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I'm doing a self-study of Axler's Linear Algebra Done Right, and am looking for some help understanding a step in the proof of Proposition 5.21, appearing on page 89 of the second edition. An abbreviated version of 5.21 and its proof is shown below, with the step I don't understand highlighted in bold:

Suppose $T \in \mathcal{L}$($V$). Let $\lambda_{1}, \ldots , \lambda_{m}$ denote the distinct eigenvalues of $T$. Then the following are equivalent:

  1. $V$ has a basis consisting of eigenvectors of T;
  2. dim V $=$ dim null ($T - \lambda_{1}I) + \cdots +$ dim null($T - \lambda_{m}I$)

Suppose that (2) holds. Choose a basis of null ($T - \lambda_{j}I$); put all these bases together to form a list ($\vec{v_{1}}, \ldots , \vec{v_{n}}$) of eigenvectors of $T$, where $n$ = dim $V$. To show this list is linearly independent, suppose

$a_{1}\vec{v_{1}} + \cdots +a_{n}\vec{v_{n}} = 0$ (1)

where $a_{1}, \ldots , a_{n} \in \mathbf{F}$. For each $j = 1,\ldots,m$, let $u_{j}$ denote the sum of all the terms of $a_{k} \vec{v_{k}}$ such that $\vec{v_{k}} \in$ null($T - \lambda_{j}I$). Thus each $u_{j}$ is an eigenvector of $T$ with eigenvalue $\lambda_{j}$, and

$u_{1} + \cdots + u_{m} = 0$

My question is this: what does it mean to "let $u_{j}$ denote the sum of all the terms of $a_{k} \vec{v_{k}}$ such that $\vec{v_{k}} \in$ null($T - \lambda_{j}I$)?" Aren't those terms already defined in equation (1), and isn't their sum already supposed to be zero? And how is it that "thus" each $u_{j}$ becomes an eigenvector of $T$ with a corresponding eigenvalue of $\lambda_{j}$?

Thanks in advance for any and all help!

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  • $\begingroup$ Axler is defining something called $u_j$, so when he tells you to get it by summing over all those terms, I think he means for you to keep $j$ fixed while varying $k$. In other words, he wants you to get $u_j$ by adding up just the terms associated with the null space of $T - \lambda_j I$, with $j$ fixed. $\endgroup$ – Vectornaut Dec 19 '14 at 19:21
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We are assuming that there is a linear combination $a_1 \vec{v_1} + \dotsb + a_n \vec{v_n} = 0$, where these $n$ vectors $\vec{v_1}, \dotsc, \vec{v_n}$ are all the basis elements we chose for the null-spaces $\operatorname{null}(T - \lambda_i I)$.

So, we collect the terms that belong to each of these $m$ sets. For instance, suppose $n = 5$ and $T$ has three eigenvalues; the first, $\lambda_1$, has an eigenspace of dimension 2 (i.e. 2 independent eigenvectors); $\lambda_2$ also has $\operatorname{dim}\operatorname{null}(T - \lambda_2 I) = 2$; and $\lambda_3$ only has a one-dimensional eigenspace. Then the layout is like $$ \underbrace{a_1 \vec{v_1} + a_2 \vec{v_2}}_{u_1} + \underbrace{a_3 \vec{v_3} + a_4 \vec{v_4}}_{u_2} + \underbrace{a_5 \vec{v_5}}_{u_3} = 0 $$ Because $u_i$ is a linear combination of eigenvectors for $\lambda_i$, it is also an eigenvector for $\lambda_i$. And it's clear that $u_1 + u_2 + u_3 = 0$.

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  • $\begingroup$ Terrific explication, @Kundor. Very much appreciated. One last request for clarification. You say that "because $u_{i}$ is a linear combination of eigenvectors for $\lambda_{i}$, it is also an eigenvector for $\lambda_{i}$." This wasn't obvious to me at all. Is this because null($T - \lambda_{i}$) is a subspace of $V$? $\endgroup$ – Scentless Apprentice Dec 19 '14 at 23:18
  • $\begingroup$ @ScentlessApprentice: Suppose $w$ and $v$ are eigenvectors for the eigenvalue $\lambda$, so $T w = \lambda w$ and $T v = \lambda v$. Then $T(w + v) = T w + T v = \lambda w + \lambda v = \lambda(w + v)$. $\endgroup$ – Kundor Dec 20 '14 at 2:19
  • $\begingroup$ That's what I thought. Thanks! $\endgroup$ – Scentless Apprentice Dec 20 '14 at 4:41

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