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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.

Maverick
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2 Answers2

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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.

  • Could we have a non-diagonal matrix which when multiplied to itself gives a diagonal matrix – Maverick May 21 '17 at 15:44
  • @PankajSinha sure. Adding an example to my answer. –  May 21 '17 at 15:52
  • But then can we have a real non-diagonal matrix which will give us the matrix in question when multiplied to itself – Maverick May 21 '17 at 16:55
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    for the record, that is not your original question. I assume by the title that the matrix you are squaring is diagonal and its square gives you the matrix that you show in the details. As per the new question: Why not? Just solve a non linear system for the 9 entries of the unknown matrix. Shouldn't be too bad, there are redundant equations and a few assumptions you can make. –  May 21 '17 at 17:02
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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.

Bernard
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    The solution isn't unique. For example $\begin{matrix}7 & 5 \ -10 & 7\end{matrix}$ is unlikely to be the answer that the question-setter expected you to find! – alephzero May 21 '17 at 21:25
  • I never said the solution was unique… – Bernard May 21 '17 at 21:30