Let $A:X\rightarrow X$ be a linear map with distinct eigenvalues $\lambda_1,\dotsc, \lambda_n$ and corresponding eigenvectors $v_1,\dotsc, v_n$.

Suppose that $f_1,\dotsc, f_n$ is any basis of eigenvectors of $A^T$. We want to show that $f_i(v_i) \neq 0$ and $f_i(v_j) = 0$ whenever $i \neq j$.

I've been playing around with this problem for a couple of days now and have reached deadends with each attempt. To make things simpler, I've been considering the case when $n=2$.

So far, what I have is that $Av_1 = \lambda_1v_1$ and $A v_2 = \lambda_2 v_2$, as well as $A^T(f_1) = f_1\circ A = \gamma_1 f_1$ $A^T(f_2) = f_2\circ A = \gamma_2 f_2$, where $\gamma_1,\gamma_2$ are scalars.

Using these equations, I focused on $f_1(v_1)$ and was able to get $$(\lambda_1-\gamma_1)f_1(v_1) = 0,$$ which clearly gives us $\lambda_1 = \gamma_1$ or $f_1(v_1) = 0$, but I don't know what to do with the assumption that $\lambda_1 = \gamma_1$, nor can I find a contradiction with assuming $f_1(v_1) = 0$.

I started looking at the nullspace of $f_1 \circ A - \lambda I$ where $I \in \operatorname{End}(X)$ is the identity map and $\lambda$ is some scalar, but I don't know where or how to proceed with this.

Any tips is greatly appreciated.

  • $\begingroup$ Just to be clear on the setup: $X$ is an $n$-dimensional space? So the eigenvalues of $A$ are all distinct, or is it that $n$ is just the number of distinct eigenvalues? $\endgroup$
    – Dave
    Feb 26 '21 at 22:47
  • $\begingroup$ @dave Thanks for the questions; I apologize for not being clear. Yes, $X$ is an $n$-dimensional space, and $A$ has $n$ distinct eigenvalues. $\endgroup$ Feb 26 '21 at 22:58

In doing this, we need to note that the eigenvalues of $A^T$ and $A$ are equal. So I assume that the $f_i$ are chosen to be eigenvectors of $A^T$ with respect to the eigenvalue $\lambda_i$ (otherwise this problem doesn't work).

For $i\neq j$ we have $$(\lambda_if_i)(v_j)=(A^T(f_i))(v_j)=f_i\circ Av_j=f_i(\lambda_jv_j)=\lambda_jf_i(v_j).$$

Spoiler to conclude the $i\neq j$ case:

The above gives $(\lambda_i-\lambda_j)f_i(v_j)=0$. When $i\neq j$ we get $\lambda_i\neq \lambda_j$, so we must have $f_i(v_j)=0$.

For the $i=j$ case: note that the $v_1,\ldots,v_n$ form a basis for $X$ since all the eigenvalues are distinct. Spoiler to conclude the $i=j$ case:

Fix $i$. Since $v_1,\ldots,v_n$ form a basis for $X$, we cannot have $f_i(v_j)=0$ for all $j$ (this would mean that $f_i=0$, but $f_i$ is nonzero as it is an eigenvector; in fact the $f_i$ are a basis for $X^*$). Thus $f_i(v_i)\neq 0$ since we already showed $f_i(v_j)=0$ for $j\neq i$.

  • $\begingroup$ Thank you for taking the time to write this answer! I have a question though. Can it be proven that $\lambda_i$ is the corresponding eigenvalue of $f_i$, or is it that something that should have been given to us in the first place? I attempted to prove this, but without knowing that $f_i(v_j) \neq 0$ and $f_i(v_i) = 0$, it's not obvious. $\endgroup$ Feb 27 '21 at 14:03
  • 1
    $\begingroup$ It needs to be the case that $f_i$ has eigenvalue $\lambda_i$ in order for the claim to hold. Otherwise the claim is not true (for example, in my solution: imagine just switching $f_1$ and $f_2$). $\endgroup$
    – Dave
    Feb 27 '21 at 16:14

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