Question

Let $a = \begin{bmatrix} \frac{1}{\sqrt{3}}\\[0.5em] \frac{-1}{\sqrt{2}}\\[0.5em] \frac{1}{\sqrt{6}}\\[0.5em] 0 \end{bmatrix}$ . Consider the following two statements.
P: The matrix $I_4-a{a}^T$ is invertible.
Q: The matrix $I_4-2a{a}^T$ is invertible.
Then, which one of the following holds?

1. P is false but Q is true
2. P is true but Q is false
3. Both P and Q are true
4. Both P and Q are false

Answer: P is false but Q is true

Solution :-

Given that $a={\left[\begin{array}{c}\frac{1}{\sqrt{3}}\\ \frac{-1}{\sqrt{2}}\\ \frac{1}{\sqrt{6}}\\ 0\end{array}\right]}_{4\times 1}\text{So,}{a}^T={\left[\begin{array}{cccc}\frac{1}{\sqrt{3}}& \frac{-1}{\sqrt{2}}& \frac{1}{\sqrt{6}}& 0\end{array}\right]}_{1\times 4}$

Now, we know that the only non-zero eigen value of $a{a}^T$ is trace( $a{a}^T$ ).
So, trace( $a{a}^T$ ) = $\frac{1}{3}+\frac{1}{2}+\frac{1}{6}+0=\frac{2+3+1}{6}=\frac{6}{6}=1.$

All eigen value of $a{a}^T$ is 0, 0, 0, 1.

Now, the eigen value of the matrix $I_4-a{a}^T$ is 1 - 0, 1 - 0, 1 - 0, 1 - 1 = 1, 1, 1, 0.
Here, 0 is an eigen value of the matrix $I_4-a{a}^T$ . So, the determinant of the matrix $I_4-a{a}^T$ is zero.

Hence, the matrix $I_4-a{a}^T$ is not invertible. statement P is false.

Now, the eigen value of the matrix $I_4-2a{a}^T$ is 1 - 2×0, 1 - 2×0, 1 - 2×0, 1 - 2×1 = 1, 1, 1, -1.
Here, 0 is not an eigen value of the matrix $I_4-2a{a}^T$ . So, the determinant of the matrix $I_4-a{a}^T$ is non-zero.

Hence, the matrix $I_4-a{a}^T$ is invertible. statement Q is true.

So, the correct answer is option-(A), $\to$ P is false but Q is true