Inverse of $A$ as a linear combination of $A$
$begingroup$
Can the matrix $A^{-1}$ be written as a linear combination of $A$ (as given below), where
$$A=begin{bmatrix}
a& b& c\
0 &a& d\
0 &0& a
end{bmatrix}$$
such that $a neq 0$?
So obviously $det(A)=a^3 neq 0$ so $A$ is invertible. Now the dimension of the $3 times 3$ matrix space over $mathbb{R}$ is $27$. So ${I, A, A^2, A^3 ldots A^{27} }$ is linearly dependent. So there exists scalars $c_i$ not all zero such that
$$c_0I+c_1A+c_2A^2 + ldots +c_{27}A^{27}=0,.$$
Now if $c_0 neq 0$ we can write
$$A^{-1}=frac{1}{c_0}(c_1+c_2A + ldots +c_{27}A^{26}=0),.$$
I cant seem to find a way to guarantee that $c_0 neq 0$. Need help.
linear-algebra matrices determinant inverse cayley-hamilton
edited Dec 14 '18 at 11:57
Batominovski
33.2k33293
asked Jun 22 '15 at 5:58
MizMiz
1,597624
$endgroup$
add a comment |
$begingroup$
Can the matrix $A^{-1}$ be written as a linear combination of $A$ (as given below), where
$$A=begin{bmatrix}
a& b& c\
0 &a& d\
0 &0& a
end{bmatrix}$$
such that $a neq 0$?
So obviously $det(A)=a^3 neq 0$ so $A$ is invertible. Now the dimension of the $3 times 3$ matrix space over $mathbb{R}$ is $27$. So ${I, A, A^2, A^3 ldots A^{27} }$ is linearly dependent. So there exists scalars $c_i$ not all zero such that
$$c_0I+c_1A+c_2A^2 + ldots +c_{27}A^{27}=0,.$$
Now if $c_0 neq 0$ we can write
$$A^{-1}=frac{1}{c_0}(c_1+c_2A + ldots +c_{27}A^{26}=0),.$$
I cant seem to find a way to guarantee that $c_0 neq 0$. Need help.
linear-algebra matrices determinant inverse cayley-hamilton
edited Dec 14 '18 at 11:57
Batominovski
33.2k33293
asked Jun 22 '15 at 5:58
MizMiz
1,597624
$endgroup$
2
$begingroup$
HINT: Cayley Hamilton.
$endgroup$
– b00n heT
Jun 22 '15 at 6:02
add a comment |
$begingroup$
Can the matrix $A^{-1}$ be written as a linear combination of $A$ (as given below), where
$$A=begin{bmatrix}
a& b& c\
0 &a& d\
0 &0& a
end{bmatrix}$$
such that $a neq 0$?
So obviously $det(A)=a^3 neq 0$ so $A$ is invertible. Now the dimension of the $3 times 3$ matrix space over $mathbb{R}$ is $27$. So ${I, A, A^2, A^3 ldots A^{27} }$ is linearly dependent. So there exists scalars $c_i$ not all zero such that
$$c_0I+c_1A+c_2A^2 + ldots +c_{27}A^{27}=0,.$$
Now if $c_0 neq 0$ we can write
$$A^{-1}=frac{1}{c_0}(c_1+c_2A + ldots +c_{27}A^{26}=0),.$$
I cant seem to find a way to guarantee that $c_0 neq 0$. Need help.
linear-algebra matrices determinant inverse cayley-hamilton
edited Dec 14 '18 at 11:57
Batominovski
33.2k33293
asked Jun 22 '15 at 5:58
MizMiz
1,597624
$endgroup$
Can the matrix $A^{-1}$ be written as a linear combination of $A$ (as given below), where
$$A=begin{bmatrix}
a& b& c\
0 &a& d\
0 &0& a
end{bmatrix}$$
such that $a neq 0$?
So obviously $det(A)=a^3 neq 0$ so $A$ is invertible. Now the dimension of the $3 times 3$ matrix space over $mathbb{R}$ is $27$. So ${I, A, A^2, A^3 ldots A^{27} }$ is linearly dependent. So there exists scalars $c_i$ not all zero such that
$$c_0I+c_1A+c_2A^2 + ldots +c_{27}A^{27}=0,.$$
Now if $c_0 neq 0$ we can write
$$A^{-1}=frac{1}{c_0}(c_1+c_2A + ldots +c_{27}A^{26}=0),.$$
I cant seem to find a way to guarantee that $c_0 neq 0$. Need help.
linear-algebra matrices determinant inverse cayley-hamilton
linear-algebra matrices determinant inverse cayley-hamilton
edited Dec 14 '18 at 11:57
Batominovski
33.2k33293
asked Jun 22 '15 at 5:58
MizMiz
1,597624
edited Dec 14 '18 at 11:57
Batominovski
33.2k33293
asked Jun 22 '15 at 5:58
MizMiz
1,597624
edited Dec 14 '18 at 11:57
Batominovski
33.2k33293
edited Dec 14 '18 at 11:57
Batominovski
33.2k33293
edited Dec 14 '18 at 11:57
Batominovski
33.2k33293
33.2k33293
asked Jun 22 '15 at 5:58
MizMiz
1,597624
asked Jun 22 '15 at 5:58
MizMiz
1,597624
asked Jun 22 '15 at 5:58
MizMiz
1,597624
1,597624
2
$begingroup$
HINT: Cayley Hamilton.
$endgroup$
– b00n heT
Jun 22 '15 at 6:02
add a comment |
2
$begingroup$
HINT: Cayley Hamilton.
$endgroup$
– b00n heT
Jun 22 '15 at 6:02
2
2
$begingroup$
HINT: Cayley Hamilton.
$endgroup$
– b00n heT
Jun 22 '15 at 6:02
$begingroup$
HINT: Cayley Hamilton.
$endgroup$
– b00n heT
Jun 22 '15 at 6:02
add a comment |
1 Answer
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$begingroup$
Well, $(A-aI)^3=0$, right? Hence, $$A^3-3aA^2+3a^2A-a^3I=0,,$$ or $$A^{-1}=a^{-3}A^2-3a^{-2}A+3a^{-1}I,.$$
answered Jun 22 '15 at 6:02
BatominovskiBatominovski
33.2k33293
$endgroup$
add a comment |
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$begingroup$
Well, $(A-aI)^3=0$, right? Hence, $$A^3-3aA^2+3a^2A-a^3I=0,,$$ or $$A^{-1}=a^{-3}A^2-3a^{-2}A+3a^{-1}I,.$$
answered Jun 22 '15 at 6:02
BatominovskiBatominovski
33.2k33293
$endgroup$
add a comment |
$begingroup$
Well, $(A-aI)^3=0$, right? Hence, $$A^3-3aA^2+3a^2A-a^3I=0,,$$ or $$A^{-1}=a^{-3}A^2-3a^{-2}A+3a^{-1}I,.$$
answered Jun 22 '15 at 6:02
BatominovskiBatominovski
33.2k33293
$endgroup$
add a comment |
$begingroup$
Well, $(A-aI)^3=0$, right? Hence, $$A^3-3aA^2+3a^2A-a^3I=0,,$$ or $$A^{-1}=a^{-3}A^2-3a^{-2}A+3a^{-1}I,.$$
answered Jun 22 '15 at 6:02
BatominovskiBatominovski
33.2k33293
$endgroup$
Well, $(A-aI)^3=0$, right? Hence, $$A^3-3aA^2+3a^2A-a^3I=0,,$$ or $$A^{-1}=a^{-3}A^2-3a^{-2}A+3a^{-1}I,.$$
answered Jun 22 '15 at 6:02
BatominovskiBatominovski
33.2k33293
edited Dec 14 '18 at 11:58
answered Jun 22 '15 at 6:02
BatominovskiBatominovski
33.2k33293
answered Jun 22 '15 at 6:02
BatominovskiBatominovski
33.2k33293
answered Jun 22 '15 at 6:02
BatominovskiBatominovski
33.2k33293
33.2k33293
add a comment |
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HINT: Cayley Hamilton.
$endgroup$
– b00n heT
Jun 22 '15 at 6:02