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Intuition underlying a linear algebra result
Linear algebra question involving principal submatriceslinear algebra linear transformation eigenvector and eigenvalueslinear map questionlinear algebra : matrix decompositionLinear Algebra True/FlaseLinear Algebra Proof for matricesLinear Algebra basis. What does it mean?Linear Algebra Tensor ProofTrue or False Linear Algebra Matrix UnderstandingLinear Algebra Ordered Basis
$begingroup$
RESULT
For any $mtimes n$ matrix $textbfA$ and $mtimes p$ matrix $textbfB$, $mathcalC(B)subsetmathcalC(A)$ if and only if there exists an $ntimes p$ matrix $textbfF$ such that $textbfB = textbfAtextbfF$, where $mathcalC(X)$ represents the column space of $textbfX$.
MY QUESTION
I would like to know if someone could provide me an intuition underlying this matrix analysis result. Any help is appreciated. Thanks in advance.
linear-algebra matrices
$endgroup$
add a comment |
$begingroup$
RESULT
For any $mtimes n$ matrix $textbfA$ and $mtimes p$ matrix $textbfB$, $mathcalC(B)subsetmathcalC(A)$ if and only if there exists an $ntimes p$ matrix $textbfF$ such that $textbfB = textbfAtextbfF$, where $mathcalC(X)$ represents the column space of $textbfX$.
MY QUESTION
I would like to know if someone could provide me an intuition underlying this matrix analysis result. Any help is appreciated. Thanks in advance.
linear-algebra matrices
$endgroup$
add a comment |
$begingroup$
RESULT
For any $mtimes n$ matrix $textbfA$ and $mtimes p$ matrix $textbfB$, $mathcalC(B)subsetmathcalC(A)$ if and only if there exists an $ntimes p$ matrix $textbfF$ such that $textbfB = textbfAtextbfF$, where $mathcalC(X)$ represents the column space of $textbfX$.
MY QUESTION
I would like to know if someone could provide me an intuition underlying this matrix analysis result. Any help is appreciated. Thanks in advance.
linear-algebra matrices
$endgroup$
RESULT
For any $mtimes n$ matrix $textbfA$ and $mtimes p$ matrix $textbfB$, $mathcalC(B)subsetmathcalC(A)$ if and only if there exists an $ntimes p$ matrix $textbfF$ such that $textbfB = textbfAtextbfF$, where $mathcalC(X)$ represents the column space of $textbfX$.
MY QUESTION
I would like to know if someone could provide me an intuition underlying this matrix analysis result. Any help is appreciated. Thanks in advance.
linear-algebra matrices
linear-algebra matrices
asked Mar 21 at 23:26
user1337user1337
47210
47210
add a comment |
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1 Answer
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$begingroup$
The condition $cal C(B)subset cal B(A)$ means that every column of $B$ is a linear combination of columns of $A$. How many scalars do you need for each column of $B$? Well, as many as $A$'s column. More succinctly,
$B=AF$
for some $ntimes p$ matrix $F$. Why $p$ columns? That's because $B$ has $p$ columns.
$endgroup$
add a comment |
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1 Answer
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1 Answer
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active
oldest
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active
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active
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votes
$begingroup$
The condition $cal C(B)subset cal B(A)$ means that every column of $B$ is a linear combination of columns of $A$. How many scalars do you need for each column of $B$? Well, as many as $A$'s column. More succinctly,
$B=AF$
for some $ntimes p$ matrix $F$. Why $p$ columns? That's because $B$ has $p$ columns.
$endgroup$
add a comment |
$begingroup$
The condition $cal C(B)subset cal B(A)$ means that every column of $B$ is a linear combination of columns of $A$. How many scalars do you need for each column of $B$? Well, as many as $A$'s column. More succinctly,
$B=AF$
for some $ntimes p$ matrix $F$. Why $p$ columns? That's because $B$ has $p$ columns.
$endgroup$
add a comment |
$begingroup$
The condition $cal C(B)subset cal B(A)$ means that every column of $B$ is a linear combination of columns of $A$. How many scalars do you need for each column of $B$? Well, as many as $A$'s column. More succinctly,
$B=AF$
for some $ntimes p$ matrix $F$. Why $p$ columns? That's because $B$ has $p$ columns.
$endgroup$
The condition $cal C(B)subset cal B(A)$ means that every column of $B$ is a linear combination of columns of $A$. How many scalars do you need for each column of $B$? Well, as many as $A$'s column. More succinctly,
$B=AF$
for some $ntimes p$ matrix $F$. Why $p$ columns? That's because $B$ has $p$ columns.
answered Mar 21 at 23:32
chhrochhro
1,442311
1,442311
add a comment |
add a comment |
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