## A square root of a 3 by 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

$\displaystyle b_{12}=\frac{a_{12}}{b_{11} + b_{22}}, \ b_{23}=\frac{a_{23}}{b_{22}+b_{33}}, \ b_{13}=\frac{a_{13} - b_{12}b_{23}}{b_{11}+b_{33}}.$

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.