Proof that $exists x in C$ that $A - xcdot I_n$ is invertible
$begingroup$
We know that $A in mathbb C^{n,n}$
Proof that $exists x in C$ that $A - xcdot I_n$ is invertible. How many such x exists?
I am thinking about this problem and generally I have only few little concepts...
We should consider two cases:
1. $A$ is reversible.
Then we can put $ x = 0 $ and $$ A - xcdot I_n = A $$ so it works.
2. $A$ is not invertible - and there I have problem. If A is invertible It means that $det_n A neq 0$. Ok, but does it give me something? We can take such $x$ to make at least one element to 0 on diagonal - But if we have $ 0 $ on diagonal, It does not guarantee that $A - xcdot I_n = A$ will be invertible... thanks for your time
linear-algebra matrices inverse
edited Dec 31 '18 at 17:08
José Carlos Santos
164k22131234
asked Dec 31 '18 at 17:00
VirtualUserVirtualUser
899114
$endgroup$
add a comment |
$begingroup$
We know that $A in mathbb C^{n,n}$
Proof that $exists x in C$ that $A - xcdot I_n$ is invertible. How many such x exists?
I am thinking about this problem and generally I have only few little concepts...
We should consider two cases:
1. $A$ is reversible.
Then we can put $ x = 0 $ and $$ A - xcdot I_n = A $$ so it works.
2. $A$ is not invertible - and there I have problem. If A is invertible It means that $det_n A neq 0$. Ok, but does it give me something? We can take such $x$ to make at least one element to 0 on diagonal - But if we have $ 0 $ on diagonal, It does not guarantee that $A - xcdot I_n = A$ will be invertible... thanks for your time
linear-algebra matrices inverse
edited Dec 31 '18 at 17:08
José Carlos Santos
164k22131234
asked Dec 31 '18 at 17:00
VirtualUserVirtualUser
899114
$endgroup$
1
$begingroup$
What is the determinant of $A - xcdot I_n$?
$endgroup$
– user3482749
Dec 31 '18 at 17:01
$begingroup$
On my lecture, I was only proof that Cauchy theorem about determinants - I have nothing about sum of matrices - unless you mean something trivial, which I do not see at the moment
$endgroup$
– VirtualUser
Dec 31 '18 at 17:03
$begingroup$
If $A$ is not invertible (i.e., singular) then its determinant is zero.
$endgroup$
– Sean Roberson
Dec 31 '18 at 17:08
$begingroup$
Yes, I mean something very simple. Just write out the determinant formula in full, if you need to do so to see it.
$endgroup$
– user3482749
Dec 31 '18 at 17:39
add a comment |
$begingroup$
We know that $A in mathbb C^{n,n}$
Proof that $exists x in C$ that $A - xcdot I_n$ is invertible. How many such x exists?
I am thinking about this problem and generally I have only few little concepts...
We should consider two cases:
1. $A$ is reversible.
Then we can put $ x = 0 $ and $$ A - xcdot I_n = A $$ so it works.
2. $A$ is not invertible - and there I have problem. If A is invertible It means that $det_n A neq 0$. Ok, but does it give me something? We can take such $x$ to make at least one element to 0 on diagonal - But if we have $ 0 $ on diagonal, It does not guarantee that $A - xcdot I_n = A$ will be invertible... thanks for your time
linear-algebra matrices inverse
edited Dec 31 '18 at 17:08
José Carlos Santos
164k22131234
asked Dec 31 '18 at 17:00
VirtualUserVirtualUser
899114
$endgroup$
We know that $A in mathbb C^{n,n}$
Proof that $exists x in C$ that $A - xcdot I_n$ is invertible. How many such x exists?
I am thinking about this problem and generally I have only few little concepts...
We should consider two cases:
1. $A$ is reversible.
Then we can put $ x = 0 $ and $$ A - xcdot I_n = A $$ so it works.
2. $A$ is not invertible - and there I have problem. If A is invertible It means that $det_n A neq 0$. Ok, but does it give me something? We can take such $x$ to make at least one element to 0 on diagonal - But if we have $ 0 $ on diagonal, It does not guarantee that $A - xcdot I_n = A$ will be invertible... thanks for your time
linear-algebra matrices inverse
linear-algebra matrices inverse
edited Dec 31 '18 at 17:08
José Carlos Santos
164k22131234
asked Dec 31 '18 at 17:00
VirtualUserVirtualUser
899114
edited Dec 31 '18 at 17:08
José Carlos Santos
164k22131234
asked Dec 31 '18 at 17:00
VirtualUserVirtualUser
899114
edited Dec 31 '18 at 17:08
José Carlos Santos
164k22131234
edited Dec 31 '18 at 17:08
José Carlos Santos
164k22131234
edited Dec 31 '18 at 17:08
José Carlos Santos
164k22131234
164k22131234
asked Dec 31 '18 at 17:00
VirtualUserVirtualUser
899114
asked Dec 31 '18 at 17:00
VirtualUserVirtualUser
899114
asked Dec 31 '18 at 17:00
VirtualUserVirtualUser
899114
899114
1
$begingroup$
What is the determinant of $A - xcdot I_n$?
$endgroup$
– user3482749
Dec 31 '18 at 17:01
$begingroup$
On my lecture, I was only proof that Cauchy theorem about determinants - I have nothing about sum of matrices - unless you mean something trivial, which I do not see at the moment
$endgroup$
– VirtualUser
Dec 31 '18 at 17:03
$begingroup$
If $A$ is not invertible (i.e., singular) then its determinant is zero.
$endgroup$
– Sean Roberson
Dec 31 '18 at 17:08
$begingroup$
Yes, I mean something very simple. Just write out the determinant formula in full, if you need to do so to see it.
$endgroup$
– user3482749
Dec 31 '18 at 17:39
add a comment |
1
$begingroup$
What is the determinant of $A - xcdot I_n$?
$endgroup$
– user3482749
Dec 31 '18 at 17:01
$begingroup$
On my lecture, I was only proof that Cauchy theorem about determinants - I have nothing about sum of matrices - unless you mean something trivial, which I do not see at the moment
$endgroup$
– VirtualUser
Dec 31 '18 at 17:03
$begingroup$
If $A$ is not invertible (i.e., singular) then its determinant is zero.
$endgroup$
– Sean Roberson
Dec 31 '18 at 17:08
$begingroup$
Yes, I mean something very simple. Just write out the determinant formula in full, if you need to do so to see it.
$endgroup$
– user3482749
Dec 31 '18 at 17:39
1
1
$begingroup$
What is the determinant of $A - xcdot I_n$?
$endgroup$
– user3482749
Dec 31 '18 at 17:01
$begingroup$
What is the determinant of $A - xcdot I_n$?
$endgroup$
– user3482749
Dec 31 '18 at 17:01
$begingroup$
On my lecture, I was only proof that Cauchy theorem about determinants - I have nothing about sum of matrices - unless you mean something trivial, which I do not see at the moment
$endgroup$
– VirtualUser
Dec 31 '18 at 17:03
$begingroup$
On my lecture, I was only proof that Cauchy theorem about determinants - I have nothing about sum of matrices - unless you mean something trivial, which I do not see at the moment
$endgroup$
– VirtualUser
Dec 31 '18 at 17:03
$begingroup$
If $A$ is not invertible (i.e., singular) then its determinant is zero.
$endgroup$
– Sean Roberson
Dec 31 '18 at 17:08
$begingroup$
If $A$ is not invertible (i.e., singular) then its determinant is zero.
$endgroup$
– Sean Roberson
Dec 31 '18 at 17:08
$begingroup$
Yes, I mean something very simple. Just write out the determinant formula in full, if you need to do so to see it.
$endgroup$
– user3482749
Dec 31 '18 at 17:39
$begingroup$
Yes, I mean something very simple. Just write out the determinant formula in full, if you need to do so to see it.
$endgroup$
– user3482749
Dec 31 '18 at 17:39
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Suppose $A-xI_n$ is not invertible. Then there exists $v_xne0$ such that $(A-xI_n)v_x=0$, which means $Av_x=xv_x$.
Suppose $xne y$ and that both $A-xI_n$ and $A-yI_n$ are not invertible. Suppose also
$$
av_x+bv_y=0
$$
Then, multiplying by $A$, we get $aAv_x+bAv_y=0$, that is, $axv_x+byv_y=0$. If we instead multiply by $y$, we obtain $ayv_x+byv_y=0$. Subtracting the two relations, we obtain
$$
axv_x-ayv_x=0
$$
and since $xne y$, we conclude $a=0$ and therefore $b=0$. In other words, $v_x$ and $v_y$ are linearly independent.
This can be generalized: suppose we have $x_1,x_2,dots,x_m$ pairwise distinct such that $A-x_kI_n$ is not invertible. With induction based on the same technique as before we conclude that the set
$$
{v_{x_1},v_{x_2},dots,v_{x_m}}
$$
is linearly independent. Since the vectors belong to $mathbb{C}^n$, we see that $mle n$.
Hence $A-xI_n$ can be non invertible for at most $n$ distinct values of $x$.
answered Dec 31 '18 at 17:17
egregegreg
183k1486205
$endgroup$
add a comment |
$begingroup$
The matrix $A-xoperatorname{Id}$ is invertible if and only if $det(A-xoperatorname{Id})neq0$. But $det(A-xoperatorname{Id})$ is a polynomial function of degree $n$ (in fact, it's the characteristic polynomial of $A$) and therefore it has at most $n$ zeros (and at least one). For all other values of $x$, the matrix is invertible.
edited Dec 31 '18 at 17:10
Jakobian
2,650721
answered Dec 31 '18 at 17:06
José Carlos SantosJosé Carlos Santos
164k22131234
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Suppose $A-xI_n$ is not invertible. Then there exists $v_xne0$ such that $(A-xI_n)v_x=0$, which means $Av_x=xv_x$.
Suppose $xne y$ and that both $A-xI_n$ and $A-yI_n$ are not invertible. Suppose also
$$
av_x+bv_y=0
$$
Then, multiplying by $A$, we get $aAv_x+bAv_y=0$, that is, $axv_x+byv_y=0$. If we instead multiply by $y$, we obtain $ayv_x+byv_y=0$. Subtracting the two relations, we obtain
$$
axv_x-ayv_x=0
$$
and since $xne y$, we conclude $a=0$ and therefore $b=0$. In other words, $v_x$ and $v_y$ are linearly independent.
This can be generalized: suppose we have $x_1,x_2,dots,x_m$ pairwise distinct such that $A-x_kI_n$ is not invertible. With induction based on the same technique as before we conclude that the set
$$
{v_{x_1},v_{x_2},dots,v_{x_m}}
$$
is linearly independent. Since the vectors belong to $mathbb{C}^n$, we see that $mle n$.
Hence $A-xI_n$ can be non invertible for at most $n$ distinct values of $x$.
answered Dec 31 '18 at 17:17
egregegreg
183k1486205
$endgroup$
add a comment |
$begingroup$
Suppose $A-xI_n$ is not invertible. Then there exists $v_xne0$ such that $(A-xI_n)v_x=0$, which means $Av_x=xv_x$.
Suppose $xne y$ and that both $A-xI_n$ and $A-yI_n$ are not invertible. Suppose also
$$
av_x+bv_y=0
$$
Then, multiplying by $A$, we get $aAv_x+bAv_y=0$, that is, $axv_x+byv_y=0$. If we instead multiply by $y$, we obtain $ayv_x+byv_y=0$. Subtracting the two relations, we obtain
$$
axv_x-ayv_x=0
$$
and since $xne y$, we conclude $a=0$ and therefore $b=0$. In other words, $v_x$ and $v_y$ are linearly independent.
This can be generalized: suppose we have $x_1,x_2,dots,x_m$ pairwise distinct such that $A-x_kI_n$ is not invertible. With induction based on the same technique as before we conclude that the set
$$
{v_{x_1},v_{x_2},dots,v_{x_m}}
$$
is linearly independent. Since the vectors belong to $mathbb{C}^n$, we see that $mle n$.
Hence $A-xI_n$ can be non invertible for at most $n$ distinct values of $x$.
answered Dec 31 '18 at 17:17
egregegreg
183k1486205
$endgroup$
add a comment |
$begingroup$
Suppose $A-xI_n$ is not invertible. Then there exists $v_xne0$ such that $(A-xI_n)v_x=0$, which means $Av_x=xv_x$.
Suppose $xne y$ and that both $A-xI_n$ and $A-yI_n$ are not invertible. Suppose also
$$
av_x+bv_y=0
$$
Then, multiplying by $A$, we get $aAv_x+bAv_y=0$, that is, $axv_x+byv_y=0$. If we instead multiply by $y$, we obtain $ayv_x+byv_y=0$. Subtracting the two relations, we obtain
$$
axv_x-ayv_x=0
$$
and since $xne y$, we conclude $a=0$ and therefore $b=0$. In other words, $v_x$ and $v_y$ are linearly independent.
This can be generalized: suppose we have $x_1,x_2,dots,x_m$ pairwise distinct such that $A-x_kI_n$ is not invertible. With induction based on the same technique as before we conclude that the set
$$
{v_{x_1},v_{x_2},dots,v_{x_m}}
$$
is linearly independent. Since the vectors belong to $mathbb{C}^n$, we see that $mle n$.
Hence $A-xI_n$ can be non invertible for at most $n$ distinct values of $x$.
answered Dec 31 '18 at 17:17
egregegreg
183k1486205
$endgroup$
Suppose $A-xI_n$ is not invertible. Then there exists $v_xne0$ such that $(A-xI_n)v_x=0$, which means $Av_x=xv_x$.
Suppose $xne y$ and that both $A-xI_n$ and $A-yI_n$ are not invertible. Suppose also
$$
av_x+bv_y=0
$$
Then, multiplying by $A$, we get $aAv_x+bAv_y=0$, that is, $axv_x+byv_y=0$. If we instead multiply by $y$, we obtain $ayv_x+byv_y=0$. Subtracting the two relations, we obtain
$$
axv_x-ayv_x=0
$$
and since $xne y$, we conclude $a=0$ and therefore $b=0$. In other words, $v_x$ and $v_y$ are linearly independent.
This can be generalized: suppose we have $x_1,x_2,dots,x_m$ pairwise distinct such that $A-x_kI_n$ is not invertible. With induction based on the same technique as before we conclude that the set
$$
{v_{x_1},v_{x_2},dots,v_{x_m}}
$$
is linearly independent. Since the vectors belong to $mathbb{C}^n$, we see that $mle n$.
Hence $A-xI_n$ can be non invertible for at most $n$ distinct values of $x$.
answered Dec 31 '18 at 17:17
egregegreg
183k1486205
answered Dec 31 '18 at 17:17
egregegreg
183k1486205
answered Dec 31 '18 at 17:17
egregegreg
183k1486205
answered Dec 31 '18 at 17:17
egregegreg
183k1486205
183k1486205
add a comment |
add a comment |
$begingroup$
The matrix $A-xoperatorname{Id}$ is invertible if and only if $det(A-xoperatorname{Id})neq0$. But $det(A-xoperatorname{Id})$ is a polynomial function of degree $n$ (in fact, it's the characteristic polynomial of $A$) and therefore it has at most $n$ zeros (and at least one). For all other values of $x$, the matrix is invertible.
edited Dec 31 '18 at 17:10
Jakobian
2,650721
answered Dec 31 '18 at 17:06
José Carlos SantosJosé Carlos Santos
164k22131234
$endgroup$
add a comment |
$begingroup$
The matrix $A-xoperatorname{Id}$ is invertible if and only if $det(A-xoperatorname{Id})neq0$. But $det(A-xoperatorname{Id})$ is a polynomial function of degree $n$ (in fact, it's the characteristic polynomial of $A$) and therefore it has at most $n$ zeros (and at least one). For all other values of $x$, the matrix is invertible.
edited Dec 31 '18 at 17:10
Jakobian
2,650721
answered Dec 31 '18 at 17:06
José Carlos SantosJosé Carlos Santos
164k22131234
$endgroup$
add a comment |
$begingroup$
The matrix $A-xoperatorname{Id}$ is invertible if and only if $det(A-xoperatorname{Id})neq0$. But $det(A-xoperatorname{Id})$ is a polynomial function of degree $n$ (in fact, it's the characteristic polynomial of $A$) and therefore it has at most $n$ zeros (and at least one). For all other values of $x$, the matrix is invertible.
edited Dec 31 '18 at 17:10
Jakobian
2,650721
answered Dec 31 '18 at 17:06
José Carlos SantosJosé Carlos Santos
164k22131234
$endgroup$
The matrix $A-xoperatorname{Id}$ is invertible if and only if $det(A-xoperatorname{Id})neq0$. But $det(A-xoperatorname{Id})$ is a polynomial function of degree $n$ (in fact, it's the characteristic polynomial of $A$) and therefore it has at most $n$ zeros (and at least one). For all other values of $x$, the matrix is invertible.
edited Dec 31 '18 at 17:10
Jakobian
2,650721
answered Dec 31 '18 at 17:06
José Carlos SantosJosé Carlos Santos
164k22131234
edited Dec 31 '18 at 17:10
Jakobian
2,650721
edited Dec 31 '18 at 17:10
Jakobian
2,650721
edited Dec 31 '18 at 17:10
Jakobian
2,650721
2,650721
answered Dec 31 '18 at 17:06
José Carlos SantosJosé Carlos Santos
164k22131234
answered Dec 31 '18 at 17:06
José Carlos SantosJosé Carlos Santos
164k22131234
answered Dec 31 '18 at 17:06
José Carlos SantosJosé Carlos Santos
164k22131234
164k22131234
add a comment |
add a comment |
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1
$begingroup$
What is the determinant of $A - xcdot I_n$?
$endgroup$
– user3482749
Dec 31 '18 at 17:01
$begingroup$
On my lecture, I was only proof that Cauchy theorem about determinants - I have nothing about sum of matrices - unless you mean something trivial, which I do not see at the moment
$endgroup$
– VirtualUser
Dec 31 '18 at 17:03
$begingroup$
If $A$ is not invertible (i.e., singular) then its determinant is zero.
$endgroup$
– Sean Roberson
Dec 31 '18 at 17:08
$begingroup$
Yes, I mean something very simple. Just write out the determinant formula in full, if you need to do so to see it.
$endgroup$
– user3482749
Dec 31 '18 at 17:39