If $AA^T=A^TA=I$ and $\det(A)=1$ then $p_A(1)=0$

If $AA^T=A^TA=I$ and $\det(A)=1$ then $p_A(1)=0$. Where $A\in M_3(\mathbb{R})$

My approach:

We known from the first equation that $A^T=A^{-1}$ and $\lambda_1\lambda_2\lambda_3=1$. Now,since $A$ and $A^T$ have the same characteristic polynomial, they have the same eigenvalues. We also know that the eigenvalues for $A^{-1}$ are $\frac{1}{\lambda_1},\frac{1}{\lambda_2},\frac{1}{\lambda_3}$ and since $A^T=A^{-1}$ this basically means that $\frac{1}{\lambda_1}=\lambda_1,\frac{1}{\lambda_2}=\lambda_2,\frac{1}{\lambda_3}=\lambda_3$

From here there are 2 possible solutions: $\lambda_1=\lambda_2=-1,\lambda_3=1$ or $\lambda_1=\lambda_2=\lambda_3=1$

Is my approach correct?

• what is $p_A{}$? Jun 11 '17 at 16:03
• Characteristic polynomial. Jun 11 '17 at 16:04
• how do you know the matrix is diagonalizable? Jun 11 '17 at 16:10
• Why should it be true that $A$ and $A'$ are similar? I agree they have the same eigenvalues but it doesn't follow they have the same eigenvectors and moreover not the same eigenvectors for the same eigenvalues. Thus the comparison of $\lambda_i$ with its inverse does not hold. Jun 11 '17 at 16:23
• @LandonCarter it is generally true that $A$ is similar to its transpose Jun 11 '17 at 16:28

Asserting that $A^TA=AA^T=\operatorname{Id}$ is the same thing as saying that $A$ is an orthogonal matrix. So, for each $v\in\mathbb{R}^3$, $\|Av\|=\|v\|$. In particular, if $v$ is an eigenvector with eigenvalue $\lambda$ (such a $v$ must exist, since $\mathbb{R}^3$ is odd-dimensional), $|\lambda|=1$, that is $\lambda=\pm1$. If $1$ is an eigenvalue,, then $P_A(1)=0$. Suppose otherwise. Then $-1$ is an eigenvalue. Let $W=(\mathbb{R}u)^\perp$. Then $A.W\subset W$ and then $A$ is an orthogonal transformation of $W$ whose determinant is $-1$. But then there is a vector $w\in W\setminus\{0\}$ such that $A.w=w$, that is, $1$ is an eigenvalue of $A$.

• Your answer seems legit,but can you tell me is my approach also correct? Jun 11 '17 at 16:15
• I don't think it is, you assumed diagonalizability Jun 11 '17 at 16:16
• No. For instance, how do you know that $A$ and $A^T$ are similar? And how do you go from $\{\lambda_1,\lambda_2,\lambda_3\}=\{\frac1{\lambda_1},\frac1{\lambda_2},\frac1{\lambda_3}\}$ to $\lambda_1=\frac1{\lambda_1}$, $\lambda_2=\frac1{\lambda_2}$, and $\lambda_3=\frac1{\lambda_3}$? Jun 11 '17 at 16:18
• This proof is faulty. Orthogonal matrices don't necessarily have real eigenvalues. You need to use the fact that the dimension of the space is $3$ to conclude that $A$ has a real eigenvalue. Jun 11 '17 at 16:21
• I know that. I shall add this to my proof. Jun 11 '17 at 16:22

$A$ is an orthogonal matrix. Hence all the eigenvalues have unit modulus.

Let $\lambda_1,\lambda_2,\lambda_3$ be the eigenvalues of the matrix.

Since, $Det(A)=1\implies\lambda_1\lambda_2\lambda_3=1$------------------(*)

Case-1:

If all $\lambda_i$'s are real, they can take values $\pm1$.

From (*) it follows that $1$ must be an eigenvalue.

Case-2:

If not all $\lambda_i$'s are real then we have a pair of complex conjugates.

From (*) it again follows that $1$ must be an eigenvalue.

Thus, $p_A(1)=0$

• Case-2 isn't legit $\lambda_1=1-i$ $\lambda_2=1+i$ from $(*)$ you get that $\lambda_3=\frac{1}{2}$ Jun 11 '17 at 17:12
• @DraganZrilić $z\overline z=|z|^2=1$ Jun 11 '17 at 17:14
• @DraganZrilić The eigenvalues must be of unit modulus(They must lie on the unit circle). The ones you've chosen do not lie there. Jun 11 '17 at 17:16