Proving that product of two matrix is Singular [closed]
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Let $A$ be a $3times 2$ matrix and $B$ be a $2times 3$ matrix. Show that $C = Acdot B$ is a singluar matrix.
Not sure how to proceed with this. How can it be always be singular if we are randomly multiplying two matrix
matrices
edited Dec 4 '18 at 10:24
Arthur
112k7109192
asked Dec 4 '18 at 10:21
ShashankShashank
31
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closed as off-topic by 5xum, Chinnapparaj R, Vidyanshu Mishra, jgon, A. Pongrácz Dec 4 '18 at 14:42
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – 5xum, Chinnapparaj R, Vidyanshu Mishra, jgon, A. Pongrácz
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Let $A$ be a $3times 2$ matrix and $B$ be a $2times 3$ matrix. Show that $C = Acdot B$ is a singluar matrix.
Not sure how to proceed with this. How can it be always be singular if we are randomly multiplying two matrix
matrices
edited Dec 4 '18 at 10:24
Arthur
112k7109192
asked Dec 4 '18 at 10:21
ShashankShashank
31
$endgroup$
closed as off-topic by 5xum, Chinnapparaj R, Vidyanshu Mishra, jgon, A. Pongrácz Dec 4 '18 at 14:42
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – 5xum, Chinnapparaj R, Vidyanshu Mishra, jgon, A. Pongrácz
If this question can be reworded to fit the rules in the help center, please edit the question.
2
$begingroup$
Hint: think about the rank.
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– GNUSupporter 8964民主女神 地下教會
Dec 4 '18 at 10:23
2
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math.meta.stackexchange.com/questions/9959/…
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– 5xum
Dec 4 '18 at 10:24
add a comment |
$begingroup$
Let $A$ be a $3times 2$ matrix and $B$ be a $2times 3$ matrix. Show that $C = Acdot B$ is a singluar matrix.
Not sure how to proceed with this. How can it be always be singular if we are randomly multiplying two matrix
matrices
edited Dec 4 '18 at 10:24
Arthur
112k7109192
asked Dec 4 '18 at 10:21
ShashankShashank
31
$endgroup$
Let $A$ be a $3times 2$ matrix and $B$ be a $2times 3$ matrix. Show that $C = Acdot B$ is a singluar matrix.
Not sure how to proceed with this. How can it be always be singular if we are randomly multiplying two matrix
matrices
matrices
edited Dec 4 '18 at 10:24
Arthur
112k7109192
asked Dec 4 '18 at 10:21
ShashankShashank
31
edited Dec 4 '18 at 10:24
Arthur
112k7109192
asked Dec 4 '18 at 10:21
ShashankShashank
31
edited Dec 4 '18 at 10:24
Arthur
112k7109192
edited Dec 4 '18 at 10:24
Arthur
112k7109192
edited Dec 4 '18 at 10:24
Arthur
112k7109192
112k7109192
asked Dec 4 '18 at 10:21
ShashankShashank
31
asked Dec 4 '18 at 10:21
ShashankShashank
31
asked Dec 4 '18 at 10:21
ShashankShashank
31
31
closed as off-topic by 5xum, Chinnapparaj R, Vidyanshu Mishra, jgon, A. Pongrácz Dec 4 '18 at 14:42
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – 5xum, Chinnapparaj R, Vidyanshu Mishra, jgon, A. Pongrácz
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by 5xum, Chinnapparaj R, Vidyanshu Mishra, jgon, A. Pongrácz Dec 4 '18 at 14:42
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – 5xum, Chinnapparaj R, Vidyanshu Mishra, jgon, A. Pongrácz
If this question can be reworded to fit the rules in the help center, please edit the question.
2
$begingroup$
Hint: think about the rank.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 4 '18 at 10:23
2
$begingroup$
math.meta.stackexchange.com/questions/9959/…
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– 5xum
Dec 4 '18 at 10:24
add a comment |
2
$begingroup$
Hint: think about the rank.
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– GNUSupporter 8964民主女神 地下教會
Dec 4 '18 at 10:23
2
$begingroup$
math.meta.stackexchange.com/questions/9959/…
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– 5xum
Dec 4 '18 at 10:24
2
2
$begingroup$
Hint: think about the rank.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 4 '18 at 10:23
$begingroup$
Hint: think about the rank.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 4 '18 at 10:23
2
2
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math.meta.stackexchange.com/questions/9959/…
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– 5xum
Dec 4 '18 at 10:24
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math.meta.stackexchange.com/questions/9959/…
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– 5xum
Dec 4 '18 at 10:24
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1 Answer
1
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$begingroup$
The beauty of linear algebra is that it gives us the language to explain the same situation in many different terms.
For example, an $p$ by $p$ matrix $D$ being singular is equivalent to
- $det(D)=0$
- $rank(D)<p$
- Their existing some vector other than the zero vector in $null(D)$
Using the second equivalence $rank(D)<p$ this tells us that to show a matrix $C=AB$ is singular, we must show that $rank(C)<3$ (since the product of $Ain R^{3x2}$ and $Bin R^{2x3}$ is in $R^{3x3}$).
Rectangular, non square matrices have a maximum rank of their smallest dimension (I mean spatial dimension rather than vector space dimension).
i.e. $$Ain R^{a x b},quad rank(A)le a quad (ale b)$$
We can use the fact that $rank(AB)le min(rank(A), rank(B))$.
We have
$$rank(A)le 2 \ rank(B)le 2$$
So
$$rank(C)=rank(AB)le 2 < 3$$
So $C$ is singular.
answered Dec 4 '18 at 10:46
Joel BiffinJoel Biffin
966
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add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The beauty of linear algebra is that it gives us the language to explain the same situation in many different terms.
For example, an $p$ by $p$ matrix $D$ being singular is equivalent to
- $det(D)=0$
- $rank(D)<p$
- Their existing some vector other than the zero vector in $null(D)$
Using the second equivalence $rank(D)<p$ this tells us that to show a matrix $C=AB$ is singular, we must show that $rank(C)<3$ (since the product of $Ain R^{3x2}$ and $Bin R^{2x3}$ is in $R^{3x3}$).
Rectangular, non square matrices have a maximum rank of their smallest dimension (I mean spatial dimension rather than vector space dimension).
i.e. $$Ain R^{a x b},quad rank(A)le a quad (ale b)$$
We can use the fact that $rank(AB)le min(rank(A), rank(B))$.
We have
$$rank(A)le 2 \ rank(B)le 2$$
So
$$rank(C)=rank(AB)le 2 < 3$$
So $C$ is singular.
answered Dec 4 '18 at 10:46
Joel BiffinJoel Biffin
966
$endgroup$
add a comment |
$begingroup$
The beauty of linear algebra is that it gives us the language to explain the same situation in many different terms.
For example, an $p$ by $p$ matrix $D$ being singular is equivalent to
- $det(D)=0$
- $rank(D)<p$
- Their existing some vector other than the zero vector in $null(D)$
Using the second equivalence $rank(D)<p$ this tells us that to show a matrix $C=AB$ is singular, we must show that $rank(C)<3$ (since the product of $Ain R^{3x2}$ and $Bin R^{2x3}$ is in $R^{3x3}$).
Rectangular, non square matrices have a maximum rank of their smallest dimension (I mean spatial dimension rather than vector space dimension).
i.e. $$Ain R^{a x b},quad rank(A)le a quad (ale b)$$
We can use the fact that $rank(AB)le min(rank(A), rank(B))$.
We have
$$rank(A)le 2 \ rank(B)le 2$$
So
$$rank(C)=rank(AB)le 2 < 3$$
So $C$ is singular.
answered Dec 4 '18 at 10:46
Joel BiffinJoel Biffin
966
$endgroup$
add a comment |
$begingroup$
The beauty of linear algebra is that it gives us the language to explain the same situation in many different terms.
For example, an $p$ by $p$ matrix $D$ being singular is equivalent to
- $det(D)=0$
- $rank(D)<p$
- Their existing some vector other than the zero vector in $null(D)$
Using the second equivalence $rank(D)<p$ this tells us that to show a matrix $C=AB$ is singular, we must show that $rank(C)<3$ (since the product of $Ain R^{3x2}$ and $Bin R^{2x3}$ is in $R^{3x3}$).
Rectangular, non square matrices have a maximum rank of their smallest dimension (I mean spatial dimension rather than vector space dimension).
i.e. $$Ain R^{a x b},quad rank(A)le a quad (ale b)$$
We can use the fact that $rank(AB)le min(rank(A), rank(B))$.
We have
$$rank(A)le 2 \ rank(B)le 2$$
So
$$rank(C)=rank(AB)le 2 < 3$$
So $C$ is singular.
answered Dec 4 '18 at 10:46
Joel BiffinJoel Biffin
966
$endgroup$
The beauty of linear algebra is that it gives us the language to explain the same situation in many different terms.
For example, an $p$ by $p$ matrix $D$ being singular is equivalent to
- $det(D)=0$
- $rank(D)<p$
- Their existing some vector other than the zero vector in $null(D)$
Using the second equivalence $rank(D)<p$ this tells us that to show a matrix $C=AB$ is singular, we must show that $rank(C)<3$ (since the product of $Ain R^{3x2}$ and $Bin R^{2x3}$ is in $R^{3x3}$).
Rectangular, non square matrices have a maximum rank of their smallest dimension (I mean spatial dimension rather than vector space dimension).
i.e. $$Ain R^{a x b},quad rank(A)le a quad (ale b)$$
We can use the fact that $rank(AB)le min(rank(A), rank(B))$.
We have
$$rank(A)le 2 \ rank(B)le 2$$
So
$$rank(C)=rank(AB)le 2 < 3$$
So $C$ is singular.
answered Dec 4 '18 at 10:46
Joel BiffinJoel Biffin
966
answered Dec 4 '18 at 10:46
Joel BiffinJoel Biffin
966
answered Dec 4 '18 at 10:46
Joel BiffinJoel Biffin
966
answered Dec 4 '18 at 10:46
Joel BiffinJoel Biffin
966
966
add a comment |
add a comment |
2
$begingroup$
Hint: think about the rank.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 4 '18 at 10:23
2
$begingroup$
math.meta.stackexchange.com/questions/9959/…
$endgroup$
– 5xum
Dec 4 '18 at 10:24