989

$f命题2:$任意方阵$A$均可分解为可逆阵$B$与对称阵$C$之积

证明：设$rleft( A ight) = r$，则存在可逆阵$P,Q$，使得
[A = Pleft( {egin{array}{*{20}{c}}
{{E_r}}&0\
0&0
end{array}} ight)Q]
从而可知
egin{align*}
A &= Pleft( {egin{array}{*{20}{c}}
{{E_r}}&0\
0&0
end{array}} ight)Q\&
= P{{Q'}^{ - 1}}Q'left( {egin{array}{*{20}{c}}
{{E_r}}&0\
0&0
end{array}} ight)Q
end{align*}
取$B = P{{Q'}^{ - 1}}$，$C = Q'left( {egin{array}{*{20}{c}}
{{E_r}}&0\
0&0
end{array}} ight)Q$，即证