Question-28
Let \(A=I_{n} +xx^{T}\), where \(I_{n}\) is the \(n\times n\) identity matrix and \(x\in \mathbb{R}^{n} ,x^{T} x=1\). Which of the following options is/are correct?
We see that \(xx^{T}\) has two eigenvalues, \(1\) and \(0\). \(x\) is an eigenvector with eigenvalue \(1\) and any vector orthogonal to \(x\) is an eigenvector with eigenvalue \(0\). The set of all vectors orthogonal to \(x\) is a subspace of dimension \(n-1\). To see why the dimension is \(n-1\), note that we can extend \(x\) to an orthonormal basis of \(\mathbb{R}^{n}\). Let us call this orthonormal basis \(\{x,x_{1} ,\cdots ,x_{n-1}\}\). Finally, note that \(1\) and \(0\) are the only eigenvalues for this matrix.
Next, notice that \(u\) is an eigenvector of a matrix \(B\) with eigenvalue \(\lambda\) if and only if \(u\) is an eigenvalue of \(B+cI\) with eigenvalue \(\lambda +c\). Using this, we see that \(x\) is an eigenvector of \(A=I+xx^{T}\) with eigenvalue \(2\) and \(x_{i}\) is an eigenvector of \(A=I+xx^{T}\) with eigenvalue \(1\). We can conclude that \(1\) and \(2\) are the only eigenvalues of \(A\).
Since \(A\) doesn’t have zero as an eigenvalue, \(Ax=0\Longrightarrow x=0\). It follows that \(A\) is invertible and full rank. The eigenvalues of \(A^{-1}\) are \(1\) and \(\frac{1}{2}\). So the last option is incorrect.