# Need a hint on showing that if $f_1,\dotsc f_n$ is any basis of eigenvectors of $A^T$, then $f_i(v_i) \neq 0$ and $f_i(v_j) = 0$ for $i \neq j$

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.

• 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?
– Dave
Feb 26 '21 at 22:47
• @dave Thanks for the questions; I apologize for not being clear. Yes, $X$ is an $n$-dimensional space, and $A$ has $n$ distinct eigenvalues. 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 $$j\neq i$$.

• 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. Feb 27 '21 at 14:03
• 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 $f_2$).
– Dave
Feb 27 '21 at 16:14