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Let $A$ be a square matrix. Prove that $A$ and $A^t$ have the same eigenvalues.

The solution my lecturer uses is: Consider the characteristic polynomial

\begin{align} P_{A^t}(x) &= \det{\left(xI-A^t\right)} \\ &= \det{\left((xI-A)^t\right)} \\ &= \det{(xI-A)} = P_{A}(x). \end{align}

Since the characteristic polynomials are equal and the eigenvalues are the roots of the characteristic polynomial, the eigenvalues will be the same.

However some of the steps he has taken seem to have come out of nowhere. How do we know $(xI-A^t) = (xI-A)^t$, also how do we know $(xI-A)^t=(xI-A)$?

user2850514
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    Be careful with your setup here -- saying "$A$ and $A^t$ have the same eigenvalues" is not the same thing as "if $\lambda$ is an eigenvalue of $A$ and $\mu$ is an eigenvalue of $A^t$, then $\lambda=\mu$". For example, what if $A$ has two different eigenvalues (say $1$ and $2$)? Nothing in your setup so far excludes the possibility that $\lambda=1$ and $\mu=2$, so you won't be able to show $\lambda=\mu$. – Kevin P. Costello May 05 '15 at 17:40

2 Answers2

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Note that for any matrix $M$, $$\det(M)=\det(M^t)$$ The eigenvalues are found by solving $$\det(A-\lambda I)=0$$ But $(A-\lambda I)^t=A^t-\lambda I^t=A^t-\lambda I$. Thus $$\det(A-\lambda I)=\det(A^t-\lambda I)$$ The two matrices therefore have the same characteristic polynomial, hence the same eigenvalues.

Matt Samuel
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The eigenvalues of $A$ are the roots of the characteristic polynomial $\operatorname{det}(A-\lambda\cdot I)$ ; keeping in mind that $A$ and $A^{T}$ have the same determinant, one can conclude that $A$ and $A^{T}$ have the same characteristic polynomial because $(A-\lambda\cdot I)^{T}=A^{T}-\lambda\cdot I$. Therefore they have the same eigenvalues

marwalix
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