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$DeclareMathOperator{rank}{rank}rank(A) = rank(B)$, Prove there exist $U, V$ invertible matrices such that:...
$begingroup$
$DeclareMathOperator{rank}{rank}$$DeclareMathOperator{Mat}{Mat}$Given two matrices $A, B in Mat_{m times n}$ , as $rank(A) = rank(B)$.
Prove there exist two invertible matrices: $$U in Mat_{m times m}, V in Mat_{n times n}$$
such that: $$A = UBV$$
My attempt:
this question essentially is to prove that multiplying a matrix of the left side is equivalent to be preforming operations on the rows, and multiplying a matrix to the right side is equivalent to be preforming operations on the columns.
I don't know how to prove this - so I tried using Linear maps and to prove that using linear mapps, which was so effective - as this does not "proves" that for every $A, B$ with an equal rank, there exist $U, V$ so that $A = UBV$.
linear-algebra matrices matrix-rank
asked Jan 14 at 10:11
JnevenJneven
867322
$endgroup$
add a comment |
$begingroup$
$DeclareMathOperator{rank}{rank}$$DeclareMathOperator{Mat}{Mat}$Given two matrices $A, B in Mat_{m times n}$ , as $rank(A) = rank(B)$.
Prove there exist two invertible matrices: $$U in Mat_{m times m}, V in Mat_{n times n}$$
such that: $$A = UBV$$
My attempt:
this question essentially is to prove that multiplying a matrix of the left side is equivalent to be preforming operations on the rows, and multiplying a matrix to the right side is equivalent to be preforming operations on the columns.
I don't know how to prove this - so I tried using Linear maps and to prove that using linear mapps, which was so effective - as this does not "proves" that for every $A, B$ with an equal rank, there exist $U, V$ so that $A = UBV$.
linear-algebra matrices matrix-rank
asked Jan 14 at 10:11
JnevenJneven
867322
$endgroup$
add a comment |
$begingroup$
$DeclareMathOperator{rank}{rank}$$DeclareMathOperator{Mat}{Mat}$Given two matrices $A, B in Mat_{m times n}$ , as $rank(A) = rank(B)$.
Prove there exist two invertible matrices: $$U in Mat_{m times m}, V in Mat_{n times n}$$
such that: $$A = UBV$$
My attempt:
this question essentially is to prove that multiplying a matrix of the left side is equivalent to be preforming operations on the rows, and multiplying a matrix to the right side is equivalent to be preforming operations on the columns.
I don't know how to prove this - so I tried using Linear maps and to prove that using linear mapps, which was so effective - as this does not "proves" that for every $A, B$ with an equal rank, there exist $U, V$ so that $A = UBV$.
linear-algebra matrices matrix-rank
asked Jan 14 at 10:11
JnevenJneven
867322
$endgroup$
$DeclareMathOperator{rank}{rank}$$DeclareMathOperator{Mat}{Mat}$Given two matrices $A, B in Mat_{m times n}$ , as $rank(A) = rank(B)$.
Prove there exist two invertible matrices: $$U in Mat_{m times m}, V in Mat_{n times n}$$
such that: $$A = UBV$$
My attempt:
this question essentially is to prove that multiplying a matrix of the left side is equivalent to be preforming operations on the rows, and multiplying a matrix to the right side is equivalent to be preforming operations on the columns.
I don't know how to prove this - so I tried using Linear maps and to prove that using linear mapps, which was so effective - as this does not "proves" that for every $A, B$ with an equal rank, there exist $U, V$ so that $A = UBV$.
linear-algebra matrices matrix-rank
linear-algebra matrices matrix-rank
asked Jan 14 at 10:11
JnevenJneven
867322
asked Jan 14 at 10:11
JnevenJneven
867322
edited Jan 14 at 18:58
Jneven
asked Jan 14 at 10:11
JnevenJneven
867322
asked Jan 14 at 10:11
JnevenJneven
867322
asked Jan 14 at 10:11
JnevenJneven
867322
867322
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Linear maps make this easier (in my opinion).
As $operatorname{rank}(A)=operatorname{rank}(B)$, the "column" subspaces $A(mathbb{R}^n)$ and $B(mathbb{R}^n)$ have the same dimension. Let $U$ be any linear isomorphism from $A(mathbb{R}^n)$ to $B(mathbb{R}^n)$. Extend $U$ to a linear isomorphism of $mathbb{R}^m$.
Now let $v_1,ldots,v_kinmathbb{R}^n$ such that $Av_1,ldots,Av_k$ form a basis for $A(mathbb{R}^n)$. Then $UAv_1,ldots,UAv_k$ form a basis for $B(mathbb{R}^n)$. Choose $w_1,ldots,w_kinmathbb{R}^n$ such that $Bw_i=UAv_i$.
Both $ker(A)$ and $ker(B)$ have dimension $n-dim(rank(A))=n-k$. Let $v_{k+1},ldots,v_n$ be a basis for $ker(A)$, and $w_{k+1},ldots,w_n$ be a basis for $ker(B)$. Then $left{v_1,ldots,v_nright}$ and $left{w_1,ldots,w_nright}$ are bases of $mathbb{R}^n$. Define a linear isomorphism $V:mathbb{R}^ntomathbb{R}^n$ on the basis as $Av_i=w_i$. Then $A=U^{-1}BV$.
answered Jan 14 at 14:02
QuestionerQuestioner
564322
$endgroup$
add a comment |
$begingroup$
If $r$ is the common rank, pushing just a little bit more the RREF you get that there are invertibile $U_{1}, V_{1}$ such that
$$
U_{1} A V_{1}
=
C_{r}
=
begin{bmatrix}
1 \
& 1\
&&&ddots\
&&&&1\
&&&&&0\
&&&&&&ddots\
&&&&&&&0& dots\
end{bmatrix},
$$
where there are $r$ ones in $C_{r}$, and unnamed entries are zero.
Similarly, there are invertibile $U_{2}, V_{2}$ such that
$$
U_{2} B V_{2}
=
C_{r}.
$$
Now the two expressions are equal, so that...
answered Jan 14 at 10:31
Andreas CarantiAndreas Caranti
56.6k34395
$endgroup$
add a comment |
$begingroup$
Hint: prove that any matrix $A$ of rank $r$ can be transformed to
$$
A=Ubegin{bmatrix}I_r & 0\0 & 0end{bmatrix}V
$$
by invertible $U$, $V$. (One way to do it quickly is SVD).
answered Jan 14 at 10:26
A.Γ.A.Γ.
22.7k32656
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Linear maps make this easier (in my opinion).
As $operatorname{rank}(A)=operatorname{rank}(B)$, the "column" subspaces $A(mathbb{R}^n)$ and $B(mathbb{R}^n)$ have the same dimension. Let $U$ be any linear isomorphism from $A(mathbb{R}^n)$ to $B(mathbb{R}^n)$. Extend $U$ to a linear isomorphism of $mathbb{R}^m$.
Now let $v_1,ldots,v_kinmathbb{R}^n$ such that $Av_1,ldots,Av_k$ form a basis for $A(mathbb{R}^n)$. Then $UAv_1,ldots,UAv_k$ form a basis for $B(mathbb{R}^n)$. Choose $w_1,ldots,w_kinmathbb{R}^n$ such that $Bw_i=UAv_i$.
Both $ker(A)$ and $ker(B)$ have dimension $n-dim(rank(A))=n-k$. Let $v_{k+1},ldots,v_n$ be a basis for $ker(A)$, and $w_{k+1},ldots,w_n$ be a basis for $ker(B)$. Then $left{v_1,ldots,v_nright}$ and $left{w_1,ldots,w_nright}$ are bases of $mathbb{R}^n$. Define a linear isomorphism $V:mathbb{R}^ntomathbb{R}^n$ on the basis as $Av_i=w_i$. Then $A=U^{-1}BV$.
answered Jan 14 at 14:02
QuestionerQuestioner
564322
$endgroup$
add a comment |
$begingroup$
Linear maps make this easier (in my opinion).
As $operatorname{rank}(A)=operatorname{rank}(B)$, the "column" subspaces $A(mathbb{R}^n)$ and $B(mathbb{R}^n)$ have the same dimension. Let $U$ be any linear isomorphism from $A(mathbb{R}^n)$ to $B(mathbb{R}^n)$. Extend $U$ to a linear isomorphism of $mathbb{R}^m$.
Now let $v_1,ldots,v_kinmathbb{R}^n$ such that $Av_1,ldots,Av_k$ form a basis for $A(mathbb{R}^n)$. Then $UAv_1,ldots,UAv_k$ form a basis for $B(mathbb{R}^n)$. Choose $w_1,ldots,w_kinmathbb{R}^n$ such that $Bw_i=UAv_i$.
Both $ker(A)$ and $ker(B)$ have dimension $n-dim(rank(A))=n-k$. Let $v_{k+1},ldots,v_n$ be a basis for $ker(A)$, and $w_{k+1},ldots,w_n$ be a basis for $ker(B)$. Then $left{v_1,ldots,v_nright}$ and $left{w_1,ldots,w_nright}$ are bases of $mathbb{R}^n$. Define a linear isomorphism $V:mathbb{R}^ntomathbb{R}^n$ on the basis as $Av_i=w_i$. Then $A=U^{-1}BV$.
answered Jan 14 at 14:02
QuestionerQuestioner
564322
$endgroup$
add a comment |
$begingroup$
Linear maps make this easier (in my opinion).
As $operatorname{rank}(A)=operatorname{rank}(B)$, the "column" subspaces $A(mathbb{R}^n)$ and $B(mathbb{R}^n)$ have the same dimension. Let $U$ be any linear isomorphism from $A(mathbb{R}^n)$ to $B(mathbb{R}^n)$. Extend $U$ to a linear isomorphism of $mathbb{R}^m$.
Now let $v_1,ldots,v_kinmathbb{R}^n$ such that $Av_1,ldots,Av_k$ form a basis for $A(mathbb{R}^n)$. Then $UAv_1,ldots,UAv_k$ form a basis for $B(mathbb{R}^n)$. Choose $w_1,ldots,w_kinmathbb{R}^n$ such that $Bw_i=UAv_i$.
Both $ker(A)$ and $ker(B)$ have dimension $n-dim(rank(A))=n-k$. Let $v_{k+1},ldots,v_n$ be a basis for $ker(A)$, and $w_{k+1},ldots,w_n$ be a basis for $ker(B)$. Then $left{v_1,ldots,v_nright}$ and $left{w_1,ldots,w_nright}$ are bases of $mathbb{R}^n$. Define a linear isomorphism $V:mathbb{R}^ntomathbb{R}^n$ on the basis as $Av_i=w_i$. Then $A=U^{-1}BV$.
answered Jan 14 at 14:02
QuestionerQuestioner
564322
$endgroup$
Linear maps make this easier (in my opinion).
As $operatorname{rank}(A)=operatorname{rank}(B)$, the "column" subspaces $A(mathbb{R}^n)$ and $B(mathbb{R}^n)$ have the same dimension. Let $U$ be any linear isomorphism from $A(mathbb{R}^n)$ to $B(mathbb{R}^n)$. Extend $U$ to a linear isomorphism of $mathbb{R}^m$.
Now let $v_1,ldots,v_kinmathbb{R}^n$ such that $Av_1,ldots,Av_k$ form a basis for $A(mathbb{R}^n)$. Then $UAv_1,ldots,UAv_k$ form a basis for $B(mathbb{R}^n)$. Choose $w_1,ldots,w_kinmathbb{R}^n$ such that $Bw_i=UAv_i$.
Both $ker(A)$ and $ker(B)$ have dimension $n-dim(rank(A))=n-k$. Let $v_{k+1},ldots,v_n$ be a basis for $ker(A)$, and $w_{k+1},ldots,w_n$ be a basis for $ker(B)$. Then $left{v_1,ldots,v_nright}$ and $left{w_1,ldots,w_nright}$ are bases of $mathbb{R}^n$. Define a linear isomorphism $V:mathbb{R}^ntomathbb{R}^n$ on the basis as $Av_i=w_i$. Then $A=U^{-1}BV$.
answered Jan 14 at 14:02
QuestionerQuestioner
564322
edited Jan 15 at 16:13
answered Jan 14 at 14:02
QuestionerQuestioner
564322
answered Jan 14 at 14:02
QuestionerQuestioner
564322
answered Jan 14 at 14:02
QuestionerQuestioner
564322
564322
add a comment |
add a comment |
$begingroup$
If $r$ is the common rank, pushing just a little bit more the RREF you get that there are invertibile $U_{1}, V_{1}$ such that
$$
U_{1} A V_{1}
=
C_{r}
=
begin{bmatrix}
1 \
& 1\
&&&ddots\
&&&&1\
&&&&&0\
&&&&&&ddots\
&&&&&&&0& dots\
end{bmatrix},
$$
where there are $r$ ones in $C_{r}$, and unnamed entries are zero.
Similarly, there are invertibile $U_{2}, V_{2}$ such that
$$
U_{2} B V_{2}
=
C_{r}.
$$
Now the two expressions are equal, so that...
answered Jan 14 at 10:31
Andreas CarantiAndreas Caranti
56.6k34395
$endgroup$
add a comment |
$begingroup$
If $r$ is the common rank, pushing just a little bit more the RREF you get that there are invertibile $U_{1}, V_{1}$ such that
$$
U_{1} A V_{1}
=
C_{r}
=
begin{bmatrix}
1 \
& 1\
&&&ddots\
&&&&1\
&&&&&0\
&&&&&&ddots\
&&&&&&&0& dots\
end{bmatrix},
$$
where there are $r$ ones in $C_{r}$, and unnamed entries are zero.
Similarly, there are invertibile $U_{2}, V_{2}$ such that
$$
U_{2} B V_{2}
=
C_{r}.
$$
Now the two expressions are equal, so that...
answered Jan 14 at 10:31
Andreas CarantiAndreas Caranti
56.6k34395
$endgroup$
add a comment |
$begingroup$
If $r$ is the common rank, pushing just a little bit more the RREF you get that there are invertibile $U_{1}, V_{1}$ such that
$$
U_{1} A V_{1}
=
C_{r}
=
begin{bmatrix}
1 \
& 1\
&&&ddots\
&&&&1\
&&&&&0\
&&&&&&ddots\
&&&&&&&0& dots\
end{bmatrix},
$$
where there are $r$ ones in $C_{r}$, and unnamed entries are zero.
Similarly, there are invertibile $U_{2}, V_{2}$ such that
$$
U_{2} B V_{2}
=
C_{r}.
$$
Now the two expressions are equal, so that...
answered Jan 14 at 10:31
Andreas CarantiAndreas Caranti
56.6k34395
$endgroup$
If $r$ is the common rank, pushing just a little bit more the RREF you get that there are invertibile $U_{1}, V_{1}$ such that
$$
U_{1} A V_{1}
=
C_{r}
=
begin{bmatrix}
1 \
& 1\
&&&ddots\
&&&&1\
&&&&&0\
&&&&&&ddots\
&&&&&&&0& dots\
end{bmatrix},
$$
where there are $r$ ones in $C_{r}$, and unnamed entries are zero.
Similarly, there are invertibile $U_{2}, V_{2}$ such that
$$
U_{2} B V_{2}
=
C_{r}.
$$
Now the two expressions are equal, so that...
answered Jan 14 at 10:31
Andreas CarantiAndreas Caranti
56.6k34395
answered Jan 14 at 10:31
Andreas CarantiAndreas Caranti
56.6k34395
answered Jan 14 at 10:31
Andreas CarantiAndreas Caranti
56.6k34395
answered Jan 14 at 10:31
Andreas CarantiAndreas Caranti
56.6k34395
56.6k34395
add a comment |
add a comment |
$begingroup$
Hint: prove that any matrix $A$ of rank $r$ can be transformed to
$$
A=Ubegin{bmatrix}I_r & 0\0 & 0end{bmatrix}V
$$
by invertible $U$, $V$. (One way to do it quickly is SVD).
answered Jan 14 at 10:26
A.Γ.A.Γ.
22.7k32656
$endgroup$
add a comment |
$begingroup$
Hint: prove that any matrix $A$ of rank $r$ can be transformed to
$$
A=Ubegin{bmatrix}I_r & 0\0 & 0end{bmatrix}V
$$
by invertible $U$, $V$. (One way to do it quickly is SVD).
answered Jan 14 at 10:26
A.Γ.A.Γ.
22.7k32656
$endgroup$
add a comment |
$begingroup$
Hint: prove that any matrix $A$ of rank $r$ can be transformed to
$$
A=Ubegin{bmatrix}I_r & 0\0 & 0end{bmatrix}V
$$
by invertible $U$, $V$. (One way to do it quickly is SVD).
answered Jan 14 at 10:26
A.Γ.A.Γ.
22.7k32656
$endgroup$
Hint: prove that any matrix $A$ of rank $r$ can be transformed to
$$
A=Ubegin{bmatrix}I_r & 0\0 & 0end{bmatrix}V
$$
by invertible $U$, $V$. (One way to do it quickly is SVD).
answered Jan 14 at 10:26
A.Γ.A.Γ.
22.7k32656
answered Jan 14 at 10:26
A.Γ.A.Γ.
22.7k32656
answered Jan 14 at 10:26
A.Γ.A.Γ.
22.7k32656
answered Jan 14 at 10:26
A.Γ.A.Γ.
22.7k32656
22.7k32656
add a comment |
add a comment |
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