Mixing
If $A=A^2$ is then $A^T A = A$?
$begingroup$
I know that for a matrix $A$:
If $A^TA = A$ then $A=A^2$
but is it if and only if? I mean:
is this true that "If $A=A^2$ then $A^TA = A$"?
linear-algebra matrices symmetric-matrices transpose
edited Jan 2 at 18:53
greedoid
38.7k114797
asked Jan 2 at 18:27
PeymanPeyman
1089
$endgroup$
add a comment |
$begingroup$
I know that for a matrix $A$:
If $A^TA = A$ then $A=A^2$
but is it if and only if? I mean:
is this true that "If $A=A^2$ then $A^TA = A$"?
linear-algebra matrices symmetric-matrices transpose
edited Jan 2 at 18:53
greedoid
38.7k114797
asked Jan 2 at 18:27
PeymanPeyman
1089
$endgroup$
add a comment |
$begingroup$
I know that for a matrix $A$:
If $A^TA = A$ then $A=A^2$
but is it if and only if? I mean:
is this true that "If $A=A^2$ then $A^TA = A$"?
linear-algebra matrices symmetric-matrices transpose
edited Jan 2 at 18:53
greedoid
38.7k114797
asked Jan 2 at 18:27
PeymanPeyman
1089
$endgroup$
I know that for a matrix $A$:
If $A^TA = A$ then $A=A^2$
but is it if and only if? I mean:
is this true that "If $A=A^2$ then $A^TA = A$"?
linear-algebra matrices symmetric-matrices transpose
linear-algebra matrices symmetric-matrices transpose
edited Jan 2 at 18:53
greedoid
38.7k114797
asked Jan 2 at 18:27
PeymanPeyman
1089
edited Jan 2 at 18:53
greedoid
38.7k114797
asked Jan 2 at 18:27
PeymanPeyman
1089
edited Jan 2 at 18:53
greedoid
38.7k114797
edited Jan 2 at 18:53
greedoid
38.7k114797
edited Jan 2 at 18:53
greedoid
38.7k114797
38.7k114797
asked Jan 2 at 18:27
PeymanPeyman
1089
asked Jan 2 at 18:27
PeymanPeyman
1089
asked Jan 2 at 18:27
PeymanPeyman
1089
1089
add a comment |
add a comment |
3 Answers
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$begingroup$
The answer is no.
Consider $A = begin{bmatrix} 1 & -1 \ 0 & 0 end{bmatrix}$. We have
$$A^2 = begin{bmatrix} 1 & -1 \ 0 & 0 end{bmatrix}begin{bmatrix} 1 & -1 \ 0 & 0 end{bmatrix} = begin{bmatrix} 1 & -1 \ 0 & 0 end{bmatrix} = A$$
but $$A^TA = begin{bmatrix} 1 & 0 \ -1 & 0 end{bmatrix}begin{bmatrix} 1 & -1 \ 0 & 0 end{bmatrix} = begin{bmatrix} 1 & -1 \ -1 & 1 end{bmatrix} ne A$$
answered Jan 2 at 18:38
mechanodroidmechanodroid
27.1k62446
$endgroup$
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$begingroup$
What about of $A=begin{bmatrix} 1&1\0&0end{bmatrix}large{?}$
answered Jan 2 at 18:37
Ángel Mario GallegosÁngel Mario Gallegos
18.5k11230
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$begingroup$
Since $$det (A)^2 = det (A)det (A) = det (A^2)= det (A) implies det (A)in{0,1}$$
So if $det (A)= 1$ then exsist $A^{-1}$ so $A = I$ and the answer is yes.
If $det(A)=0$ then examples show that answer is negative.
answered Jan 2 at 18:44
greedoidgreedoid
38.7k114797
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The answer is no.
Consider $A = begin{bmatrix} 1 & -1 \ 0 & 0 end{bmatrix}$. We have
$$A^2 = begin{bmatrix} 1 & -1 \ 0 & 0 end{bmatrix}begin{bmatrix} 1 & -1 \ 0 & 0 end{bmatrix} = begin{bmatrix} 1 & -1 \ 0 & 0 end{bmatrix} = A$$
but $$A^TA = begin{bmatrix} 1 & 0 \ -1 & 0 end{bmatrix}begin{bmatrix} 1 & -1 \ 0 & 0 end{bmatrix} = begin{bmatrix} 1 & -1 \ -1 & 1 end{bmatrix} ne A$$
answered Jan 2 at 18:38
mechanodroidmechanodroid
27.1k62446
$endgroup$
add a comment |
$begingroup$
The answer is no.
Consider $A = begin{bmatrix} 1 & -1 \ 0 & 0 end{bmatrix}$. We have
$$A^2 = begin{bmatrix} 1 & -1 \ 0 & 0 end{bmatrix}begin{bmatrix} 1 & -1 \ 0 & 0 end{bmatrix} = begin{bmatrix} 1 & -1 \ 0 & 0 end{bmatrix} = A$$
but $$A^TA = begin{bmatrix} 1 & 0 \ -1 & 0 end{bmatrix}begin{bmatrix} 1 & -1 \ 0 & 0 end{bmatrix} = begin{bmatrix} 1 & -1 \ -1 & 1 end{bmatrix} ne A$$
answered Jan 2 at 18:38
mechanodroidmechanodroid
27.1k62446
$endgroup$
add a comment |
$begingroup$
The answer is no.
Consider $A = begin{bmatrix} 1 & -1 \ 0 & 0 end{bmatrix}$. We have
$$A^2 = begin{bmatrix} 1 & -1 \ 0 & 0 end{bmatrix}begin{bmatrix} 1 & -1 \ 0 & 0 end{bmatrix} = begin{bmatrix} 1 & -1 \ 0 & 0 end{bmatrix} = A$$
but $$A^TA = begin{bmatrix} 1 & 0 \ -1 & 0 end{bmatrix}begin{bmatrix} 1 & -1 \ 0 & 0 end{bmatrix} = begin{bmatrix} 1 & -1 \ -1 & 1 end{bmatrix} ne A$$
answered Jan 2 at 18:38
mechanodroidmechanodroid
27.1k62446
$endgroup$
The answer is no.
Consider $A = begin{bmatrix} 1 & -1 \ 0 & 0 end{bmatrix}$. We have
$$A^2 = begin{bmatrix} 1 & -1 \ 0 & 0 end{bmatrix}begin{bmatrix} 1 & -1 \ 0 & 0 end{bmatrix} = begin{bmatrix} 1 & -1 \ 0 & 0 end{bmatrix} = A$$
but $$A^TA = begin{bmatrix} 1 & 0 \ -1 & 0 end{bmatrix}begin{bmatrix} 1 & -1 \ 0 & 0 end{bmatrix} = begin{bmatrix} 1 & -1 \ -1 & 1 end{bmatrix} ne A$$
answered Jan 2 at 18:38
mechanodroidmechanodroid
27.1k62446
answered Jan 2 at 18:38
mechanodroidmechanodroid
27.1k62446
answered Jan 2 at 18:38
mechanodroidmechanodroid
27.1k62446
answered Jan 2 at 18:38
mechanodroidmechanodroid
27.1k62446
27.1k62446
add a comment |
add a comment |
$begingroup$
What about of $A=begin{bmatrix} 1&1\0&0end{bmatrix}large{?}$
answered Jan 2 at 18:37
Ángel Mario GallegosÁngel Mario Gallegos
18.5k11230
$endgroup$
add a comment |
$begingroup$
What about of $A=begin{bmatrix} 1&1\0&0end{bmatrix}large{?}$
answered Jan 2 at 18:37
Ángel Mario GallegosÁngel Mario Gallegos
18.5k11230
$endgroup$
add a comment |
$begingroup$
What about of $A=begin{bmatrix} 1&1\0&0end{bmatrix}large{?}$
answered Jan 2 at 18:37
Ángel Mario GallegosÁngel Mario Gallegos
18.5k11230
$endgroup$
What about of $A=begin{bmatrix} 1&1\0&0end{bmatrix}large{?}$
answered Jan 2 at 18:37
Ángel Mario GallegosÁngel Mario Gallegos
18.5k11230
answered Jan 2 at 18:37
Ángel Mario GallegosÁngel Mario Gallegos
18.5k11230
answered Jan 2 at 18:37
Ángel Mario GallegosÁngel Mario Gallegos
18.5k11230
answered Jan 2 at 18:37
Ángel Mario GallegosÁngel Mario Gallegos
18.5k11230
18.5k11230
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add a comment |
$begingroup$
Since $$det (A)^2 = det (A)det (A) = det (A^2)= det (A) implies det (A)in{0,1}$$
So if $det (A)= 1$ then exsist $A^{-1}$ so $A = I$ and the answer is yes.
If $det(A)=0$ then examples show that answer is negative.
answered Jan 2 at 18:44
greedoidgreedoid
38.7k114797
$endgroup$
add a comment |
$begingroup$
Since $$det (A)^2 = det (A)det (A) = det (A^2)= det (A) implies det (A)in{0,1}$$
So if $det (A)= 1$ then exsist $A^{-1}$ so $A = I$ and the answer is yes.
If $det(A)=0$ then examples show that answer is negative.
answered Jan 2 at 18:44
greedoidgreedoid
38.7k114797
$endgroup$
add a comment |
$begingroup$
Since $$det (A)^2 = det (A)det (A) = det (A^2)= det (A) implies det (A)in{0,1}$$
So if $det (A)= 1$ then exsist $A^{-1}$ so $A = I$ and the answer is yes.
If $det(A)=0$ then examples show that answer is negative.
answered Jan 2 at 18:44
greedoidgreedoid
38.7k114797
$endgroup$
Since $$det (A)^2 = det (A)det (A) = det (A^2)= det (A) implies det (A)in{0,1}$$
So if $det (A)= 1$ then exsist $A^{-1}$ so $A = I$ and the answer is yes.
If $det(A)=0$ then examples show that answer is negative.
answered Jan 2 at 18:44
greedoidgreedoid
38.7k114797
answered Jan 2 at 18:44
greedoidgreedoid
38.7k114797
answered Jan 2 at 18:44
greedoidgreedoid
38.7k114797
answered Jan 2 at 18:44
greedoidgreedoid
38.7k114797
38.7k114797
add a comment |
add a comment |
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