Question-50
Let \(\{x_{1} ,x_{2} ,x_{3} ,x_{4} ,x_{5}\}\) be a set of orthonormal vectors in \(\mathbb{R}^{10}\). Consider the matrix \(A=x_{1} x_{1}^{T} +\cdots +x_{5} x_{5}^{T}\). Which of the following statements is/are correct?
Let us extend \(\{x_{1} ,\cdots ,x_{5}\}\) to an orthonormal basis \(\{x_{1} ,\cdots ,x_{5} ,x_{6} ,\cdots ,x_{10}\}\) for \(\mathbb{R}^{10}\). Now, we can express \(A\) as:
\[ A=1\cdot x_{1} x_{1}^{T} +\cdots +1\cdot x_{5} x_{5}^{T} +0\cdot x_{6} x_{6}^{T} +\cdots +0\cdot x_{10} x_{10}^{T} \]
Setting \(Q=\begin{bmatrix} | & & |\\ x_{1} & \cdots & x_{10}\\ | & & | \end{bmatrix}\) and \(D=\begin{bmatrix} I & 0\\ 0 & 0 \end{bmatrix}\), where \(D\) is a block diagonal matrix with each entry being a square matrix of order \(5\), we see that \(A=QDQ^{T}\). Next, note that each \(x_{i}\) is an eigenvector of \(A\) because \(Ax_{i} =\lambda _{i} x_{i}\), where \(\lambda =1\) or \(\lambda =0\). Therefore, what we have is an eigen-decomposition of \(A\). In fact, we also see that \(A^{T} A=AA^{T} =A\). We can also conclude that \(A=QDA^{T}\) is the singular value decomposition of \(A\). In summary:
\(A\) has the same singular/eigenvalues which are either \(1\) or \(0\)
Since \(0\) is an eigenvalue, \(A\) is neither invertible nor is its determinant \(1\).
As an additional note, we see that \(A\) is actually a projection matrix. It projects points in \(\mathbb{R}^{10}\) onto the span of the vectors \(\{x_{1} ,\cdots ,x_{5}\}\).