Column Space, Rank and Matrix Concatenation
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I have the following question:
Given Matrices $A$ and $B$, the following relation exists between their column spaces:
$$text{col}(B) subseteq text{col}(A)$$
Then, which of the following is true for Matrix $C=[A,,,,,B]$?
A) $text{rank}(C)=text{rank}(A)$
B) $text{rank}(C)=text{rank}(B)$
C) It is not possible to specify $text{rank}(C)$ in terms of $text{rank}(A)$ and $text{rank}(B)$
My guess, and it seems a reasonable one, would be alternative (A), but I don't know how to solve/express it mathematically.
Is my guess correct? Could you walk me through the steps to prove it?
Thanks in advance
linear-algebra matrices matrix-rank
asked Jan 4 at 22:16
bertozzijrbertozzijr
5641618
$endgroup$
add a comment |
$begingroup$
I have the following question:
Given Matrices $A$ and $B$, the following relation exists between their column spaces:
$$text{col}(B) subseteq text{col}(A)$$
Then, which of the following is true for Matrix $C=[A,,,,,B]$?
A) $text{rank}(C)=text{rank}(A)$
B) $text{rank}(C)=text{rank}(B)$
C) It is not possible to specify $text{rank}(C)$ in terms of $text{rank}(A)$ and $text{rank}(B)$
My guess, and it seems a reasonable one, would be alternative (A), but I don't know how to solve/express it mathematically.
Is my guess correct? Could you walk me through the steps to prove it?
Thanks in advance
linear-algebra matrices matrix-rank
asked Jan 4 at 22:16
bertozzijrbertozzijr
5641618
$endgroup$
$begingroup$
It is true that $mathrm{rank}(C) = mathrm{rank}(A)$, since $mathrm{col}(C)subsetmathrm{col}(A)$ and vice versa.
$endgroup$
– Math1000
Jan 4 at 22:19
1
$begingroup$
What is the column space of C?
$endgroup$
– Doug M
Jan 4 at 22:20
$begingroup$
$C binom{x}{y} = Ax + By$. Hence ${cal R}C = {cal R}A + {cal R} B$. You are given ${cal R}B subset {cal R} A$.
$endgroup$
– copper.hat
Jan 4 at 22:31
$begingroup$
Is it correct that if $text{col}(B) subseteq text{col}(A)$, then $text{rank}(B) leq text{rank}(A)$ and the rank of a concatenated matrix is equal to the maximum rank among the concatenating matrices? Is this a way to solve it?
$endgroup$
– bertozzijr
Jan 4 at 23:11
add a comment |
$begingroup$
I have the following question:
Given Matrices $A$ and $B$, the following relation exists between their column spaces:
$$text{col}(B) subseteq text{col}(A)$$
Then, which of the following is true for Matrix $C=[A,,,,,B]$?
A) $text{rank}(C)=text{rank}(A)$
B) $text{rank}(C)=text{rank}(B)$
C) It is not possible to specify $text{rank}(C)$ in terms of $text{rank}(A)$ and $text{rank}(B)$
My guess, and it seems a reasonable one, would be alternative (A), but I don't know how to solve/express it mathematically.
Is my guess correct? Could you walk me through the steps to prove it?
Thanks in advance
linear-algebra matrices matrix-rank
asked Jan 4 at 22:16
bertozzijrbertozzijr
5641618
$endgroup$
I have the following question:
Given Matrices $A$ and $B$, the following relation exists between their column spaces:
$$text{col}(B) subseteq text{col}(A)$$
Then, which of the following is true for Matrix $C=[A,,,,,B]$?
A) $text{rank}(C)=text{rank}(A)$
B) $text{rank}(C)=text{rank}(B)$
C) It is not possible to specify $text{rank}(C)$ in terms of $text{rank}(A)$ and $text{rank}(B)$
My guess, and it seems a reasonable one, would be alternative (A), but I don't know how to solve/express it mathematically.
Is my guess correct? Could you walk me through the steps to prove it?
Thanks in advance
linear-algebra matrices matrix-rank
linear-algebra matrices matrix-rank
asked Jan 4 at 22:16
bertozzijrbertozzijr
5641618
asked Jan 4 at 22:16
bertozzijrbertozzijr
5641618
asked Jan 4 at 22:16
bertozzijrbertozzijr
5641618
asked Jan 4 at 22:16
bertozzijrbertozzijr
5641618
asked Jan 4 at 22:16
bertozzijrbertozzijr
5641618
5641618
$begingroup$
It is true that $mathrm{rank}(C) = mathrm{rank}(A)$, since $mathrm{col}(C)subsetmathrm{col}(A)$ and vice versa.
$endgroup$
– Math1000
Jan 4 at 22:19
1
$begingroup$
What is the column space of C?
$endgroup$
– Doug M
Jan 4 at 22:20
$begingroup$
$C binom{x}{y} = Ax + By$. Hence ${cal R}C = {cal R}A + {cal R} B$. You are given ${cal R}B subset {cal R} A$.
$endgroup$
– copper.hat
Jan 4 at 22:31
$begingroup$
Is it correct that if $text{col}(B) subseteq text{col}(A)$, then $text{rank}(B) leq text{rank}(A)$ and the rank of a concatenated matrix is equal to the maximum rank among the concatenating matrices? Is this a way to solve it?
$endgroup$
– bertozzijr
Jan 4 at 23:11
add a comment |
$begingroup$
It is true that $mathrm{rank}(C) = mathrm{rank}(A)$, since $mathrm{col}(C)subsetmathrm{col}(A)$ and vice versa.
$endgroup$
– Math1000
Jan 4 at 22:19
1
$begingroup$
What is the column space of C?
$endgroup$
– Doug M
Jan 4 at 22:20
$begingroup$
$C binom{x}{y} = Ax + By$. Hence ${cal R}C = {cal R}A + {cal R} B$. You are given ${cal R}B subset {cal R} A$.
$endgroup$
– copper.hat
Jan 4 at 22:31
$begingroup$
Is it correct that if $text{col}(B) subseteq text{col}(A)$, then $text{rank}(B) leq text{rank}(A)$ and the rank of a concatenated matrix is equal to the maximum rank among the concatenating matrices? Is this a way to solve it?
$endgroup$
– bertozzijr
Jan 4 at 23:11
$begingroup$
It is true that $mathrm{rank}(C) = mathrm{rank}(A)$, since $mathrm{col}(C)subsetmathrm{col}(A)$ and vice versa.
$endgroup$
– Math1000
Jan 4 at 22:19
$begingroup$
It is true that $mathrm{rank}(C) = mathrm{rank}(A)$, since $mathrm{col}(C)subsetmathrm{col}(A)$ and vice versa.
$endgroup$
– Math1000
Jan 4 at 22:19
1
1
$begingroup$
What is the column space of C?
$endgroup$
– Doug M
Jan 4 at 22:20
$begingroup$
What is the column space of C?
$endgroup$
– Doug M
Jan 4 at 22:20
$begingroup$
$C binom{x}{y} = Ax + By$. Hence ${cal R}C = {cal R}A + {cal R} B$. You are given ${cal R}B subset {cal R} A$.
$endgroup$
– copper.hat
Jan 4 at 22:31
$begingroup$
$C binom{x}{y} = Ax + By$. Hence ${cal R}C = {cal R}A + {cal R} B$. You are given ${cal R}B subset {cal R} A$.
$endgroup$
– copper.hat
Jan 4 at 22:31
$begingroup$
Is it correct that if $text{col}(B) subseteq text{col}(A)$, then $text{rank}(B) leq text{rank}(A)$ and the rank of a concatenated matrix is equal to the maximum rank among the concatenating matrices? Is this a way to solve it?
$endgroup$
– bertozzijr
Jan 4 at 23:11
$begingroup$
Is it correct that if $text{col}(B) subseteq text{col}(A)$, then $text{rank}(B) leq text{rank}(A)$ and the rank of a concatenated matrix is equal to the maximum rank among the concatenating matrices? Is this a way to solve it?
$endgroup$
– bertozzijr
Jan 4 at 23:11
add a comment |
1 Answer
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$begingroup$
Yes, that's correct.
The columns of $C$ are just the columns of $A$ followed by the columns of $B$, which are all included in $mathrm{col}(A)$, hence
$$mathrm{col}(C) = mathrm{col}(A)$$
so their dimensions - the ranks - coincide.
answered Jan 4 at 22:52
BerciBerci
60.5k23672
$endgroup$
add a comment |
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$begingroup$
Yes, that's correct.
The columns of $C$ are just the columns of $A$ followed by the columns of $B$, which are all included in $mathrm{col}(A)$, hence
$$mathrm{col}(C) = mathrm{col}(A)$$
so their dimensions - the ranks - coincide.
answered Jan 4 at 22:52
BerciBerci
60.5k23672
$endgroup$
add a comment |
$begingroup$
Yes, that's correct.
The columns of $C$ are just the columns of $A$ followed by the columns of $B$, which are all included in $mathrm{col}(A)$, hence
$$mathrm{col}(C) = mathrm{col}(A)$$
so their dimensions - the ranks - coincide.
answered Jan 4 at 22:52
BerciBerci
60.5k23672
$endgroup$
add a comment |
$begingroup$
Yes, that's correct.
The columns of $C$ are just the columns of $A$ followed by the columns of $B$, which are all included in $mathrm{col}(A)$, hence
$$mathrm{col}(C) = mathrm{col}(A)$$
so their dimensions - the ranks - coincide.
answered Jan 4 at 22:52
BerciBerci
60.5k23672
$endgroup$
Yes, that's correct.
The columns of $C$ are just the columns of $A$ followed by the columns of $B$, which are all included in $mathrm{col}(A)$, hence
$$mathrm{col}(C) = mathrm{col}(A)$$
so their dimensions - the ranks - coincide.
answered Jan 4 at 22:52
BerciBerci
60.5k23672
answered Jan 4 at 22:52
BerciBerci
60.5k23672
answered Jan 4 at 22:52
BerciBerci
60.5k23672
answered Jan 4 at 22:52
BerciBerci
60.5k23672
60.5k23672
add a comment |
add a comment |
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$begingroup$
It is true that $mathrm{rank}(C) = mathrm{rank}(A)$, since $mathrm{col}(C)subsetmathrm{col}(A)$ and vice versa.
$endgroup$
– Math1000
Jan 4 at 22:19
1
$begingroup$
What is the column space of C?
$endgroup$
– Doug M
Jan 4 at 22:20
$begingroup$
$C binom{x}{y} = Ax + By$. Hence ${cal R}C = {cal R}A + {cal R} B$. You are given ${cal R}B subset {cal R} A$.
$endgroup$
– copper.hat
Jan 4 at 22:31
$begingroup$
Is it correct that if $text{col}(B) subseteq text{col}(A)$, then $text{rank}(B) leq text{rank}(A)$ and the rank of a concatenated matrix is equal to the maximum rank among the concatenating matrices? Is this a way to solve it?
$endgroup$
– bertozzijr
Jan 4 at 23:11