On a special kind of $6$-dimensional vector subspace of $mathbb C^9$
$begingroup$
Let $V subseteq mathbb C^9$ be a vector subspace of dimension $6$. Suppose that there exists $A,B in M_{3 times 6} (mathbb C)$ such that
$V={(x_1,...,x_9)in mathbb C^9 : (x_3,x_6,x_9)^T = A (x_1,x_2,x_4,x_5,x_7,x_8)^T }={(x_1,...,x_9)in mathbb C^9 : (x_3,x_6,x_9)^T = B(x_1,x_2,x_4,x_5,x_7,x_8)^T }$
; then is it true that $A=B$ ?
linear-algebra matrices
asked Dec 13 '18 at 8:38
user521337user521337
1,1551417
$endgroup$
add a comment |
$begingroup$
Let $V subseteq mathbb C^9$ be a vector subspace of dimension $6$. Suppose that there exists $A,B in M_{3 times 6} (mathbb C)$ such that
$V={(x_1,...,x_9)in mathbb C^9 : (x_3,x_6,x_9)^T = A (x_1,x_2,x_4,x_5,x_7,x_8)^T }={(x_1,...,x_9)in mathbb C^9 : (x_3,x_6,x_9)^T = B(x_1,x_2,x_4,x_5,x_7,x_8)^T }$
; then is it true that $A=B$ ?
linear-algebra matrices
asked Dec 13 '18 at 8:38
user521337user521337
1,1551417
$endgroup$
add a comment |
$begingroup$
Let $V subseteq mathbb C^9$ be a vector subspace of dimension $6$. Suppose that there exists $A,B in M_{3 times 6} (mathbb C)$ such that
$V={(x_1,...,x_9)in mathbb C^9 : (x_3,x_6,x_9)^T = A (x_1,x_2,x_4,x_5,x_7,x_8)^T }={(x_1,...,x_9)in mathbb C^9 : (x_3,x_6,x_9)^T = B(x_1,x_2,x_4,x_5,x_7,x_8)^T }$
; then is it true that $A=B$ ?
linear-algebra matrices
asked Dec 13 '18 at 8:38
user521337user521337
1,1551417
$endgroup$
Let $V subseteq mathbb C^9$ be a vector subspace of dimension $6$. Suppose that there exists $A,B in M_{3 times 6} (mathbb C)$ such that
$V={(x_1,...,x_9)in mathbb C^9 : (x_3,x_6,x_9)^T = A (x_1,x_2,x_4,x_5,x_7,x_8)^T }={(x_1,...,x_9)in mathbb C^9 : (x_3,x_6,x_9)^T = B(x_1,x_2,x_4,x_5,x_7,x_8)^T }$
; then is it true that $A=B$ ?
linear-algebra matrices
linear-algebra matrices
asked Dec 13 '18 at 8:38
user521337user521337
1,1551417
asked Dec 13 '18 at 8:38
user521337user521337
1,1551417
asked Dec 13 '18 at 8:38
user521337user521337
1,1551417
asked Dec 13 '18 at 8:38
user521337user521337
1,1551417
asked Dec 13 '18 at 8:38
user521337user521337
1,1551417
1,1551417
add a comment |
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
Yes, this is true. Up to the numbering of coordinates, you just write $V={(y,Ay):yinmathbb C^6}$ and $V={(y,By):yinmathbb C^6}$. Now if you insert the basis vectors $e_1,dots,e_6$, you readily see that the vectors $(e_1,Ae_1),dots,(e_6,Ae_6)$ are linearly indent. Since $dim(V)=6$, they have to be a basis of $V$. But this implies that any element $vin V$ is determined by the coordinates $v_1,dots,v_6$, so given $yinmathbb C^6$, there is at most one $zinmathbb C^3$ such that $(y,z)in V$, and this implies $A=B$.
answered Dec 13 '18 at 10:46
Andreas CapAndreas Cap
11.2k923
$endgroup$
$begingroup$
thanks ... could you also please take a look at the related question math.stackexchange.com/questions/3037774/… ?
$endgroup$
– user521337
Dec 13 '18 at 23:58
add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
Yes, this is true. Up to the numbering of coordinates, you just write $V={(y,Ay):yinmathbb C^6}$ and $V={(y,By):yinmathbb C^6}$. Now if you insert the basis vectors $e_1,dots,e_6$, you readily see that the vectors $(e_1,Ae_1),dots,(e_6,Ae_6)$ are linearly indent. Since $dim(V)=6$, they have to be a basis of $V$. But this implies that any element $vin V$ is determined by the coordinates $v_1,dots,v_6$, so given $yinmathbb C^6$, there is at most one $zinmathbb C^3$ such that $(y,z)in V$, and this implies $A=B$.
answered Dec 13 '18 at 10:46
Andreas CapAndreas Cap
11.2k923
$endgroup$
$begingroup$
thanks ... could you also please take a look at the related question math.stackexchange.com/questions/3037774/… ?
$endgroup$
– user521337
Dec 13 '18 at 23:58
add a comment |
$begingroup$
Yes, this is true. Up to the numbering of coordinates, you just write $V={(y,Ay):yinmathbb C^6}$ and $V={(y,By):yinmathbb C^6}$. Now if you insert the basis vectors $e_1,dots,e_6$, you readily see that the vectors $(e_1,Ae_1),dots,(e_6,Ae_6)$ are linearly indent. Since $dim(V)=6$, they have to be a basis of $V$. But this implies that any element $vin V$ is determined by the coordinates $v_1,dots,v_6$, so given $yinmathbb C^6$, there is at most one $zinmathbb C^3$ such that $(y,z)in V$, and this implies $A=B$.
answered Dec 13 '18 at 10:46
Andreas CapAndreas Cap
11.2k923
$endgroup$
$begingroup$
thanks ... could you also please take a look at the related question math.stackexchange.com/questions/3037774/… ?
$endgroup$
– user521337
Dec 13 '18 at 23:58
add a comment |
$begingroup$
Yes, this is true. Up to the numbering of coordinates, you just write $V={(y,Ay):yinmathbb C^6}$ and $V={(y,By):yinmathbb C^6}$. Now if you insert the basis vectors $e_1,dots,e_6$, you readily see that the vectors $(e_1,Ae_1),dots,(e_6,Ae_6)$ are linearly indent. Since $dim(V)=6$, they have to be a basis of $V$. But this implies that any element $vin V$ is determined by the coordinates $v_1,dots,v_6$, so given $yinmathbb C^6$, there is at most one $zinmathbb C^3$ such that $(y,z)in V$, and this implies $A=B$.
answered Dec 13 '18 at 10:46
Andreas CapAndreas Cap
11.2k923
$endgroup$
Yes, this is true. Up to the numbering of coordinates, you just write $V={(y,Ay):yinmathbb C^6}$ and $V={(y,By):yinmathbb C^6}$. Now if you insert the basis vectors $e_1,dots,e_6$, you readily see that the vectors $(e_1,Ae_1),dots,(e_6,Ae_6)$ are linearly indent. Since $dim(V)=6$, they have to be a basis of $V$. But this implies that any element $vin V$ is determined by the coordinates $v_1,dots,v_6$, so given $yinmathbb C^6$, there is at most one $zinmathbb C^3$ such that $(y,z)in V$, and this implies $A=B$.
answered Dec 13 '18 at 10:46
Andreas CapAndreas Cap
11.2k923
answered Dec 13 '18 at 10:46
Andreas CapAndreas Cap
11.2k923
answered Dec 13 '18 at 10:46
Andreas CapAndreas Cap
11.2k923
answered Dec 13 '18 at 10:46
Andreas CapAndreas Cap
11.2k923
11.2k923
$begingroup$
thanks ... could you also please take a look at the related question math.stackexchange.com/questions/3037774/… ?
$endgroup$
– user521337
Dec 13 '18 at 23:58
add a comment |
$begingroup$
thanks ... could you also please take a look at the related question math.stackexchange.com/questions/3037774/… ?
$endgroup$
– user521337
Dec 13 '18 at 23:58
$begingroup$
thanks ... could you also please take a look at the related question math.stackexchange.com/questions/3037774/… ?
$endgroup$
– user521337
Dec 13 '18 at 23:58
$begingroup$
thanks ... could you also please take a look at the related question math.stackexchange.com/questions/3037774/… ?
$endgroup$
– user521337
Dec 13 '18 at 23:58
add a comment |
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