Mixing
what is the equivalent of $xy=a$, then $y→∞$ as $x→0$ for matrices
3
$begingroup$
For scalars, given the equation $xy=a$, then $y→∞$ as $x→0$, i.e. as $x$ tends to being non-invertible.
I wanted to find an equivalent theorem for matrices. I came up with:
For matrices and vectors, given the equation $Xy=a$, then $y→∞$ as $X$ tends to being non-invertible (singular), i.e. as $X→A$, where $A$ is such that $Az=0$ has a non trivial solution.
Is this correct? Where can I find this in a book?
More interestingly, I have an idea that there are as many infinite entries in $y→∞$ as there are non-zero entries in $z$. Is this correct too? If so, how could I prove it?
Please help.
matrices
asked Jan 10 at 1:03
a.giannela.giannel
163
$endgroup$
|
show 2 more comments
3
$begingroup$
For scalars, given the equation $xy=a$, then $y→∞$ as $x→0$, i.e. as $x$ tends to being non-invertible.
I wanted to find an equivalent theorem for matrices. I came up with:
For matrices and vectors, given the equation $Xy=a$, then $y→∞$ as $X$ tends to being non-invertible (singular), i.e. as $X→A$, where $A$ is such that $Az=0$ has a non trivial solution.
Is this correct? Where can I find this in a book?
More interestingly, I have an idea that there are as many infinite entries in $y→∞$ as there are non-zero entries in $z$. Is this correct too? If so, how could I prove it?
Please help.
matrices
asked Jan 10 at 1:03
a.giannela.giannel
163
$endgroup$
1
$begingroup$
I appreciate the curiosity behind this question, but I'm really struggling to make sense of it. When you say, "$x$ tends to being invertible", at one point you mean it tending to $0$ (the only non-invertible scalar), and at another, where $X$ tends to a singular matrix $A$. Do you mean "singular" (non-invertible) when you say "invertible"?
$endgroup$
– Theo Bendit
Jan 10 at 1:30
$begingroup$
Thank you for the corrections.
$endgroup$
– a.giannel
Jan 10 at 1:36
$begingroup$
Suppose $X$ is $ntimes m$. Then consider the $n$ rows of the matrix equation $Xy=a$. The $i$th row has the form $$sum_{j=1}^{m}x_{i,j}cdot y_j=a_i$$ If even one entry of $y$ explodes, say $y_kto infty$, then we can rewrite $$x_{i,k}cdot y_k=a_i-sum_{substack{j=1\jneq k}}^{m}x_{i,j}cdot y_j$$ so that $x_{i,j}to 0$ for every row $i$. But a matrix with a column of zeros is singular, and so not invertible.
$endgroup$
– nathan.j.mcdougall
Jan 10 at 1:45
$begingroup$
Very interesting and cleaver. Thank you. Do you believe that with that I can work on to find a solution the second part of the question, regarding the number of infinite entries for $y$ and zero entries for $z$?
$endgroup$
– a.giannel
Jan 10 at 1:50
1
$begingroup$
The analogy with scalars breaks down pretty quickly. For any vector $y$ you can construct a nonzero matrix $A$ such that $Ay=0$. On the flip side, you can construct $A$ such that the $i$th element of $y$ has no effect on the value of $Ay$, so you’ll need to sharpen up what you mean by $ytoinfty$.
$endgroup$
– amd
Jan 10 at 3:19
|
show 2 more comments
3
3
3
$begingroup$
For scalars, given the equation $xy=a$, then $y→∞$ as $x→0$, i.e. as $x$ tends to being non-invertible.
I wanted to find an equivalent theorem for matrices. I came up with:
For matrices and vectors, given the equation $Xy=a$, then $y→∞$ as $X$ tends to being non-invertible (singular), i.e. as $X→A$, where $A$ is such that $Az=0$ has a non trivial solution.
Is this correct? Where can I find this in a book?
More interestingly, I have an idea that there are as many infinite entries in $y→∞$ as there are non-zero entries in $z$. Is this correct too? If so, how could I prove it?
Please help.
matrices
asked Jan 10 at 1:03
a.giannela.giannel
163
$endgroup$
For scalars, given the equation $xy=a$, then $y→∞$ as $x→0$, i.e. as $x$ tends to being non-invertible.
I wanted to find an equivalent theorem for matrices. I came up with:
For matrices and vectors, given the equation $Xy=a$, then $y→∞$ as $X$ tends to being non-invertible (singular), i.e. as $X→A$, where $A$ is such that $Az=0$ has a non trivial solution.
Is this correct? Where can I find this in a book?
More interestingly, I have an idea that there are as many infinite entries in $y→∞$ as there are non-zero entries in $z$. Is this correct too? If so, how could I prove it?
Please help.
matrices
matrices
asked Jan 10 at 1:03
a.giannela.giannel
163
asked Jan 10 at 1:03
a.giannela.giannel
163
edited Jan 10 at 1:35
a.giannel
asked Jan 10 at 1:03
a.giannela.giannel
163
asked Jan 10 at 1:03
a.giannela.giannel
163
asked Jan 10 at 1:03
a.giannela.giannel
163
163
1
$begingroup$
I appreciate the curiosity behind this question, but I'm really struggling to make sense of it. When you say, "$x$ tends to being invertible", at one point you mean it tending to $0$ (the only non-invertible scalar), and at another, where $X$ tends to a singular matrix $A$. Do you mean "singular" (non-invertible) when you say "invertible"?
$endgroup$
– Theo Bendit
Jan 10 at 1:30
$begingroup$
Thank you for the corrections.
$endgroup$
– a.giannel
Jan 10 at 1:36
$begingroup$
Suppose $X$ is $ntimes m$. Then consider the $n$ rows of the matrix equation $Xy=a$. The $i$th row has the form $$sum_{j=1}^{m}x_{i,j}cdot y_j=a_i$$ If even one entry of $y$ explodes, say $y_kto infty$, then we can rewrite $$x_{i,k}cdot y_k=a_i-sum_{substack{j=1\jneq k}}^{m}x_{i,j}cdot y_j$$ so that $x_{i,j}to 0$ for every row $i$. But a matrix with a column of zeros is singular, and so not invertible.
$endgroup$
– nathan.j.mcdougall
Jan 10 at 1:45
$begingroup$
Very interesting and cleaver. Thank you. Do you believe that with that I can work on to find a solution the second part of the question, regarding the number of infinite entries for $y$ and zero entries for $z$?
$endgroup$
– a.giannel
Jan 10 at 1:50
1
$begingroup$
The analogy with scalars breaks down pretty quickly. For any vector $y$ you can construct a nonzero matrix $A$ such that $Ay=0$. On the flip side, you can construct $A$ such that the $i$th element of $y$ has no effect on the value of $Ay$, so you’ll need to sharpen up what you mean by $ytoinfty$.
$endgroup$
– amd
Jan 10 at 3:19
|
show 2 more comments
1
$begingroup$
I appreciate the curiosity behind this question, but I'm really struggling to make sense of it. When you say, "$x$ tends to being invertible", at one point you mean it tending to $0$ (the only non-invertible scalar), and at another, where $X$ tends to a singular matrix $A$. Do you mean "singular" (non-invertible) when you say "invertible"?
$endgroup$
– Theo Bendit
Jan 10 at 1:30
$begingroup$
Thank you for the corrections.
$endgroup$
– a.giannel
Jan 10 at 1:36
$begingroup$
Suppose $X$ is $ntimes m$. Then consider the $n$ rows of the matrix equation $Xy=a$. The $i$th row has the form $$sum_{j=1}^{m}x_{i,j}cdot y_j=a_i$$ If even one entry of $y$ explodes, say $y_kto infty$, then we can rewrite $$x_{i,k}cdot y_k=a_i-sum_{substack{j=1\jneq k}}^{m}x_{i,j}cdot y_j$$ so that $x_{i,j}to 0$ for every row $i$. But a matrix with a column of zeros is singular, and so not invertible.
$endgroup$
– nathan.j.mcdougall
Jan 10 at 1:45
$begingroup$
Very interesting and cleaver. Thank you. Do you believe that with that I can work on to find a solution the second part of the question, regarding the number of infinite entries for $y$ and zero entries for $z$?
$endgroup$
– a.giannel
Jan 10 at 1:50
1
$begingroup$
The analogy with scalars breaks down pretty quickly. For any vector $y$ you can construct a nonzero matrix $A$ such that $Ay=0$. On the flip side, you can construct $A$ such that the $i$th element of $y$ has no effect on the value of $Ay$, so you’ll need to sharpen up what you mean by $ytoinfty$.
$endgroup$
– amd
Jan 10 at 3:19
1
1
$begingroup$
I appreciate the curiosity behind this question, but I'm really struggling to make sense of it. When you say, "$x$ tends to being invertible", at one point you mean it tending to $0$ (the only non-invertible scalar), and at another, where $X$ tends to a singular matrix $A$. Do you mean "singular" (non-invertible) when you say "invertible"?
$endgroup$
– Theo Bendit
Jan 10 at 1:30
$begingroup$
I appreciate the curiosity behind this question, but I'm really struggling to make sense of it. When you say, "$x$ tends to being invertible", at one point you mean it tending to $0$ (the only non-invertible scalar), and at another, where $X$ tends to a singular matrix $A$. Do you mean "singular" (non-invertible) when you say "invertible"?
$endgroup$
– Theo Bendit
Jan 10 at 1:30
$begingroup$
Thank you for the corrections.
$endgroup$
– a.giannel
Jan 10 at 1:36
$begingroup$
Thank you for the corrections.
$endgroup$
– a.giannel
Jan 10 at 1:36
$begingroup$
Suppose $X$ is $ntimes m$. Then consider the $n$ rows of the matrix equation $Xy=a$. The $i$th row has the form $$sum_{j=1}^{m}x_{i,j}cdot y_j=a_i$$ If even one entry of $y$ explodes, say $y_kto infty$, then we can rewrite $$x_{i,k}cdot y_k=a_i-sum_{substack{j=1\jneq k}}^{m}x_{i,j}cdot y_j$$ so that $x_{i,j}to 0$ for every row $i$. But a matrix with a column of zeros is singular, and so not invertible.
$endgroup$
– nathan.j.mcdougall
Jan 10 at 1:45
$begingroup$
Suppose $X$ is $ntimes m$. Then consider the $n$ rows of the matrix equation $Xy=a$. The $i$th row has the form $$sum_{j=1}^{m}x_{i,j}cdot y_j=a_i$$ If even one entry of $y$ explodes, say $y_kto infty$, then we can rewrite $$x_{i,k}cdot y_k=a_i-sum_{substack{j=1\jneq k}}^{m}x_{i,j}cdot y_j$$ so that $x_{i,j}to 0$ for every row $i$. But a matrix with a column of zeros is singular, and so not invertible.
$endgroup$
– nathan.j.mcdougall
Jan 10 at 1:45
$begingroup$
Very interesting and cleaver. Thank you. Do you believe that with that I can work on to find a solution the second part of the question, regarding the number of infinite entries for $y$ and zero entries for $z$?
$endgroup$
– a.giannel
Jan 10 at 1:50
$begingroup$
Very interesting and cleaver. Thank you. Do you believe that with that I can work on to find a solution the second part of the question, regarding the number of infinite entries for $y$ and zero entries for $z$?
$endgroup$
– a.giannel
Jan 10 at 1:50
1
1
$begingroup$
The analogy with scalars breaks down pretty quickly. For any vector $y$ you can construct a nonzero matrix $A$ such that $Ay=0$. On the flip side, you can construct $A$ such that the $i$th element of $y$ has no effect on the value of $Ay$, so you’ll need to sharpen up what you mean by $ytoinfty$.
$endgroup$
– amd
Jan 10 at 3:19
$begingroup$
The analogy with scalars breaks down pretty quickly. For any vector $y$ you can construct a nonzero matrix $A$ such that $Ay=0$. On the flip side, you can construct $A$ such that the $i$th element of $y$ has no effect on the value of $Ay$, so you’ll need to sharpen up what you mean by $ytoinfty$.
$endgroup$
– amd
Jan 10 at 3:19
|
show 2 more comments
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1
$begingroup$
I appreciate the curiosity behind this question, but I'm really struggling to make sense of it. When you say, "$x$ tends to being invertible", at one point you mean it tending to $0$ (the only non-invertible scalar), and at another, where $X$ tends to a singular matrix $A$. Do you mean "singular" (non-invertible) when you say "invertible"?
$endgroup$
– Theo Bendit
Jan 10 at 1:30
$begingroup$
Thank you for the corrections.
$endgroup$
– a.giannel
Jan 10 at 1:36
$begingroup$
Suppose $X$ is $ntimes m$. Then consider the $n$ rows of the matrix equation $Xy=a$. The $i$th row has the form $$sum_{j=1}^{m}x_{i,j}cdot y_j=a_i$$ If even one entry of $y$ explodes, say $y_kto infty$, then we can rewrite $$x_{i,k}cdot y_k=a_i-sum_{substack{j=1\jneq k}}^{m}x_{i,j}cdot y_j$$ so that $x_{i,j}to 0$ for every row $i$. But a matrix with a column of zeros is singular, and so not invertible.
$endgroup$
– nathan.j.mcdougall
Jan 10 at 1:45
$begingroup$
Very interesting and cleaver. Thank you. Do you believe that with that I can work on to find a solution the second part of the question, regarding the number of infinite entries for $y$ and zero entries for $z$?
$endgroup$
– a.giannel
Jan 10 at 1:50
1
$begingroup$
The analogy with scalars breaks down pretty quickly. For any vector $y$ you can construct a nonzero matrix $A$ such that $Ay=0$. On the flip side, you can construct $A$ such that the $i$th element of $y$ has no effect on the value of $Ay$, so you’ll need to sharpen up what you mean by $ytoinfty$.
$endgroup$
– amd
Jan 10 at 3:19