Express $A^{-1}$ in the form $alpha I + beta A$.
$begingroup$
Let $A in Bbb {M_3} (Bbb R)$ whose eigen values are $1,1,3$. Then express $A^{-1}$ in the form $alpha I + beta A$, $alpha,beta in Bbb R$.
What I found is that $A^{-1} = frac {1} {3} (A^2 -5A+7I)$. How do I express it in the desired form? Please help me in this regard.
Thank you very much.
linear-algebra eigenvalues-eigenvectors
asked Dec 4 '18 at 19:01
Dbchatto67Dbchatto67
1,210218
$endgroup$
add a comment |
$begingroup$
Let $A in Bbb {M_3} (Bbb R)$ whose eigen values are $1,1,3$. Then express $A^{-1}$ in the form $alpha I + beta A$, $alpha,beta in Bbb R$.
What I found is that $A^{-1} = frac {1} {3} (A^2 -5A+7I)$. How do I express it in the desired form? Please help me in this regard.
Thank you very much.
linear-algebra eigenvalues-eigenvectors
asked Dec 4 '18 at 19:01
Dbchatto67Dbchatto67
1,210218
$endgroup$
$begingroup$
Did you mean "express $A^{-1}$ in the form $alpha I + beta A$"?
$endgroup$
– Omnomnomnom
Dec 4 '18 at 19:05
$begingroup$
Yes @Omnomnomnom.
$endgroup$
– Dbchatto67
Dec 4 '18 at 19:06
add a comment |
$begingroup$
Let $A in Bbb {M_3} (Bbb R)$ whose eigen values are $1,1,3$. Then express $A^{-1}$ in the form $alpha I + beta A$, $alpha,beta in Bbb R$.
What I found is that $A^{-1} = frac {1} {3} (A^2 -5A+7I)$. How do I express it in the desired form? Please help me in this regard.
Thank you very much.
linear-algebra eigenvalues-eigenvectors
asked Dec 4 '18 at 19:01
Dbchatto67Dbchatto67
1,210218
$endgroup$
Let $A in Bbb {M_3} (Bbb R)$ whose eigen values are $1,1,3$. Then express $A^{-1}$ in the form $alpha I + beta A$, $alpha,beta in Bbb R$.
What I found is that $A^{-1} = frac {1} {3} (A^2 -5A+7I)$. How do I express it in the desired form? Please help me in this regard.
Thank you very much.
linear-algebra eigenvalues-eigenvectors
linear-algebra eigenvalues-eigenvectors
asked Dec 4 '18 at 19:01
Dbchatto67Dbchatto67
1,210218
asked Dec 4 '18 at 19:01
Dbchatto67Dbchatto67
1,210218
edited Dec 4 '18 at 19:06
Dbchatto67
asked Dec 4 '18 at 19:01
Dbchatto67Dbchatto67
1,210218
asked Dec 4 '18 at 19:01
Dbchatto67Dbchatto67
1,210218
asked Dec 4 '18 at 19:01
Dbchatto67Dbchatto67
1,210218
1,210218
$begingroup$
Did you mean "express $A^{-1}$ in the form $alpha I + beta A$"?
$endgroup$
– Omnomnomnom
Dec 4 '18 at 19:05
$begingroup$
Yes @Omnomnomnom.
$endgroup$
– Dbchatto67
Dec 4 '18 at 19:06
add a comment |
$begingroup$
Did you mean "express $A^{-1}$ in the form $alpha I + beta A$"?
$endgroup$
– Omnomnomnom
Dec 4 '18 at 19:05
$begingroup$
Yes @Omnomnomnom.
$endgroup$
– Dbchatto67
Dec 4 '18 at 19:06
$begingroup$
Did you mean "express $A^{-1}$ in the form $alpha I + beta A$"?
$endgroup$
– Omnomnomnom
Dec 4 '18 at 19:05
$begingroup$
Did you mean "express $A^{-1}$ in the form $alpha I + beta A$"?
$endgroup$
– Omnomnomnom
Dec 4 '18 at 19:05
$begingroup$
Yes @Omnomnomnom.
$endgroup$
– Dbchatto67
Dec 4 '18 at 19:06
$begingroup$
Yes @Omnomnomnom.
$endgroup$
– Dbchatto67
Dec 4 '18 at 19:06
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Your formula (computed using the characteristic polynomial) is correct. If $A$ fails to be diagonalizable, then $A$ is non-derogatory and, as my post here explains, no further reduction will be possible. In particular, we can conclude that since ${I,A,A^2}$ is a linearly independent set, the set ${A^{-1},I,A}$ will also be linearly independent.
However, if $A$ is diagonalizable, then its minimal polynomial will be $(x-1)(x-3) = x^2 - 4x + 3$, which is to say that $A$ will satisfy
$$
A^2 - 4A + 3I = 0
$$
which we can rearrange to find that $A^{-1} = frac 13(-A + 4I)$.
answered Dec 4 '18 at 19:12
OmnomnomnomOmnomnomnom
128k791184
$endgroup$
$begingroup$
Very nice idea @Omnomnomnom. Thank you very much for your help.
$endgroup$
– Dbchatto67
Dec 4 '18 at 19:15
add a comment |
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1 Answer
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active
oldest
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1 Answer
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active
oldest
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$begingroup$
Your formula (computed using the characteristic polynomial) is correct. If $A$ fails to be diagonalizable, then $A$ is non-derogatory and, as my post here explains, no further reduction will be possible. In particular, we can conclude that since ${I,A,A^2}$ is a linearly independent set, the set ${A^{-1},I,A}$ will also be linearly independent.
However, if $A$ is diagonalizable, then its minimal polynomial will be $(x-1)(x-3) = x^2 - 4x + 3$, which is to say that $A$ will satisfy
$$
A^2 - 4A + 3I = 0
$$
which we can rearrange to find that $A^{-1} = frac 13(-A + 4I)$.
answered Dec 4 '18 at 19:12
OmnomnomnomOmnomnomnom
128k791184
$endgroup$
$begingroup$
Very nice idea @Omnomnomnom. Thank you very much for your help.
$endgroup$
– Dbchatto67
Dec 4 '18 at 19:15
add a comment |
$begingroup$
Your formula (computed using the characteristic polynomial) is correct. If $A$ fails to be diagonalizable, then $A$ is non-derogatory and, as my post here explains, no further reduction will be possible. In particular, we can conclude that since ${I,A,A^2}$ is a linearly independent set, the set ${A^{-1},I,A}$ will also be linearly independent.
However, if $A$ is diagonalizable, then its minimal polynomial will be $(x-1)(x-3) = x^2 - 4x + 3$, which is to say that $A$ will satisfy
$$
A^2 - 4A + 3I = 0
$$
which we can rearrange to find that $A^{-1} = frac 13(-A + 4I)$.
answered Dec 4 '18 at 19:12
OmnomnomnomOmnomnomnom
128k791184
$endgroup$
$begingroup$
Very nice idea @Omnomnomnom. Thank you very much for your help.
$endgroup$
– Dbchatto67
Dec 4 '18 at 19:15
add a comment |
$begingroup$
Your formula (computed using the characteristic polynomial) is correct. If $A$ fails to be diagonalizable, then $A$ is non-derogatory and, as my post here explains, no further reduction will be possible. In particular, we can conclude that since ${I,A,A^2}$ is a linearly independent set, the set ${A^{-1},I,A}$ will also be linearly independent.
However, if $A$ is diagonalizable, then its minimal polynomial will be $(x-1)(x-3) = x^2 - 4x + 3$, which is to say that $A$ will satisfy
$$
A^2 - 4A + 3I = 0
$$
which we can rearrange to find that $A^{-1} = frac 13(-A + 4I)$.
answered Dec 4 '18 at 19:12
OmnomnomnomOmnomnomnom
128k791184
$endgroup$
Your formula (computed using the characteristic polynomial) is correct. If $A$ fails to be diagonalizable, then $A$ is non-derogatory and, as my post here explains, no further reduction will be possible. In particular, we can conclude that since ${I,A,A^2}$ is a linearly independent set, the set ${A^{-1},I,A}$ will also be linearly independent.
However, if $A$ is diagonalizable, then its minimal polynomial will be $(x-1)(x-3) = x^2 - 4x + 3$, which is to say that $A$ will satisfy
$$
A^2 - 4A + 3I = 0
$$
which we can rearrange to find that $A^{-1} = frac 13(-A + 4I)$.
answered Dec 4 '18 at 19:12
OmnomnomnomOmnomnomnom
128k791184
answered Dec 4 '18 at 19:12
OmnomnomnomOmnomnomnom
128k791184
answered Dec 4 '18 at 19:12
OmnomnomnomOmnomnomnom
128k791184
answered Dec 4 '18 at 19:12
OmnomnomnomOmnomnomnom
128k791184
128k791184
$begingroup$
Very nice idea @Omnomnomnom. Thank you very much for your help.
$endgroup$
– Dbchatto67
Dec 4 '18 at 19:15
add a comment |
$begingroup$
Very nice idea @Omnomnomnom. Thank you very much for your help.
$endgroup$
– Dbchatto67
Dec 4 '18 at 19:15
$begingroup$
Very nice idea @Omnomnomnom. Thank you very much for your help.
$endgroup$
– Dbchatto67
Dec 4 '18 at 19:15
$begingroup$
Very nice idea @Omnomnomnom. Thank you very much for your help.
$endgroup$
– Dbchatto67
Dec 4 '18 at 19:15
add a comment |
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$begingroup$
Did you mean "express $A^{-1}$ in the form $alpha I + beta A$"?
$endgroup$
– Omnomnomnom
Dec 4 '18 at 19:05
$begingroup$
Yes @Omnomnomnom.
$endgroup$
– Dbchatto67
Dec 4 '18 at 19:06