Question-52
An \(n\times n\) matrix with real entries satisfies the property \(||Ax||^{2} =||x||^{2}\), for all \(x\in \mathbb{R}^{n}\), where \(||\cdot ||\) denotes the Euclidean norm. Which of the following statements is/are ALWAYS correct?
We take up the first option which is true.
To see why \(A\) is orthogonal, let \(A=\begin{bmatrix} | & & |\\ v_{1} & \cdots & v_{n}\\ | & & | \end{bmatrix}\). If \(\{e_{1} ,\cdots ,e_{n}\}\) is the standard ordered basis for \(\mathbb{R}^{n}\), then \(Ae_{i} =v_{i}\). Also, since \(||Ae_{i} ||^{2} =||e_{i} ||^{2} =1\), we have \(||v_{i} ||^{2} =1\). This shows that \(v_{i}\) is a unit-norm vector.
Next, let \(u=a_{1} e_{1} +\cdots +a_{n} e_{n}\) be any vector in \(\mathbb{R}^{n}\), then,
\[ Au=a_{1} v_{1} +\cdots +a_{n} v_{n}$ \]
Now:
\[ \begin{aligned} ||Au||^{2} & =a_{1}^{2} +\cdots +a_{n}^{2} +2\sum\limits _{i=1}^{n}\sum\limits _{j=i+1}^{n} a_{i} a_{j} v_{i}^{T} v_{j} \end{aligned} \] Since \(||Au||^{2} =||u||^{2}\), we have:
\[ \sum\limits _{i=1}^{n}\sum\limits _{j=i+1}^{n} a_{i} a_{j} v_{i}^{T} v_{j} =0 \]
This is true for every \(u\in \mathbb{R}^{n}\). Setting \(a_{i} =a_{j} =1\) and \(a_{k} =0\) for all other components, we get \(v_{i}^{T} v_{j} =0\). This shows that \(\{v_{1} ,\cdots ,v_{n}\}\) is an orthogonal set. Together with the earlier result, we see that the columns of \(A\) are orthonormal. Therefore, \(A\) is an orthogonal matrix.
For the other three options:
It follows that \(A\) is invertible. In fact, \(A^{-1} =A^{T}\) and \(A\) has full rank.
\(A=-I\) is another orthogonal matrix. So this is a counter-example for option-(B).
Finally, an orthogonal matrix has complex eigenvalues whose magnitude is always \(1\). If the eigenvalues are real, then it is indeed true that they can only be \(1\) or \(-1\). An example of an orthogonal matrix that doesn’t have real eigenvalues is the following rotation matrix:
\[ A=\frac{1}{\sqrt{2}}\begin{bmatrix} 1 & -1\\ 1 & 1 \end{bmatrix} \]
This is a counter-example for option-(C).