**Theorem:** Suppose the $n \times n$ matrix $A$ has $n$ linearly independent eigenvectors. If these eigenvectors are the columns of a matrix $S,$ then $S^{-1}AS$ is a diagonal matrix $\Lambda$. The eigenvalues of $A$ are on the diagonal of $\Lambda$.

$S^{-1}AS = \Lambda$ (A diagonal Matrix with diagonal values representing eigen values of A)$ = \begin{bmatrix} \lambda _{1} & & & & \\ & \lambda _{2} & & & \\ & & . & & \\ & & & . & \lambda _{n}\\ & & & & \end{bmatrix}$

Now if $A$ is diagonalizable, $S^{-1}$ must exist. What is $S?$ $S$ is a matrix whose columns are eigen-vectors of matrix $A.$

Now $S^{-1}$ would only exist if $S$ is invertible. And $S$ would be invertible if all rows and columns of $S$ are independent.

If we would have same eigen-vectors, then $S^{-1}$ won't exist and hence $A$ won't be diagonalizable.

Even if a matrix $A$ has same eigen values, it does not mean that it is not diagonizable. Take the trivial example of the Identity Matrix $I.$

$\begin{bmatrix} 1 & & \\ & 1& \\ & & 1 \end{bmatrix}$

The Eigen values are $1,1,1$ and Eigen vectors are $\begin{bmatrix} 1\\ 0\\ 0\end{bmatrix}$ , $\begin{bmatrix} 0\\ 1\\ 0\end{bmatrix}$, $\begin{bmatrix} 0\\ 0\\ 1\end{bmatrix}$

We form $S=\begin{bmatrix} 1 & 0 &0 \\ 0 & 1 &0 \\ 0 & 0& 1 \end{bmatrix}$ and this is invertible.

So, this makes $S^{-1} I S$ as $I =\Lambda$ which is in-fact a diagonal matrix.

So, even if we have same eigenvalues the matrix may or may not be diagonalizable. But yes, we need full $n$ set of linearly independent eigenvectors for this matrix $A$ of size $n \times n$ to be diagonalizable.

Now our problem says that we have a matrix $P$ whose only eigenvectors are multiples of $\begin{bmatrix} 1\\ 4 \end{bmatrix}$.

Means we have only $\begin{bmatrix} 1\\ 4 \end{bmatrix}$ as independent eigen vector. Surely, this matrix is not diagonalizable.

Since, eigen vectors are multiples of $\begin{bmatrix} 1\\ 4 \end{bmatrix}$, we have repeated eigen values.

Let us assume $\lambda _{1}$ and $\lambda _{2}$ are different eigen values.

Since, we have only eigen vectors multiple of $\begin{bmatrix} 1\\ 4 \end{bmatrix}$.

Let this vector be $x_1.$

So, $Px_1=\lambda _{1}x_1$

$\implies Px_1=\lambda _{2}x_1$

$\implies \lambda _{1}x_1=\lambda _{2}x_1$

Multiply by $x_1-1$ we get $\lambda _{1} =\lambda _{2}.$ Means, same Eigen vector gives same Eigen Value.

So yes, $P$ has repeated Eigen Values. (II) statement is true.

Now Statement (I) is not true. We cannot say this statement with exact surety.

Consider a matrix $A=\begin{bmatrix} 0 & 1\\ 0 & 0 \end{bmatrix}$

It's eigenvales are $\lambda _{1}=\lambda _{2}=0$.

All the eigenvectors of this $A$ are multiples of vector $(1,0)$

$\begin{bmatrix} 0 & 1\\ 0 & 0 \end{bmatrix}. x=\begin{bmatrix} 0\\ 0 \end{bmatrix}$

or $x=\begin{bmatrix} c\\ 0 \end{bmatrix}$.

This matrix surely is not diagonalizable but this matrix $A$ has determinant $= -1.$

Since determinant is not $0,$ so $A^{-1}$ exists.

Hence, (II) and (III) are surely valid under all cases of this question.

For (I) part, Matrix $P$ will not have an inverse when $\det(P)=0$ and this implies one of the eigenvalue of $P$ is zero. But in question, no where it is mentioned about what are the eigen values. So, **I is not necessarily true**.

Answer-(D) (**Remember options says "necessarily true"**)