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$Ker(A)=0 Leftrightarrow rank(A)=n$

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0

1

$begingroup$

Let $A: ntimes n$, I want to prove that if $N(A)=0$ if, and only if, $A$ has $n$ linearly independent columns.

Important: You cannot use $dim(N(A)) + dim(R(A)) = n$.

I thought of doing so, as $N(A)=0Rightarrow Ax = 0 Rightarrow x = 0$ so A is invertible, so there is a single inverse matrix $A^-1$ and the system $Ax=b Rightarrow x=A^-1 b$ soon, this system has a single solution so each column of A has a pivot so the $n$ columns are linearly independent and $rank(A)=n$.

I need to formalize better but I would like to know if the ideas are good.

Any tips on how to start the other implication?

linear-algebra matrices matrix-rank

share|cite|improve this question

asked Mar 21 at 19:04

Juliana de SouzaJuliana de Souza

726

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Hint: It is easier to prove that $operatornameKer(A)neq 0iff operatornamerank(A)<n.$
  $endgroup$
  – Thomas Andrews
  Mar 21 at 19:09
  

  
  
  
  

add a comment | 

0

1

$begingroup$

Let $A: ntimes n$, I want to prove that if $N(A)=0$ if, and only if, $A$ has $n$ linearly independent columns.

Important: You cannot use $dim(N(A)) + dim(R(A)) = n$.

I thought of doing so, as $N(A)=0Rightarrow Ax = 0 Rightarrow x = 0$ so A is invertible, so there is a single inverse matrix $A^-1$ and the system $Ax=b Rightarrow x=A^-1 b$ soon, this system has a single solution so each column of A has a pivot so the $n$ columns are linearly independent and $rank(A)=n$.

I need to formalize better but I would like to know if the ideas are good.

Any tips on how to start the other implication?

linear-algebra matrices matrix-rank

share|cite|improve this question

asked Mar 21 at 19:04

Juliana de SouzaJuliana de Souza

726

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Hint: It is easier to prove that $operatornameKer(A)neq 0iff operatornamerank(A)<n.$
  $endgroup$
  – Thomas Andrews
  Mar 21 at 19:09
  

  
  
  
  

add a comment | 

$begingroup$

Let $A: ntimes n$, I want to prove that if $N(A)=0$ if, and only if, $A$ has $n$ linearly independent columns.

Important: You cannot use $dim(N(A)) + dim(R(A)) = n$.

I thought of doing so, as $N(A)=0Rightarrow Ax = 0 Rightarrow x = 0$ so A is invertible, so there is a single inverse matrix $A^-1$ and the system $Ax=b Rightarrow x=A^-1 b$ soon, this system has a single solution so each column of A has a pivot so the $n$ columns are linearly independent and $rank(A)=n$.

I need to formalize better but I would like to know if the ideas are good.

Any tips on how to start the other implication?

linear-algebra matrices matrix-rank

share|cite|improve this question

asked Mar 21 at 19:04

Juliana de SouzaJuliana de Souza

726

$endgroup$

share|cite|improve this question

asked Mar 21 at 19:04

Juliana de SouzaJuliana de Souza

726

share|cite|improve this question

asked Mar 21 at 19:04

Juliana de SouzaJuliana de Souza

726

asked Mar 21 at 19:04

Juliana de SouzaJuliana de Souza

726

asked Mar 21 at 19:04

Juliana de SouzaJuliana de Souza

726

1 Answer
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active

oldest

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1

$begingroup$

Hint: If the columns of $A$ are, in order, $mathbf c_1,mathbf c_2,dots, mathbf c_n$ then:

$$Abeginpmatrixa_1\a_2\vdots\a_nendpmatrix = a_1mathbf c_1+a_2mathbf c_2+cdots + a_nmathbf c_n.$$

Now use the definition for "linearly dependent" to show the equivalent:
$$operatornameKer(A)neq 0iff operatornamerank(A)<n.$$

That is there is a nonzero vector $mathbf x$ such that $Amathbf x=0$ if and only if the columns are not linearly dependent.

share|cite|improve this answer

answered Mar 21 at 19:20

Thomas AndrewsThomas Andrews

131k12147298

$endgroup$

add a comment | 

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1

$begingroup$

Hint: If the columns of $A$ are, in order, $mathbf c_1,mathbf c_2,dots, mathbf c_n$ then:

$$Abeginpmatrixa_1\a_2\vdots\a_nendpmatrix = a_1mathbf c_1+a_2mathbf c_2+cdots + a_nmathbf c_n.$$

Now use the definition for "linearly dependent" to show the equivalent:
$$operatornameKer(A)neq 0iff operatornamerank(A)<n.$$

That is there is a nonzero vector $mathbf x$ such that $Amathbf x=0$ if and only if the columns are not linearly dependent.

share|cite|improve this answer

answered Mar 21 at 19:20

Thomas AndrewsThomas Andrews

131k12147298

$endgroup$

add a comment | 

1

$begingroup$

Hint: If the columns of $A$ are, in order, $mathbf c_1,mathbf c_2,dots, mathbf c_n$ then:

$$Abeginpmatrixa_1\a_2\vdots\a_nendpmatrix = a_1mathbf c_1+a_2mathbf c_2+cdots + a_nmathbf c_n.$$

Now use the definition for "linearly dependent" to show the equivalent:
$$operatornameKer(A)neq 0iff operatornamerank(A)<n.$$

That is there is a nonzero vector $mathbf x$ such that $Amathbf x=0$ if and only if the columns are not linearly dependent.

share|cite|improve this answer

answered Mar 21 at 19:20

Thomas AndrewsThomas Andrews

131k12147298

$endgroup$

add a comment | 

$begingroup$

Hint: If the columns of $A$ are, in order, $mathbf c_1,mathbf c_2,dots, mathbf c_n$ then:

$$Abeginpmatrixa_1\a_2\vdots\a_nendpmatrix = a_1mathbf c_1+a_2mathbf c_2+cdots + a_nmathbf c_n.$$

Now use the definition for "linearly dependent" to show the equivalent:
$$operatornameKer(A)neq 0iff operatornamerank(A)<n.$$

That is there is a nonzero vector $mathbf x$ such that $Amathbf x=0$ if and only if the columns are not linearly dependent.

share|cite|improve this answer

answered Mar 21 at 19:20

Thomas AndrewsThomas Andrews

131k12147298

$endgroup$

share|cite|improve this answer

answered Mar 21 at 19:20

Thomas AndrewsThomas Andrews

131k12147298

answered Mar 21 at 19:20

Thomas AndrewsThomas Andrews

131k12147298

answered Mar 21 at 19:20

Thomas AndrewsThomas Andrews

131k12147298