Square of a diagonal matrix with real entries Is there a $3\times3 $ matrix with real entries which when multiplied with itself gives the matrix
 \begin{pmatrix}   1&    0&    0\\
    0  & -1 &   0\\
    0&    0  &  -1\end{pmatrix}
Can such a matrix be found.
My Attempt:
I tried finding the square root which came out to be \begin{pmatrix}   1&    0&    0\\
    0  & i &   0\\
    0&    0  &  i\end{pmatrix}
But I think we are not required to find the square root.Moreover this matrix does not have all real entries.
 A: EDIT: What follows is my answer before the OP fixed a typo by which the element of the third row, first column of the product was $\neq 0$.
My answer to the edited question is below.
ORIGINAL ANSWER:
No, there isn't. 
Take a look at the element in the third row, first column. If there were such a diagonal matrix, that element would be zero, since the product of diagonal matrices is a diagonal matrix.
EDIT: I know this is not part of the question but the OP asked me in the comments.
Is there a non diagonal matrix that multiplied by itself gives a diagonal matrix? 
Yes. Example:
$$
        \begin{pmatrix}
        0 & 0 \\
        1 & 0 \\
        \end{pmatrix}
$$
ANSWER TO THE EDITED QUESTION: No, there isn't a real diagonal matrix that multiplied by itself gives the matrix that you show because the diagonal elements of such matrix to the square are equal to the (respective) diagonal elements of the product matrix.
A: Hint:
Using block multiplication, you see one  can't find a diagonal matrix satisfying this condition. However I think you didn't formulate the question in a proper way.
You can find a square matrix satisfying this condition: this amounts to finding a $2\times2$ matrix $\;A=\begin{pmatrix}a&b\\c&d\end{pmatrix}$ such that
$$A^2=\begin{pmatrix}-1&0\\0&-1\end{pmatrix}.$$
This shouldn't be too hard. Knowing the matrix definition of complex numbers may help.
