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$Ker(A)=0 Leftrightarrow rank(A)=n$
Linearly independent rows and matricesMatrix columns and independenceWhat are the relations between the columns, rows and rank of a matrix A, in order to show that A has an inverse?Proofs for matrix rows and columns as basies, linearly independent, etcMaximize rank of Gramian kernel matrixLet $L$ a linear operator. Show that if $L$ is injective, then $det(L)neq 0$If $A$ is of rank $n$ then why is it non-singular?Why is dimension of solution space of homogeneous equations n-r?Pivots and linear independenceRelationship between rank and determinant rank
$begingroup$
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
asked Mar 21 at 19:04
Juliana de SouzaJuliana de Souza
726
$endgroup$
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
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
asked Mar 21 at 19:04
Juliana de SouzaJuliana de Souza
726
$endgroup$
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
linear-algebra matrices matrix-rank
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
asked Mar 21 at 19:04
Juliana de SouzaJuliana de Souza
726
asked Mar 21 at 19:04
Juliana de SouzaJuliana de Souza
726
726
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 |
1
$begingroup$
Hint: It is easier to prove that $operatornameKer(A)neq 0iff operatornamerank(A)<n.$
$endgroup$
– Thomas Andrews
Mar 21 at 19:09
1
1
$begingroup$
Hint: It is easier to prove that $operatornameKer(A)neq 0iff operatornamerank(A)<n.$
$endgroup$
– Thomas Andrews
Mar 21 at 19:09
$begingroup$
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 |
1 Answer
1
active
oldest
votes
$begingroup$
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.
answered Mar 21 at 19:20
Thomas AndrewsThomas Andrews
131k12147298
$endgroup$
add a comment |
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$begingroup$
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.
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.
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.
answered Mar 21 at 19:20
Thomas AndrewsThomas Andrews
131k12147298
$endgroup$
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.
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
answered Mar 21 at 19:20
Thomas AndrewsThomas Andrews
131k12147298
131k12147298
add a comment |
add a comment |
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Hint: It is easier to prove that $operatornameKer(A)neq 0iff operatornamerank(A)<n.$
$endgroup$
– Thomas Andrews
Mar 21 at 19:09