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Suppose $V$ is finite-dimensional, $T \in L(V)$, and $\lambda \in \mathbf{F}$. Show that $\lambda$ is an eigenvalue of $T$ if and only if $\lambda$ is an eigenvalue of the dual operator $T' \in L(V')$.

This is question 15 of 5A of Sheldon Axler's Linear Algebra Done Right 4th Edition.

  • $L(V)$: Space of linear maps from $V$ to $V$.
  • $\mathbf{F}$: Field associated with $V$.
  • $V'$: Space of linear functionals on $V$.

I think that a proof for the "only if" direction could be something like this: $\lambda$ is an eigenvalue of $T$ associated with a non-zero $u$ in $V$. Can extend $u$ to a basis (since it's non-zero) of the range of $T$ and then of $V$. Define the functional that sends $u$ to $\lambda$ and all other basis vectors to $0$. This functional could be our eigenvector?

The proof for the other direction could be that $T'$ has some non-zero eigenvector $\varphi$. Since $\varphi$ is non-zero, there is at least one non-zero $u$ in $V$ such that $\varphi(u) \neq 0$. Then $$ \varphi(T(u)) = T'(\varphi)(u) = \lambda \varphi(u) = \varphi(\lambda u), $$ but $\varphi$ need not be injective so I'm stuck.

md2perpe
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Shivansh Raman
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    As an FYI: MSE is not a "do my work for me" site. Such posts (e.g. "PSQs") are in poor taste, downvoted, & closed; we expect users to put in effort! If you want meaningful help, [edit] your post to add relevant context (more), e.g. relevant definitions, your work/attempts, the problem's source, where specifically you're stuck, what you do/don't understand & have learned recently, & so on. If needed: this is useful for formatting LaTeX, & Approach0 helps find past questions. – PrincessEev Aug 11 '24 at 21:05
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    Do you mean transpose of a linear map instead of dual of a linear map? You want to prove that $T\colon V\to V$ has same eigenvalues as $T^t\colon V^* \to V^$ where for $f\in V^$, $T^t(f)$ is a functional which sends a vector $v\in V$ to $f(v)$? – Nothing special Aug 11 '24 at 21:11
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    @Nothingspecial dual is the language used in the Axler textbook. I want to prove the second statement about functionals – Shivansh Raman Aug 11 '24 at 21:36
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    Does this answer your question? Eigenvalue of an operator implies eigenvalue of the dual? – azif00 Aug 11 '24 at 22:01
  • That is helpful but I think I didn't want to use the characteristic polynomial bit since they hadn't introduced the (characteristic polynomial uniquely pins down eigenvalues) bit in the textbook yet so there must be something I'm missing – Shivansh Raman Aug 11 '24 at 22:07
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    @ShivanshRaman perhaps you could look at https://math.stackexchange.com/questions/2787875/the-dual-transformation-has-the-same-eigenvalues/2787952#2787952 – Osama Ghani Aug 12 '24 at 00:03

1 Answers1

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Your attempt for "only if" doesn't work, even if you set $\varphi(u)=1$ instead of $\varphi(u)=\lambda$: you'll thus get $T'(\varphi)(u)=\lambda\varphi(u)$, but for the other vectors $u_2,\dots,u_n$ of your basis, $T'(\varphi)(u_k)=\varphi(T(u_k))$ has no reason to be equal to $\lambda\varphi(u_k)$ i.e. to $0$, i.e. $T$ has no reason to map $\ker\varphi=\operatorname{span}(u_2,\dots,u_n)$ into itself.

So, let's start from scratch. Since $(T-\lambda\mathrm{id}_V)'=T'-\lambda\mathrm{id}_{V'}$, it suffices to prove the case $\lambda=0$, i.e. to show that $\ker T\ne\{0\}$ iff $\ker T'\ne\{0\}$.

$\ker T\ne\{0\}$ is equivalent, by the rank-nullity theorem, to $\operatorname{im}T$ being a proper subspace of $V$, i.e. $\operatorname{im}T\subseteq\ker\varphi$ for some non zero $\varphi\in V'$, i.e. $T'(\varphi)=\varphi\circ T=0$ for some non zero $\varphi\in V'$, i.e. $\ker T'\ne\{0\}$. To summarize: $$\ker(T')=(V/\operatorname{im}T)'\cong(\ker T)'.$$

Anne Bauval
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  • I think you can also show that $((T-\lambda \text{id}V)^k)'=(T'-\lambda \text{id}{V'})^k$ for every positive integer $k$ and make conclusion about algebraic multiplicity of $\lambda$ in $T$ and $T'$ being same. – Nothing special Aug 12 '24 at 11:29
  • @Nothingspecial $((T-\lambda \text{id}V)^k)'=(T'-\lambda \text{id}{V'})^k$ is easy but I don't see how you conclude about algebraic multiplicities. They are indeed the same by https://math.stackexchange.com/a/3080741, but the OP doesn't want to use the characteristic polynomial. – Anne Bauval Aug 12 '24 at 11:43
  • The generalized eigenspaces of $\lambda$ i.e., $\ker(T'-\lambda \text{id}{V'})^k$ and $\ker(T-\lambda \text{id}{V})^k$ have same dimension. Now see Connection between algebraic multiplicity and dimension of generalized eigenspace. Actually, this gives a stronger result. In a closed field, we can show that $T$ and $T'$ have the same Jordan canonical form. See this. – Nothing special Aug 12 '24 at 12:13
  • @Nothingspecial I agree, since I essentially proved that $\dim\ker (T')=\dim\ker T$ and this can be applied to $(T-\lambda \text{id}_V)^k$ for every $k$ (in particular for $k\ge\dim V$, whence the equality of dimensions of generalized eigenspaces). But I prefer to keep my answer simple and fitting the question. – Anne Bauval Aug 12 '24 at 13:33
  • Did I discover an basis free and abstract way of proving that a square matrix is similar to its transpose? I haven't seen it anywhere. – Nothing special Aug 12 '24 at 13:42
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    @Nothingspecial https://math.stackexchange.com/a/2689625 – Anne Bauval Aug 12 '24 at 13:51