Any vector in the Null space of a matrix A is orthogonal to any vector in the Row space of A












0












$begingroup$


Let $A$ be a $m times n$ matrix thus $A$ can be represented by the linear transformation $T: mathbb{R}^n to mathbb{R}^m$



Let $v in R(A) implies v in mathbb{R}^n$



If $w in N(A) implies Tw=0$
I want to prove that the inner product $langle v,wrangle=0$ but I don't know how to proceed






linear-algebra functional-analysis proof-writing linear-transformations







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$endgroup$












  • $begingroup$
    Hint: Write out in sum form what $A cdot X = 0$ means e.g. like $a_{11} x_1 + ... + a_{1n} x_n = 0$. This is an inner product that is equal to $0$.
    $endgroup$
    – Story123
    Dec 24 '18 at 1:35


















0












$begingroup$


Let $A$ be a $m times n$ matrix thus $A$ can be represented by the linear transformation $T: mathbb{R}^n to mathbb{R}^m$



Let $v in R(A) implies v in mathbb{R}^n$



If $w in N(A) implies Tw=0$
I want to prove that the inner product $langle v,wrangle=0$ but I don't know how to proceed






linear-algebra functional-analysis proof-writing linear-transformations







share|cite|improve this question











$endgroup$












  • $begingroup$
    Hint: Write out in sum form what $A cdot X = 0$ means e.g. like $a_{11} x_1 + ... + a_{1n} x_n = 0$. This is an inner product that is equal to $0$.
    $endgroup$
    – Story123
    Dec 24 '18 at 1:35
















0












0








0


0



$begingroup$


Let $A$ be a $m times n$ matrix thus $A$ can be represented by the linear transformation $T: mathbb{R}^n to mathbb{R}^m$



Let $v in R(A) implies v in mathbb{R}^n$



If $w in N(A) implies Tw=0$
I want to prove that the inner product $langle v,wrangle=0$ but I don't know how to proceed






linear-algebra functional-analysis proof-writing linear-transformations







share|cite|improve this question











$endgroup$




Let $A$ be a $m times n$ matrix thus $A$ can be represented by the linear transformation $T: mathbb{R}^n to mathbb{R}^m$



Let $v in R(A) implies v in mathbb{R}^n$



If $w in N(A) implies Tw=0$
I want to prove that the inner product $langle v,wrangle=0$ but I don't know how to proceed






linear-algebra functional-analysis proof-writing linear-transformations




linear-algebra functional-analysis proof-writing linear-transformations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 24 '18 at 2:02









Santana Afton

2,7192629




2,7192629










asked Dec 24 '18 at 1:32









Dreamer123Dreamer123

32329




32329












  • $begingroup$
    Hint: Write out in sum form what $A cdot X = 0$ means e.g. like $a_{11} x_1 + ... + a_{1n} x_n = 0$. This is an inner product that is equal to $0$.
    $endgroup$
    – Story123
    Dec 24 '18 at 1:35




















  • $begingroup$
    Hint: Write out in sum form what $A cdot X = 0$ means e.g. like $a_{11} x_1 + ... + a_{1n} x_n = 0$. This is an inner product that is equal to $0$.
    $endgroup$
    – Story123
    Dec 24 '18 at 1:35


















$begingroup$
Hint: Write out in sum form what $A cdot X = 0$ means e.g. like $a_{11} x_1 + ... + a_{1n} x_n = 0$. This is an inner product that is equal to $0$.
$endgroup$
– Story123
Dec 24 '18 at 1:35






$begingroup$
Hint: Write out in sum form what $A cdot X = 0$ means e.g. like $a_{11} x_1 + ... + a_{1n} x_n = 0$. This is an inner product that is equal to $0$.
$endgroup$
– Story123
Dec 24 '18 at 1:35












1 Answer
1






active

oldest

votes


















1












$begingroup$

As you know when we multiply a matrix by a vector we find the dot product of each row of the matrix and the given vector.



Since $T(w)= Aw$ and $w$ is in the null space of $T$, we have $Aw=0.$



Now in order to find $Aw$ you need to find the dot product of each row of $A$ and $w$.



Since $Aw=0$, each element $Aw$ is zero, so the dot product of each row of $A$ and $w$ must be zero.



That implies each vector in the row space is orthogonal to $w$ as well,because vectors in row space are simply linear combinations of row vectors of $A$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    yes, that is correct
    $endgroup$
    – Mohammad Riazi-Kermani
    Dec 24 '18 at 4:35










  • $begingroup$
    @ Ted Shifrin Done, thanks for the comment.
    $endgroup$
    – Mohammad Riazi-Kermani
    Dec 24 '18 at 5:05











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$begingroup$

As you know when we multiply a matrix by a vector we find the dot product of each row of the matrix and the given vector.



Since $T(w)= Aw$ and $w$ is in the null space of $T$, we have $Aw=0.$



Now in order to find $Aw$ you need to find the dot product of each row of $A$ and $w$.



Since $Aw=0$, each element $Aw$ is zero, so the dot product of each row of $A$ and $w$ must be zero.



That implies each vector in the row space is orthogonal to $w$ as well,because vectors in row space are simply linear combinations of row vectors of $A$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    yes, that is correct
    $endgroup$
    – Mohammad Riazi-Kermani
    Dec 24 '18 at 4:35










  • $begingroup$
    @ Ted Shifrin Done, thanks for the comment.
    $endgroup$
    – Mohammad Riazi-Kermani
    Dec 24 '18 at 5:05
















1












$begingroup$

As you know when we multiply a matrix by a vector we find the dot product of each row of the matrix and the given vector.



Since $T(w)= Aw$ and $w$ is in the null space of $T$, we have $Aw=0.$



Now in order to find $Aw$ you need to find the dot product of each row of $A$ and $w$.



Since $Aw=0$, each element $Aw$ is zero, so the dot product of each row of $A$ and $w$ must be zero.



That implies each vector in the row space is orthogonal to $w$ as well,because vectors in row space are simply linear combinations of row vectors of $A$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    yes, that is correct
    $endgroup$
    – Mohammad Riazi-Kermani
    Dec 24 '18 at 4:35










  • $begingroup$
    @ Ted Shifrin Done, thanks for the comment.
    $endgroup$
    – Mohammad Riazi-Kermani
    Dec 24 '18 at 5:05














1












1








1





$begingroup$

As you know when we multiply a matrix by a vector we find the dot product of each row of the matrix and the given vector.



Since $T(w)= Aw$ and $w$ is in the null space of $T$, we have $Aw=0.$



Now in order to find $Aw$ you need to find the dot product of each row of $A$ and $w$.



Since $Aw=0$, each element $Aw$ is zero, so the dot product of each row of $A$ and $w$ must be zero.



That implies each vector in the row space is orthogonal to $w$ as well,because vectors in row space are simply linear combinations of row vectors of $A$






share|cite|improve this answer











$endgroup$



As you know when we multiply a matrix by a vector we find the dot product of each row of the matrix and the given vector.



Since $T(w)= Aw$ and $w$ is in the null space of $T$, we have $Aw=0.$



Now in order to find $Aw$ you need to find the dot product of each row of $A$ and $w$.



Since $Aw=0$, each element $Aw$ is zero, so the dot product of each row of $A$ and $w$ must be zero.



That implies each vector in the row space is orthogonal to $w$ as well,because vectors in row space are simply linear combinations of row vectors of $A$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 24 '18 at 5:03

























answered Dec 24 '18 at 2:17









Mohammad Riazi-KermaniMohammad Riazi-Kermani

41.6k42061




41.6k42061












  • $begingroup$
    yes, that is correct
    $endgroup$
    – Mohammad Riazi-Kermani
    Dec 24 '18 at 4:35










  • $begingroup$
    @ Ted Shifrin Done, thanks for the comment.
    $endgroup$
    – Mohammad Riazi-Kermani
    Dec 24 '18 at 5:05


















  • $begingroup$
    yes, that is correct
    $endgroup$
    – Mohammad Riazi-Kermani
    Dec 24 '18 at 4:35










  • $begingroup$
    @ Ted Shifrin Done, thanks for the comment.
    $endgroup$
    – Mohammad Riazi-Kermani
    Dec 24 '18 at 5:05
















$begingroup$
yes, that is correct
$endgroup$
– Mohammad Riazi-Kermani
Dec 24 '18 at 4:35




$begingroup$
yes, that is correct
$endgroup$
– Mohammad Riazi-Kermani
Dec 24 '18 at 4:35












$begingroup$
@ Ted Shifrin Done, thanks for the comment.
$endgroup$
– Mohammad Riazi-Kermani
Dec 24 '18 at 5:05




$begingroup$
@ Ted Shifrin Done, thanks for the comment.
$endgroup$
– Mohammad Riazi-Kermani
Dec 24 '18 at 5:05


















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