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

Question

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 $A$.

Answer 1

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 $A^{T}$.

Answer 2

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 $A^t$.

Answer 3

$$ \operatorname{det}(A-tI) = \operatorname{det}((A-tI)^T) = \operatorname{det}(A^T-tI)$$ 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.

Answer 4

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.