# A matrix and its transpose have the same set of eigenvalues/other version: $A$ and $A^T$ have the same spectrum

Let $$\sigma(A)$$ be the set of all eigenvalues of $$A$$. Show that $$\sigma(A) = \sigma\left(A^T\right)$$ where $$A^T$$ is the transpose matrix of $$A$$.

• This is a bit more advanced than what you need, but: an interesting article. Mar 24, 2012 at 13:10
• I guess your work in an algebraically closed field. In this case, use the fact that $r$ is an eigenvalue of $A$ if and only if $r$ is an eigenvalue of $A^T$. In fact, it can be shown that $A$ and $A^T$ are similar. Mar 24, 2012 at 13:11
• Here's one possible simpler problem that will get you started on the right path. If $A$ is an n by n singular matrix, can you show that $A^T$ is also singular? Mar 24, 2012 at 13:25
• Please don't post your questions in the imperative; please tells us what your thoughts are about the question, so that people don't tell you things you already know; please tell us the context in which you encountered the question, so that people can write their answers at an appropriate level. Mar 24, 2012 at 21:03
• I edited your title so as to cut down the number of duplicates since this is a FAQ. May 20, 2020 at 8:54

The matrix $$(A - \lambda I)^{T}$$ is the same as the matrix $$\left(A^{T} - \lambda I\right)$$, since the identity matrix is symmetric.

Thus:

$$\det\left(A^{T} - \lambda I\right) = \det\left((A - \lambda I)^{T}\right) = \det (A - \lambda I)$$

From this it is obvious that the eigenvalues are the same for both $$A$$ and $$A^{T}$$.

• Do they have same minimal polynomial ? Oct 6, 2015 at 10:49
• Any polynomial satisfied by $A$ is also satisfied by $A^T$ so yeah. Oct 6, 2015 at 10:51
• I suppose that @Invisible is saying that because $A^T$ is similar to $A$ considering a special similarity transformation on Jordan blocks. That way is possible to biject explicitly the eigenvalues and eigenvector of $A$ with $A^T$. The analysis follows this link: math.stackexchange.com/questions/94599/…. Quite interesting to know that. I have never heard of that result in my linear algebra courses. Jul 30, 2021 at 21:05

I'm going to work a little bit more generally.

Let $V$ be a finite dimensional vector space over some field $K$, and let $\langle\cdot,\cdot\rangle$ be a nondegenerate bilinear form on $V$.

We then have for every linear endomorphism $A$ of $V$, that there is a unique endomorphism $A^*$ of $V$ such that $$\langle Ax,y\rangle=\langle x,A^*y\rangle$$ for all $x$ and $y\in V$.

The existence and uniqueness of such an $A^*$ requires some explanation, but I will take it for granted.

Proposition: Given an endomorphism $A$ of a finite dimensional vector space $V$ equipped with a nondegenerate bilinear form $\langle\cdot,\cdot\rangle$, the endomorphisms $A$ and $A^*$ have the same set of eigenvalues.

Proof: Let $\lambda$ be an eigenvalue of $A$. And let $v$ be an eigenvector of $A$ corresponding to $\lambda$ (in particular, $v$ is nonzero). Let $w$ be another arbitrary vector. We then have that: $$\langle v,\lambda w\rangle=\langle\lambda v,w\rangle=\langle Av,w\rangle=\langle v,A^*w\rangle$$ This implies that $\langle v,\lambda w-A^*w\rangle =0$ for all $w\in V$. Now either $\lambda$ is an eigenvalue of $A^*$ or not. If it isn't, the operator $\lambda I -A^*$ is an automorphism of $V$ since $\lambda I-A^*$ being singular is equivalent to $\lambda$ being an eigenvalue of $A^*$. In particular, this means that $\langle v, z\rangle = 0$ for all $z\in V$. But since $\langle\cdot,\cdot\rangle$ is nondegenerate, this implies that $v=0$. A contradiction. $\lambda$ must have been an eigenvalue of $A^*$ to begin with. Thus every eigenvalue of $A$ is an eigenvalue of $A^*$. The other inclusion can be derived similarly.

How can we use this in your case? I believe you're working over a real vector space and considering the dot product as your bilinear form. Now consider an endomorphism $T$ of $\Bbb R^n$ which is given by $T(x)=Ax$ for some $n\times n$ matrix $A$. It just so happens that for all $y\in\Bbb R^n$ we have $T^*(y)=A^t y$. Since $T$ and $T^*$ have the same eigenvalues, so do $A$ and $A^t$.

• For an explanation on the things I took for granted, I suggest you read these excellent lecture notes: dpmms.cam.ac.uk/study/IB/LinearAlgebra/2008-2009/…
– user123641
Jul 24, 2014 at 23:54
• I have a questions: First, why did you assume surjectivity of A ?
– Our
Oct 5, 2017 at 8:16
• Plus, where did we used the fact that V is finite dimensional. (I'm only interested with the fact that $A$ and $A^*$ has the same eigenvalues.)
– Our
Oct 5, 2017 at 10:00
• @onurcanbektas Where have I used surjectivity? The finite dimension is used in my reference to construct $A^*$. If you have $A^*$ to start with, construction isn't required. I would have to double check whether $\lambda I-A^*$ is an automorphism (bounded) cause I don't have infinite dimensional facts at my fingertips. But you might.
– user123641
Oct 5, 2017 at 13:12
• Why do you write $⟨v,\lambda w⟩=⟨\lambda v,w⟩$? In a general case we have $⟨v,\bar\lambda w⟩=⟨\lambda v,w⟩$ and your computation should be corrected accordingly. Apr 14, 2020 at 11:13

$$\operatorname{det}(A-tI) = \operatorname{det}((A-tI)^T) = \operatorname{det}(A^T-tI)$$ A matrix and its transpose have the same determinant. If you apply properties of transposition, you get that both $$A$$ and its transpose have the same characteristic polynomial.

Here is another proof: Suppose that $$v$$ is an eigenvector of $$A$$ with eigenvalue $$\lambda$$, i.e. $$Av = \lambda v$$. Then $$v^T A^T = (Av)^T = \lambda v^T$$. This means that $$v^T(A^T - \lambda I) = 0$$. Thus, $$v^T$$ is a left-eigenvector of $$A^T$$. If $$A^T - \lambda I$$ was invertible then multiplying from the right with the inverse leads to $$v=0$$ which is a contradiction.