How to find square root of a matrix if its eigenvalues are same.

Question

I came across a question in Strang's book stated:

Since it is a matrix with same eigenvalues that is $0$ and eigenvector $[1\ 0]^t$

In second part, even with replacing diagonal entries that is $0$ to $4$ the eigenvalues remains equal that is $4$ and eigenvector $[1\ 0]^t$

Since this can't be diagonalized as $S^{-1}$ doesn't exist as matrix have dependent eigenvector

How to find its square root ?

Sorry for the formatting and thanks in advance.

Answer 1

If $S$ is a $2\times2$ matrix whose only eigenvalue is $0$, then $S$ is similar to a matrix of the form$$\begin{pmatrix}0&a\\0&0\end{pmatrix}.$$Therefore, $S^2$ is the null matrix.

On the other hand,$$\begin{pmatrix}2&a\\0&2\end{pmatrix}^2=\begin{pmatrix}4&4a\\0&4\end{pmatrix}.$$So, take $a=\frac14$.

Answer 2

Assume $S^2 = A$. We have $S^2 = A \ne 0$ but $S^4 = A^2 = 0$.

Therefore $S$ is nilpotent but $x^2$ does not annihilate $S$, which is impossible because the degree of the minimal polynomial of $S$ is $\le 2$.

Answer 3

The minimal polynomial of $A$ is $p(\lambda)=\lambda^2$. If $A=B^2$, then the minimal polynomial $q(\lambda)$ of $B$ must divide $\lambda^4$, and must be of order $\le 2$, which forces $B^2=0$, which is impossible because $B^2=A\ne 0$.