A square root of a 3×3 matrix

Posted: December 6, 2009 in Elementary Algebra; Problems & Solutions, Linear Algebra
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Finding a square root of a 3 \times 3 upper triangular matrix A=[a_{ij}] is not hard.  Assuming that the entries on the diagonal of A are real and positive, define the 3 \times 3 matrix B=[b_{ij}] as follows. For i=1,2,3, let b_{ii}=\sqrt{a_{ii}}. Also define b_{21}=b_{31}=b_{32}=0. Let

\displaystyle b_{12}=\frac{a_{12}}{b_{11} + b_{22}}, \ b_{23}=\frac{a_{23}}{b_{22}+b_{33}}

and

\displaystyle b_{13}=\frac{a_{13} - b_{12}b_{23}}{b_{11}+b_{33}}.

See that B^2=A. Of course, this will also work for matrices with complex entries if you choose a square root for each a_{ii} and if the denominators in b_{12}, \ b_{23} and b_{13} are non-zero. Since every square matrix with complex entries is similar to an upper triangular matrix, you can use the above to find a square root of an arbitrary 3 \times 3 matrix.

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