Question-90

Let \(\displaystyle A\) be a real idempotent matrix of order \(\displaystyle n\). Then which of the following are true?

• \(\displaystyle I-A\) is also an idempotent matrix.
• \(\displaystyle A\) is always invertible
• \(\displaystyle A\) has \(\displaystyle n\) real eigenvalues (including multiplicity)
• All non-zero vectors in the column space of \(\displaystyle A\) are eigenvectors of \(\displaystyle A\)
• The rank of \(\displaystyle A\) is equal to its trace

Answer

• \(\displaystyle I-A\) is also an idempotent matrix.
• \(\displaystyle A\) is always invertible
• \(\displaystyle A\) has \(\displaystyle n\) real eigenvalues (including multiplicity)
• All non-zero vectors in the column space of \(\displaystyle A\) are eigenvectors of \(\displaystyle A\)
• The rank of \(\displaystyle A\) is equal to its trace

Solution

Option-1

We have:

\[ \begin{equation*} \begin{aligned} ( I-A)^{2} & =I^{2} +A^{2} -2IA\\ & =I+A-2A\\ & =I-A \end{aligned} \end{equation*} \]

Option-2

This is not true. A simple counter-example: \(\displaystyle \begin{bmatrix} 0 & 0\\ 1 & 1 \end{bmatrix}\)

Option-3

If \(\displaystyle ( \lambda ,v)\) is an eigenpair of \(\displaystyle A\), we have:

\[ \begin{equation*} \begin{aligned} Av & =\lambda v \end{aligned} ,\ \ \begin{aligned} A^{2} v & =\lambda ^{2} v\\ Av & =\lambda ^{2} v \end{aligned} \end{equation*} \]

From this:

\[ \begin{equation*} \lambda ^{2} v=\lambda v\Longrightarrow \lambda ( \lambda -1) v=0 \end{equation*} \]

It follows that \(\displaystyle \lambda =0,1\). It is known that every matrix has \(\displaystyle n\) complex eigenvalues. The above argument shows that every eigenvalue of an idempotent has to be either \(\displaystyle 1\) or \(\displaystyle 0\). Therefore, \(\displaystyle A\) has \(\displaystyle n\) real eigenvalues.

Option-4

Let \(\displaystyle C( A) =\text{span}\{v_{1} ,\cdots ,v_{r}\}\) be the column space of \(\displaystyle A\) whose rank is \(\displaystyle r\). For each \(\displaystyle v_{i}\), there is some \(\displaystyle u_{i} \in \mathbb{R}^{n}\) such that \(\displaystyle v_{i} =Au_{i}\). This is from the definition of the column space. With this:

\[ \begin{equation*} \begin{aligned} Av_{i} & =A^{2} u_{i}\\ & =Au_{i}\\ & =v_{i} \end{aligned} \end{equation*} \]

We see that \(\displaystyle ( 1,v_{i})\) is an eigenpair of \(\displaystyle A\). Therefore, every eigenvector in the column space of \(\displaystyle A\) is an eigenvector with eigenvalue \(\displaystyle 1\).

Option-5

Here is an outline of the proof. In the previous option, we have shown that \(\displaystyle A\) (of rank \(\displaystyle r\)) has \(\displaystyle r\) linearly independent eigenvectors with eigenvalue \(\displaystyle 1\). With some more work, we can show that the characteristic polynomial of \(\displaystyle A\) exactly factors as \(\displaystyle ( \lambda -1)^{r} \lambda ^{n-r}\). The sum of the eigenvalues is the trace which is \(\displaystyle r\).