spaces. From 5.16 where have proven that $T$ is not invertible iff one of the $\lambda_k$'s equals $0$. Hence $T - λI$ is not invertible if and only if $λ$ equals one of the $λj$'s. In other words, $λ$ is an eigenvalue of $T$ if and only if $λ$ equals one of the $λj$s, as desired.

Question:

This only showed that one of the diagonal entries is en eigenvalue but not all of them as the theorem claimed.

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    $\begingroup$ $\lambda$ is arbitrary, as long as it is one of the $\lambda_j$. It shows one of them, but it is any one of them, thus they are all the eigenvalues. $\endgroup$ – adam W Dec 25 '12 at 15:38
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I understand why this idiom (which is common in math writing) might seem confusing, but what the author is saying is correct. When he says that

$\lambda$ is an eigenvalue of $T$ if and only if $\lambda$ equals one of the $\lambda_j$'s

he means that

$\lambda$ is an eigenvalue of $T$ if and only if $\lambda\in\{\lambda_1,\ldots,\lambda_n\}$

or, to phrase it another way,

$\lambda$ is an eigenvalue of $T$ if and only if $\lambda=\lambda_1$, or $\lambda=\lambda_2$, ..., or $\lambda=\lambda_n$

Thus, if I set $\lambda$ equal to $\lambda_1$, the right side of the biconditional is true, so that $\lambda$ is an eigenvalue of $T$ when $\lambda=\lambda_1$; and similarly with all of the diagonal entries $\lambda_1,\ldots,\lambda_n$.

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  • $\begingroup$ Would you please explain how $\lambda = \lambda_1$ and $\lambda = \lambda_2$ and $\lambda = \lambda_i$ for all $1 \le i \le n$? I still don't apprehend this idea? $\endgroup$ – Greek - Area 51 Proposal May 26 '14 at 16:30

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