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How to show that if $x^T x = 0$, then $x=0$ for any column vector $x$?
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$begingroup$
I did the proof and got it wrong, it'd be really helpful if someone could point me in the right direction or show me how to do it. The $T$ in $(x)^Tx=0$ means transpose; I wasn't sure if that was clear in the title.
linear-algebra vectors norm
edited 2 days ago
J. W. Tanner
3,0681320
asked 2 days ago
gmb12gmb12
31
New contributor
gmb12 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$
add a comment |
$begingroup$
I did the proof and got it wrong, it'd be really helpful if someone could point me in the right direction or show me how to do it. The $T$ in $(x)^Tx=0$ means transpose; I wasn't sure if that was clear in the title.
linear-algebra vectors norm
edited 2 days ago
J. W. Tanner
3,0681320
asked 2 days ago
gmb12gmb12
31
New contributor
gmb12 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
7
$begingroup$
$(x)^Tx$ is a sum of squares of entries of $x$; if any entry were non-zero the sum of squares would be positive
$endgroup$
– J. W. Tanner
2 days ago
3
$begingroup$
Another way to see it: ${bf x}^T{bf x}$ is the squared length of $bf x$; if the squared length is $0$, then ${bf x} = {bf 0}$.
$endgroup$
– Théophile
2 days ago
$begingroup$
Also, $x^Tx$ is simply the scalar product $<x,x> = sum_{i=1}^{n}x_{i}^2$ which is only zero in the case of $x = 0$
$endgroup$
– Victoria M
2 days ago
add a comment |
$begingroup$
I did the proof and got it wrong, it'd be really helpful if someone could point me in the right direction or show me how to do it. The $T$ in $(x)^Tx=0$ means transpose; I wasn't sure if that was clear in the title.
linear-algebra vectors norm
edited 2 days ago
J. W. Tanner
3,0681320
asked 2 days ago
gmb12gmb12
31
New contributor
gmb12 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I did the proof and got it wrong, it'd be really helpful if someone could point me in the right direction or show me how to do it. The $T$ in $(x)^Tx=0$ means transpose; I wasn't sure if that was clear in the title.
linear-algebra vectors norm
linear-algebra vectors norm
edited 2 days ago
J. W. Tanner
3,0681320
asked 2 days ago
gmb12gmb12
31
New contributor
gmb12 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 days ago
J. W. Tanner
3,0681320
asked 2 days ago
gmb12gmb12
31
New contributor
gmb12 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 days ago
J. W. Tanner
3,0681320
edited 2 days ago
J. W. Tanner
3,0681320
edited 2 days ago
J. W. Tanner
3,0681320
3,0681320
asked 2 days ago
gmb12gmb12
31
New contributor
gmb12 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
gmb12gmb12
31
asked 2 days ago
gmb12gmb12
31
31
New contributor
gmb12 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
gmb12 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
gmb12 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
7
$begingroup$
$(x)^Tx$ is a sum of squares of entries of $x$; if any entry were non-zero the sum of squares would be positive
$endgroup$
– J. W. Tanner
2 days ago
3
$begingroup$
Another way to see it: ${bf x}^T{bf x}$ is the squared length of $bf x$; if the squared length is $0$, then ${bf x} = {bf 0}$.
$endgroup$
– Théophile
2 days ago
$begingroup$
Also, $x^Tx$ is simply the scalar product $<x,x> = sum_{i=1}^{n}x_{i}^2$ which is only zero in the case of $x = 0$
$endgroup$
– Victoria M
2 days ago
add a comment |
7
$begingroup$
$(x)^Tx$ is a sum of squares of entries of $x$; if any entry were non-zero the sum of squares would be positive
$endgroup$
– J. W. Tanner
2 days ago
3
$begingroup$
Another way to see it: ${bf x}^T{bf x}$ is the squared length of $bf x$; if the squared length is $0$, then ${bf x} = {bf 0}$.
$endgroup$
– Théophile
2 days ago
$begingroup$
Also, $x^Tx$ is simply the scalar product $<x,x> = sum_{i=1}^{n}x_{i}^2$ which is only zero in the case of $x = 0$
$endgroup$
– Victoria M
2 days ago
7
7
$begingroup$
$(x)^Tx$ is a sum of squares of entries of $x$; if any entry were non-zero the sum of squares would be positive
$endgroup$
– J. W. Tanner
2 days ago
$begingroup$
$(x)^Tx$ is a sum of squares of entries of $x$; if any entry were non-zero the sum of squares would be positive
$endgroup$
– J. W. Tanner
2 days ago
3
3
$begingroup$
Another way to see it: ${bf x}^T{bf x}$ is the squared length of $bf x$; if the squared length is $0$, then ${bf x} = {bf 0}$.
$endgroup$
– Théophile
2 days ago
$begingroup$
Another way to see it: ${bf x}^T{bf x}$ is the squared length of $bf x$; if the squared length is $0$, then ${bf x} = {bf 0}$.
$endgroup$
– Théophile
2 days ago
$begingroup$
Also, $x^Tx$ is simply the scalar product $<x,x> = sum_{i=1}^{n}x_{i}^2$ which is only zero in the case of $x = 0$
$endgroup$
– Victoria M
2 days ago
$begingroup$
Also, $x^Tx$ is simply the scalar product $<x,x> = sum_{i=1}^{n}x_{i}^2$ which is only zero in the case of $x = 0$
$endgroup$
– Victoria M
2 days ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let $$x=left(
begin{matrix}
x_1 \
x_2 \
. \
.\
.\
x_n
end{matrix}
right).
$$
Then $x^Tx$ = $x_1^2+x_2^2+...+x_n^2$.
Since for all real $x_i$, $x_i^2 = 0$ if $x_i=0$ and $x_i^2>0$ otherwise,
$x^Tx=0$ implies $x_i=0$ for all $i$, i.e., $x=mathbf 0.$
answered 2 days ago
J. W. TannerJ. W. Tanner
3,0681320
$endgroup$
add a comment |
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1 Answer
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active
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1 Answer
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oldest
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active
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votes
$begingroup$
Let $$x=left(
begin{matrix}
x_1 \
x_2 \
. \
.\
.\
x_n
end{matrix}
right).
$$
Then $x^Tx$ = $x_1^2+x_2^2+...+x_n^2$.
Since for all real $x_i$, $x_i^2 = 0$ if $x_i=0$ and $x_i^2>0$ otherwise,
$x^Tx=0$ implies $x_i=0$ for all $i$, i.e., $x=mathbf 0.$
answered 2 days ago
J. W. TannerJ. W. Tanner
3,0681320
$endgroup$
add a comment |
$begingroup$
Let $$x=left(
begin{matrix}
x_1 \
x_2 \
. \
.\
.\
x_n
end{matrix}
right).
$$
Then $x^Tx$ = $x_1^2+x_2^2+...+x_n^2$.
Since for all real $x_i$, $x_i^2 = 0$ if $x_i=0$ and $x_i^2>0$ otherwise,
$x^Tx=0$ implies $x_i=0$ for all $i$, i.e., $x=mathbf 0.$
answered 2 days ago
J. W. TannerJ. W. Tanner
3,0681320
$endgroup$
add a comment |
$begingroup$
Let $$x=left(
begin{matrix}
x_1 \
x_2 \
. \
.\
.\
x_n
end{matrix}
right).
$$
Then $x^Tx$ = $x_1^2+x_2^2+...+x_n^2$.
Since for all real $x_i$, $x_i^2 = 0$ if $x_i=0$ and $x_i^2>0$ otherwise,
$x^Tx=0$ implies $x_i=0$ for all $i$, i.e., $x=mathbf 0.$
answered 2 days ago
J. W. TannerJ. W. Tanner
3,0681320
$endgroup$
Let $$x=left(
begin{matrix}
x_1 \
x_2 \
. \
.\
.\
x_n
end{matrix}
right).
$$
Then $x^Tx$ = $x_1^2+x_2^2+...+x_n^2$.
Since for all real $x_i$, $x_i^2 = 0$ if $x_i=0$ and $x_i^2>0$ otherwise,
$x^Tx=0$ implies $x_i=0$ for all $i$, i.e., $x=mathbf 0.$
answered 2 days ago
J. W. TannerJ. W. Tanner
3,0681320
answered 2 days ago
J. W. TannerJ. W. Tanner
3,0681320
answered 2 days ago
J. W. TannerJ. W. Tanner
3,0681320
answered 2 days ago
J. W. TannerJ. W. Tanner
3,0681320
3,0681320
add a comment |
add a comment |
gmb12 is a new contributor. Be nice, and check out our Code of Conduct.
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7
$begingroup$
$(x)^Tx$ is a sum of squares of entries of $x$; if any entry were non-zero the sum of squares would be positive
$endgroup$
– J. W. Tanner
2 days ago
3
$begingroup$
Another way to see it: ${bf x}^T{bf x}$ is the squared length of $bf x$; if the squared length is $0$, then ${bf x} = {bf 0}$.
$endgroup$
– Théophile
2 days ago
$begingroup$
Also, $x^Tx$ is simply the scalar product $<x,x> = sum_{i=1}^{n}x_{i}^2$ which is only zero in the case of $x = 0$
$endgroup$
– Victoria M
2 days ago