Question-61

Consider the following square matrix of order \(\displaystyle n\) that is given by:

\[ \begin{equation*} A=( x-1) I+vv^{T} \end{equation*} \]

Here, \(\displaystyle x\in ( 1-\epsilon ,1+\epsilon )\) for \(\displaystyle \epsilon \ll 1\), \(\displaystyle v=( 1,\cdots ,1)\) is a vector of ones and \(\displaystyle I\) is the identity matrix. Which of the following are true?

• If \(\displaystyle x=1\), the rank of \(\displaystyle A\) is \(\displaystyle 1\)
• If \(\displaystyle x=1\), the rank of \(\displaystyle A\) is equal to \(\displaystyle n\)
• If \(\displaystyle x\neq 1\), the rank of \(\displaystyle A\) is equal to \(\displaystyle 1\)
• If \(\displaystyle x\neq 1\), the rank of \(\displaystyle A\) is equal to \(\displaystyle n\)

Hint

Take a simple \(3 \times 3\) or \(4 \times 4\) matrix and perform row reduction. Generalise.

Answer

• If \(\displaystyle x=1\), the rank of \(\displaystyle A\) is \(\displaystyle 1\)
• If \(\displaystyle x=1\), the rank of \(\displaystyle A\) is equal to \(\displaystyle n\)
• If \(\displaystyle x\neq 1\), the rank of \(\displaystyle A\) is equal to \(\displaystyle 1\)
• If \(\displaystyle x\neq 1\), the rank of \(\displaystyle A\) is equal to \(\displaystyle n\)

Solution

Let us take a look at a simple case of \(\displaystyle n=4\). \(\displaystyle vv^{T}\) is just a matrix of ones.

\[ \begin{equation*} \begin{bmatrix} x-1 & 0 & 0 & 0\\ 0 & x-1 & 0 & 0\\ 0 & 0 & x-1 & 0\\ 0 & 0 & 0 & x-1 \end{bmatrix} +\begin{bmatrix} 1 & 1 & 1 & 1\\ 1 & 1 & 1 & 1\\ 1 & 1 & 1 & 1\\ 1 & 1 & 1 & 1 \end{bmatrix} =\begin{bmatrix} x & 1 & 1 & 1\\ 1 & x & 1 & 1\\ 1 & 1 & x & 1\\ 1 & 1 & 1 & x \end{bmatrix} \end{equation*} \]

If \(\displaystyle x=1\), this is clearly a matrix of rank \(\displaystyle 1\). So let us assume that \(\displaystyle x\neq 1\). We now proceed with row reduction:

Step-1

\[ \begin{equation*} \begin{bmatrix} x & 1 & 1 & 1\\ 1 & x & 1 & 1\\ 1 & 1 & x & 1\\ 1 & 1 & 1 & x \end{bmatrix}\xrightarrow{R_{4}\rightarrow R_{4} +R_{3} +R_{2} +R_{1}}\begin{bmatrix} x & 1 & 1 & 1\\ 1 & x & 1 & 1\\ 1 & 1 & x & 1\\ x+3 & x+3 & x+3 & x+3 \end{bmatrix} \end{equation*} \]

Step-2

Since \(\displaystyle x\in ( 1-\epsilon ,1+\epsilon )\) for a small \(\displaystyle \epsilon\), \(\displaystyle x+3\neq 0\). This allows us to do the following row operation:

\[ \begin{equation*} \begin{bmatrix} x & 1 & 1 & 1\\ 1 & x & 1 & 1\\ 1 & 1 & x & 1\\ x+3 & x+3 & x+3 & x+3 \end{bmatrix}\xrightarrow{R_{4}\rightarrow \frac{1}{x+3} R_{4}}\begin{bmatrix} x & 1 & 1 & 1\\ 1 & x & 1 & 1\\ 1 & 1 & x & 1\\ 1 & 1 & 1 & 1 \end{bmatrix} \end{equation*} \]

Step-3

We can use the last row to eliminate a lot of entries in the first three rows:

\[ \begin{equation*} \begin{bmatrix} x & 1 & 1 & 1\\ 1 & x & 1 & 1\\ 1 & 1 & x & 1\\ 1 & 1 & 1 & 1 \end{bmatrix}\rightarrow \begin{bmatrix} x-1 & 0 & 0 & 0\\ 0 & x-1 & 0 & 0\\ 0 & 0 & x-1 & 0\\ 1 & 1 & 1 & 1 \end{bmatrix} \end{equation*} \]

Step-4

Since \(\displaystyle x\neq 1\), we can divide by \(\displaystyle x-1\) in each of the first three rows:

\[ \begin{equation*} \begin{bmatrix} x-1 & 0 & 0 & 0\\ 0 & x-1 & 0 & 0\\ 0 & 0 & x-1 & 0\\ 1 & 1 & 1 & 1 \end{bmatrix}\rightarrow \begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 1 & 1 & 1 & 1 \end{bmatrix} \end{equation*} \]

After a few more operations, we get the identity matrix. This shows that \(\displaystyle A\) has rank \(\displaystyle 4\) when \(\displaystyle x\neq 1\). These steps readily generalise to any \(\displaystyle n\).