There do not exist $m*n$ and $n*m$ matrices A and B such that $AB=I_m$ (The $m*m$ Identity matrix)
$begingroup$
Let $m,ninmathbb{N}$ be positive with $m>n$. Prove that there do not
exist $m*n$ and $n*m$ matrices A and B such that $AB=I_m$ (The $m*m$
Identity matrix)
I have noticed that this is not true if the matrices are multiplied the other way. That is, say $B=begin{pmatrix}
1 &0 &0 \
0& 1& 0
end{pmatrix}$ and $A=begin{pmatrix}
1 &0 \
0&1 \
0&0
end{pmatrix}$
Then the product $BA$ yields $I_2$, But not the other way. Here this does not contradicts with the statement in the question because of the order of the matrix mentioned in the question (that is the one with a higher number of rows should be multiplied first).
So a proof for the given question is highly appreciated
linear-algebra
asked Jan 7 at 19:27
DD90DD90
2648
$endgroup$
add a comment |
$begingroup$
Let $m,ninmathbb{N}$ be positive with $m>n$. Prove that there do not
exist $m*n$ and $n*m$ matrices A and B such that $AB=I_m$ (The $m*m$
Identity matrix)
I have noticed that this is not true if the matrices are multiplied the other way. That is, say $B=begin{pmatrix}
1 &0 &0 \
0& 1& 0
end{pmatrix}$ and $A=begin{pmatrix}
1 &0 \
0&1 \
0&0
end{pmatrix}$
Then the product $BA$ yields $I_2$, But not the other way. Here this does not contradicts with the statement in the question because of the order of the matrix mentioned in the question (that is the one with a higher number of rows should be multiplied first).
So a proof for the given question is highly appreciated
linear-algebra
asked Jan 7 at 19:27
DD90DD90
2648
$endgroup$
$begingroup$
Hint: consider ranks
$endgroup$
– Wojowu
Jan 7 at 19:30
$begingroup$
Hi, so in this case the maximum rank of A=m and Maximum rank of B=n (Obtained by considering the dimension of row space). Thus can you please tell whether there is a relationship with the rank of AB?
$endgroup$
– DD90
Jan 7 at 19:33
add a comment |
$begingroup$
Let $m,ninmathbb{N}$ be positive with $m>n$. Prove that there do not
exist $m*n$ and $n*m$ matrices A and B such that $AB=I_m$ (The $m*m$
Identity matrix)
I have noticed that this is not true if the matrices are multiplied the other way. That is, say $B=begin{pmatrix}
1 &0 &0 \
0& 1& 0
end{pmatrix}$ and $A=begin{pmatrix}
1 &0 \
0&1 \
0&0
end{pmatrix}$
Then the product $BA$ yields $I_2$, But not the other way. Here this does not contradicts with the statement in the question because of the order of the matrix mentioned in the question (that is the one with a higher number of rows should be multiplied first).
So a proof for the given question is highly appreciated
linear-algebra
asked Jan 7 at 19:27
DD90DD90
2648
$endgroup$
Let $m,ninmathbb{N}$ be positive with $m>n$. Prove that there do not
exist $m*n$ and $n*m$ matrices A and B such that $AB=I_m$ (The $m*m$
Identity matrix)
I have noticed that this is not true if the matrices are multiplied the other way. That is, say $B=begin{pmatrix}
1 &0 &0 \
0& 1& 0
end{pmatrix}$ and $A=begin{pmatrix}
1 &0 \
0&1 \
0&0
end{pmatrix}$
Then the product $BA$ yields $I_2$, But not the other way. Here this does not contradicts with the statement in the question because of the order of the matrix mentioned in the question (that is the one with a higher number of rows should be multiplied first).
So a proof for the given question is highly appreciated
linear-algebra
linear-algebra
asked Jan 7 at 19:27
DD90DD90
2648
asked Jan 7 at 19:27
DD90DD90
2648
asked Jan 7 at 19:27
DD90DD90
2648
asked Jan 7 at 19:27
DD90DD90
2648
asked Jan 7 at 19:27
DD90DD90
2648
2648
$begingroup$
Hint: consider ranks
$endgroup$
– Wojowu
Jan 7 at 19:30
$begingroup$
Hi, so in this case the maximum rank of A=m and Maximum rank of B=n (Obtained by considering the dimension of row space). Thus can you please tell whether there is a relationship with the rank of AB?
$endgroup$
– DD90
Jan 7 at 19:33
add a comment |
$begingroup$
Hint: consider ranks
$endgroup$
– Wojowu
Jan 7 at 19:30
$begingroup$
Hi, so in this case the maximum rank of A=m and Maximum rank of B=n (Obtained by considering the dimension of row space). Thus can you please tell whether there is a relationship with the rank of AB?
$endgroup$
– DD90
Jan 7 at 19:33
$begingroup$
Hint: consider ranks
$endgroup$
– Wojowu
Jan 7 at 19:30
$begingroup$
Hint: consider ranks
$endgroup$
– Wojowu
Jan 7 at 19:30
$begingroup$
Hi, so in this case the maximum rank of A=m and Maximum rank of B=n (Obtained by considering the dimension of row space). Thus can you please tell whether there is a relationship with the rank of AB?
$endgroup$
– DD90
Jan 7 at 19:33
$begingroup$
Hi, so in this case the maximum rank of A=m and Maximum rank of B=n (Obtained by considering the dimension of row space). Thus can you please tell whether there is a relationship with the rank of AB?
$endgroup$
– DD90
Jan 7 at 19:33
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
The column space of an $mtimes n$ matrix $A$ is the span of the columns of $A$ (considered as elements in $mathbb{R}^m$ or whatever field your matrices are on). We can describe it as
$$
{Ax:xinmathbb{R}^n}
$$
Therefore the column space of $AB$ is a subspace of the column space of $A$, so it cannot have larger dimension.
Since the column space of $A$ has dimension at most $n$, the column space of $AB$ also has dimension at most $n$. The column space of the identity matrix $I_m$ has dimension $m>n$. Therefore $ABne I_m$.
The dimension of the column space of $A$ is commonly known as the rank of $A$, $operatorname{rk}(A)$. The above argument can then be stated as $operatorname{rk}(AB)leoperatorname{rk}(A)$. It also holds that $operatorname{rk}(AB)leoperatorname{rk}(B)$ (not needed for the above proof).
answered Jan 7 at 21:17
egregegreg
180k1485202
$endgroup$
add a comment |
$begingroup$
Observe that $B$ has more columns than rows, so kernel of $B$ is non trivial (a free column will exist). So a nonzero vector $x$ exists such that $Bx=0$. Thus $AB$ cannot be a one-one map (because both $0$ and $x$ map to $0$), hence not equal to $I$.
answered Jan 7 at 19:35
Anurag AAnurag A
25.8k12249
$endgroup$
add a comment |
$begingroup$
Let $B$ be a matrix with more columns $B_1,ldots, B_n$ than rows. Then there exists a $l$ s.t. $B_l = sum_{i not = l} c_iB_i$, where the $c_i$s are scalars.
However, if there is a matrix $A$ such that $AB = I$ (so $A$ has $m$ rows) then for each $k$, the following must hold: $A_k cdot B_k = 1$ and $A_k cdot B_i = 0$ for each $i not = k$. Thus, letting $A_l$ be the $l$-th row of $A$, the following must hold: $A_lcdot B_l = 1$ and $A_l cdot B_i = 0$ for all $i not = l$. However, $A_l cdot B_i = 0$ for all $i not = l$ would imply $A_l cdot B_l = sum_{i not = l} c_iA_l cdot B_i = 0$, contradicting the equation $A_l cdot B_l = 1$.
answered Jan 7 at 19:36
MikeMike
3,363311
$endgroup$
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
$begingroup$
The column space of an $mtimes n$ matrix $A$ is the span of the columns of $A$ (considered as elements in $mathbb{R}^m$ or whatever field your matrices are on). We can describe it as
$$
{Ax:xinmathbb{R}^n}
$$
Therefore the column space of $AB$ is a subspace of the column space of $A$, so it cannot have larger dimension.
Since the column space of $A$ has dimension at most $n$, the column space of $AB$ also has dimension at most $n$. The column space of the identity matrix $I_m$ has dimension $m>n$. Therefore $ABne I_m$.
The dimension of the column space of $A$ is commonly known as the rank of $A$, $operatorname{rk}(A)$. The above argument can then be stated as $operatorname{rk}(AB)leoperatorname{rk}(A)$. It also holds that $operatorname{rk}(AB)leoperatorname{rk}(B)$ (not needed for the above proof).
answered Jan 7 at 21:17
egregegreg
180k1485202
$endgroup$
add a comment |
$begingroup$
The column space of an $mtimes n$ matrix $A$ is the span of the columns of $A$ (considered as elements in $mathbb{R}^m$ or whatever field your matrices are on). We can describe it as
$$
{Ax:xinmathbb{R}^n}
$$
Therefore the column space of $AB$ is a subspace of the column space of $A$, so it cannot have larger dimension.
Since the column space of $A$ has dimension at most $n$, the column space of $AB$ also has dimension at most $n$. The column space of the identity matrix $I_m$ has dimension $m>n$. Therefore $ABne I_m$.
The dimension of the column space of $A$ is commonly known as the rank of $A$, $operatorname{rk}(A)$. The above argument can then be stated as $operatorname{rk}(AB)leoperatorname{rk}(A)$. It also holds that $operatorname{rk}(AB)leoperatorname{rk}(B)$ (not needed for the above proof).
answered Jan 7 at 21:17
egregegreg
180k1485202
$endgroup$
add a comment |
$begingroup$
The column space of an $mtimes n$ matrix $A$ is the span of the columns of $A$ (considered as elements in $mathbb{R}^m$ or whatever field your matrices are on). We can describe it as
$$
{Ax:xinmathbb{R}^n}
$$
Therefore the column space of $AB$ is a subspace of the column space of $A$, so it cannot have larger dimension.
Since the column space of $A$ has dimension at most $n$, the column space of $AB$ also has dimension at most $n$. The column space of the identity matrix $I_m$ has dimension $m>n$. Therefore $ABne I_m$.
The dimension of the column space of $A$ is commonly known as the rank of $A$, $operatorname{rk}(A)$. The above argument can then be stated as $operatorname{rk}(AB)leoperatorname{rk}(A)$. It also holds that $operatorname{rk}(AB)leoperatorname{rk}(B)$ (not needed for the above proof).
answered Jan 7 at 21:17
egregegreg
180k1485202
$endgroup$
The column space of an $mtimes n$ matrix $A$ is the span of the columns of $A$ (considered as elements in $mathbb{R}^m$ or whatever field your matrices are on). We can describe it as
$$
{Ax:xinmathbb{R}^n}
$$
Therefore the column space of $AB$ is a subspace of the column space of $A$, so it cannot have larger dimension.
Since the column space of $A$ has dimension at most $n$, the column space of $AB$ also has dimension at most $n$. The column space of the identity matrix $I_m$ has dimension $m>n$. Therefore $ABne I_m$.
The dimension of the column space of $A$ is commonly known as the rank of $A$, $operatorname{rk}(A)$. The above argument can then be stated as $operatorname{rk}(AB)leoperatorname{rk}(A)$. It also holds that $operatorname{rk}(AB)leoperatorname{rk}(B)$ (not needed for the above proof).
answered Jan 7 at 21:17
egregegreg
180k1485202
answered Jan 7 at 21:17
egregegreg
180k1485202
answered Jan 7 at 21:17
egregegreg
180k1485202
answered Jan 7 at 21:17
egregegreg
180k1485202
180k1485202
add a comment |
add a comment |
$begingroup$
Observe that $B$ has more columns than rows, so kernel of $B$ is non trivial (a free column will exist). So a nonzero vector $x$ exists such that $Bx=0$. Thus $AB$ cannot be a one-one map (because both $0$ and $x$ map to $0$), hence not equal to $I$.
answered Jan 7 at 19:35
Anurag AAnurag A
25.8k12249
$endgroup$
add a comment |
$begingroup$
Observe that $B$ has more columns than rows, so kernel of $B$ is non trivial (a free column will exist). So a nonzero vector $x$ exists such that $Bx=0$. Thus $AB$ cannot be a one-one map (because both $0$ and $x$ map to $0$), hence not equal to $I$.
answered Jan 7 at 19:35
Anurag AAnurag A
25.8k12249
$endgroup$
add a comment |
$begingroup$
Observe that $B$ has more columns than rows, so kernel of $B$ is non trivial (a free column will exist). So a nonzero vector $x$ exists such that $Bx=0$. Thus $AB$ cannot be a one-one map (because both $0$ and $x$ map to $0$), hence not equal to $I$.
answered Jan 7 at 19:35
Anurag AAnurag A
25.8k12249
$endgroup$
Observe that $B$ has more columns than rows, so kernel of $B$ is non trivial (a free column will exist). So a nonzero vector $x$ exists such that $Bx=0$. Thus $AB$ cannot be a one-one map (because both $0$ and $x$ map to $0$), hence not equal to $I$.
answered Jan 7 at 19:35
Anurag AAnurag A
25.8k12249
answered Jan 7 at 19:35
Anurag AAnurag A
25.8k12249
answered Jan 7 at 19:35
Anurag AAnurag A
25.8k12249
answered Jan 7 at 19:35
Anurag AAnurag A
25.8k12249
25.8k12249
add a comment |
add a comment |
$begingroup$
Let $B$ be a matrix with more columns $B_1,ldots, B_n$ than rows. Then there exists a $l$ s.t. $B_l = sum_{i not = l} c_iB_i$, where the $c_i$s are scalars.
However, if there is a matrix $A$ such that $AB = I$ (so $A$ has $m$ rows) then for each $k$, the following must hold: $A_k cdot B_k = 1$ and $A_k cdot B_i = 0$ for each $i not = k$. Thus, letting $A_l$ be the $l$-th row of $A$, the following must hold: $A_lcdot B_l = 1$ and $A_l cdot B_i = 0$ for all $i not = l$. However, $A_l cdot B_i = 0$ for all $i not = l$ would imply $A_l cdot B_l = sum_{i not = l} c_iA_l cdot B_i = 0$, contradicting the equation $A_l cdot B_l = 1$.
answered Jan 7 at 19:36
MikeMike
3,363311
$endgroup$
add a comment |
$begingroup$
Let $B$ be a matrix with more columns $B_1,ldots, B_n$ than rows. Then there exists a $l$ s.t. $B_l = sum_{i not = l} c_iB_i$, where the $c_i$s are scalars.
However, if there is a matrix $A$ such that $AB = I$ (so $A$ has $m$ rows) then for each $k$, the following must hold: $A_k cdot B_k = 1$ and $A_k cdot B_i = 0$ for each $i not = k$. Thus, letting $A_l$ be the $l$-th row of $A$, the following must hold: $A_lcdot B_l = 1$ and $A_l cdot B_i = 0$ for all $i not = l$. However, $A_l cdot B_i = 0$ for all $i not = l$ would imply $A_l cdot B_l = sum_{i not = l} c_iA_l cdot B_i = 0$, contradicting the equation $A_l cdot B_l = 1$.
answered Jan 7 at 19:36
MikeMike
3,363311
$endgroup$
add a comment |
$begingroup$
Let $B$ be a matrix with more columns $B_1,ldots, B_n$ than rows. Then there exists a $l$ s.t. $B_l = sum_{i not = l} c_iB_i$, where the $c_i$s are scalars.
However, if there is a matrix $A$ such that $AB = I$ (so $A$ has $m$ rows) then for each $k$, the following must hold: $A_k cdot B_k = 1$ and $A_k cdot B_i = 0$ for each $i not = k$. Thus, letting $A_l$ be the $l$-th row of $A$, the following must hold: $A_lcdot B_l = 1$ and $A_l cdot B_i = 0$ for all $i not = l$. However, $A_l cdot B_i = 0$ for all $i not = l$ would imply $A_l cdot B_l = sum_{i not = l} c_iA_l cdot B_i = 0$, contradicting the equation $A_l cdot B_l = 1$.
answered Jan 7 at 19:36
MikeMike
3,363311
$endgroup$
Let $B$ be a matrix with more columns $B_1,ldots, B_n$ than rows. Then there exists a $l$ s.t. $B_l = sum_{i not = l} c_iB_i$, where the $c_i$s are scalars.
However, if there is a matrix $A$ such that $AB = I$ (so $A$ has $m$ rows) then for each $k$, the following must hold: $A_k cdot B_k = 1$ and $A_k cdot B_i = 0$ for each $i not = k$. Thus, letting $A_l$ be the $l$-th row of $A$, the following must hold: $A_lcdot B_l = 1$ and $A_l cdot B_i = 0$ for all $i not = l$. However, $A_l cdot B_i = 0$ for all $i not = l$ would imply $A_l cdot B_l = sum_{i not = l} c_iA_l cdot B_i = 0$, contradicting the equation $A_l cdot B_l = 1$.
answered Jan 7 at 19:36
MikeMike
3,363311
answered Jan 7 at 19:36
MikeMike
3,363311
answered Jan 7 at 19:36
MikeMike
3,363311
answered Jan 7 at 19:36
MikeMike
3,363311
3,363311
add a comment |
add a comment |
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$begingroup$
Hint: consider ranks
$endgroup$
– Wojowu
Jan 7 at 19:30
$begingroup$
Hi, so in this case the maximum rank of A=m and Maximum rank of B=n (Obtained by considering the dimension of row space). Thus can you please tell whether there is a relationship with the rank of AB?
$endgroup$
– DD90
Jan 7 at 19:33