# If $\lambda$ is the eigen-value of a $n\times n$ non-singular orthogonal matrix $A$, then prove that $\frac{1}{\lambda}$ is also an eigen-value.

QUESTION:

If $\lambda$ is the eigen-value of a $n\times n$ non-singular matrix $A$ and $A$ is a real orthogonal matrix, then prove that $\frac{1}{\lambda}$ is an eigen-value of the matrix $A$.

MY ATTEMPT:

Since $\lambda$ is the eigen-value of a $n\times n$ matrix $A$, we have $$|A-\lambda I_n|=0$$ Also since $A$ is a real orthogonal matrix,we have $$AA^T=A^TA=I_n$$ So we can conclude that $$|A-\lambda( AA^T)|=0$$ Or, $$|\lambda A\left(\frac{1}{\lambda}I_n- A^T\right)|=0$$ Or, $$\left|\lambda A\right|\cdot \left|\frac{1}{\lambda}I_n- A^T\right|=0$$ Or, since $A$ is non-singular, $$\left|A^T-\frac{1}{\lambda} I_n\right|=0$$

So, we can conclude that $\frac{1}{\lambda}$ is an eigen-value of the matrix $A^T$.

But how do I prove that $\frac{1}{\lambda}$ is an eigen-value of the matrix $A$?

Is my working faulty? Or is there a mistake in the question?

• Everything is fine. You just need to know that $A$ and $A^T$ has same characteristic polynomial and hence same eigen values. – Kushal Bhuyan May 30 '16 at 5:56

Suppose that $v\neq 0$ and $Av=\lambda v$. Then, $$\frac{1}{\lambda}v=\frac{1}{\lambda}Iv=\frac{1}{\lambda}(A'A)v=\frac{1}{\lambda}A'(Av)=\frac{1}{\lambda}A'\lambda v=A'v.$$ So $\lambda^{-1}$ is an eigenvalue of $A'$. But $A'$ and $A$ have the same eigenvalues (because $\det(I-\zeta A)=\det[(I-\zeta A)']=\det(I-\zeta A')$ for all real $\zeta$), and so $\lambda^{-1}$ is an eigenvalue of $A$.

Argument that doesn't use determinant: suppose that $v\neq 0$ and $Av=\lambda v$. We have $$\lambda^2||v||^2=(\lambda v)'(\lambda v)=(Av)'(Av)=v'A'Av=v'(I)v=||v||^2.$$ This implies $\lambda^2=1$ and so $\lambda=\pm 1$. In either case, $\lambda=\lambda^{-1}$ and so of course if $\lambda$ is an eigenvalue of $A$ then it's trivial that $\lambda^{-1}$ ($=\lambda$) is also an eigenvalue of $A$.

Hints:

1. if $v$ is eigenvalue for $\lambda$ of an invertible matrix $A$, then it is also eigenvalue for $\lambda^{-1}$ of $A^{-1}$,

2. $A$ and $A^T$ have the same characteristic polynomials, hence the same eigenvalues.

3. If $A$ is orthogonal then it is invertible, and $A^{-1}=A^T$.

• I got your first hint, but not what you meant by your second hint. Can you please explain? – SchrodingersCat May 30 '16 at 5:36
• @SchrodingersCat The second hint means that $A$ and $A^T$ have same characteristic polynomial and hence same eigen values. Since $\det(A-\lambda I)=\det((A-\lambda I)^T)=\det(A^T-\lambda I)$, follows from the fact that $\det(A)=\det(A^T)$ – Kushal Bhuyan May 30 '16 at 6:02
• I've added a final hint; I hope the argument is obvious now. – Marc van Leeuwen May 30 '16 at 7:25

Let's prove that for any matrix $A$ we have $\det{A}=\det{A^T}$. Once this is done it is obvious that

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

and this proves that $A$ and $A^T$ have the same characteristic polynomial and therefore the same eigenvalues.

Let's now start wit the Leibniz formula for a determinant

$$\det{A}=\sum_{\sigma\in\Sigma_n}\epsilon(\sigma)a_{\sigma(1),1}\cdots a_{\sigma(n),n}$$

Transposing we get

$$\det{A^T}=\sum_{\sigma\in\Sigma_n}\epsilon(\sigma)a_{1,\sigma(1)}\cdots a_{n,\sigma(n)}$$

Because all the $\sigma$ are bijections

$$\det{A^T}=\sum_{\sigma\in\Sigma_n}\epsilon(\sigma)a_{\sigma^{-1}(1),1}\cdots a_{\sigma^{-1}(n),n}$$

Now reindexing with $\tau=\sigma^{-1}$ we get

$$\det{A^T}=\sum_{\tau\in\Sigma_n}\epsilon(\tau)a_{\tau(1),1}\cdots a_{\tau(n),n}=\det{A}$$

• So, the question is wrong? And what about my work? Is it wrong too? Can you tell me where I made the mistake? – SchrodingersCat May 30 '16 at 5:48
• No the question is correct and your work as well ! It is just incomplete. In a nutshell you have proven that if $\lambda$ is an eigenvalue of an orthogonal matrix $1/\lambda$ is an eigenvalue of its transpose. I am adding that in general a matrix and its transpose have the same set of eigenvalues and therefore $1/\lambda$ is an eigenvalue of the original matrix – marwalix May 30 '16 at 6:00