How do I prove this equality holds? If A is similar to a diagonal matrix with diagonal $a_1,…,a_n$ then...
$begingroup$
I'm not quite sure how to prove whats written in the title. We have a field $F$ and a matrix $A$ that is similar to the diagonal matrix $D=operatorname{diag}(a_1,...,a_n)=begin{bmatrix} a_1& 0 &0&. . .& 0 \ 0& a_2 &0&...&0 \ . \. \.\0&0&0 &...&a_nend{bmatrix}$
I need to show that $(A-a_1I) cdot (A-a_2I) cdot ... cdot (A-a_nI)=0$.
I tried this for long time but I am not sure how to do this. I know there is a matrix $P$ such that $A=P^{-1}DP$, but I'm not sure how that helps. It was written that it is enough to prove this for n=2 (n is the number of rows and columns of A) and I was able to do this for n=2 but I don't know how that helps me to prove this generally. If anyone can provide any help (I'd love to get some hints first, and only then the full solution if I'm still stuck), It will help me a lot. Thanks!
**All I know about similar matrices is that there exists $P$ such that $A=P^{-1}BP$, so the solution should be quite basic, using that fact only.
linear-algebra matrices
asked Dec 28 '18 at 20:56
OmerOmer
535110
$endgroup$
add a comment |
$begingroup$
I'm not quite sure how to prove whats written in the title. We have a field $F$ and a matrix $A$ that is similar to the diagonal matrix $D=operatorname{diag}(a_1,...,a_n)=begin{bmatrix} a_1& 0 &0&. . .& 0 \ 0& a_2 &0&...&0 \ . \. \.\0&0&0 &...&a_nend{bmatrix}$
I need to show that $(A-a_1I) cdot (A-a_2I) cdot ... cdot (A-a_nI)=0$.
I tried this for long time but I am not sure how to do this. I know there is a matrix $P$ such that $A=P^{-1}DP$, but I'm not sure how that helps. It was written that it is enough to prove this for n=2 (n is the number of rows and columns of A) and I was able to do this for n=2 but I don't know how that helps me to prove this generally. If anyone can provide any help (I'd love to get some hints first, and only then the full solution if I'm still stuck), It will help me a lot. Thanks!
**All I know about similar matrices is that there exists $P$ such that $A=P^{-1}BP$, so the solution should be quite basic, using that fact only.
linear-algebra matrices
asked Dec 28 '18 at 20:56
OmerOmer
535110
$endgroup$
1
$begingroup$
Hint: Prove that similar matrices has the same characteristic polynomial and the fact a matrix satisfies its own characteristic polynomial.
$endgroup$
– dezdichado
Dec 28 '18 at 21:01
$begingroup$
@dezdichado Hi, I haven't learnt about characterisitc polynomial yet, all I know about similar matrices is that there exists P such that A=P^-1*B*P, and that's it. They said in the question that it is enough to show this for n=2 and than explain how that also implies it for all values of n. sorry for not mentioning that, I'll edit my post now.
$endgroup$
– Omer
Dec 28 '18 at 21:07
add a comment |
$begingroup$
I'm not quite sure how to prove whats written in the title. We have a field $F$ and a matrix $A$ that is similar to the diagonal matrix $D=operatorname{diag}(a_1,...,a_n)=begin{bmatrix} a_1& 0 &0&. . .& 0 \ 0& a_2 &0&...&0 \ . \. \.\0&0&0 &...&a_nend{bmatrix}$
I need to show that $(A-a_1I) cdot (A-a_2I) cdot ... cdot (A-a_nI)=0$.
I tried this for long time but I am not sure how to do this. I know there is a matrix $P$ such that $A=P^{-1}DP$, but I'm not sure how that helps. It was written that it is enough to prove this for n=2 (n is the number of rows and columns of A) and I was able to do this for n=2 but I don't know how that helps me to prove this generally. If anyone can provide any help (I'd love to get some hints first, and only then the full solution if I'm still stuck), It will help me a lot. Thanks!
**All I know about similar matrices is that there exists $P$ such that $A=P^{-1}BP$, so the solution should be quite basic, using that fact only.
linear-algebra matrices
asked Dec 28 '18 at 20:56
OmerOmer
535110
$endgroup$
I'm not quite sure how to prove whats written in the title. We have a field $F$ and a matrix $A$ that is similar to the diagonal matrix $D=operatorname{diag}(a_1,...,a_n)=begin{bmatrix} a_1& 0 &0&. . .& 0 \ 0& a_2 &0&...&0 \ . \. \.\0&0&0 &...&a_nend{bmatrix}$
I need to show that $(A-a_1I) cdot (A-a_2I) cdot ... cdot (A-a_nI)=0$.
I tried this for long time but I am not sure how to do this. I know there is a matrix $P$ such that $A=P^{-1}DP$, but I'm not sure how that helps. It was written that it is enough to prove this for n=2 (n is the number of rows and columns of A) and I was able to do this for n=2 but I don't know how that helps me to prove this generally. If anyone can provide any help (I'd love to get some hints first, and only then the full solution if I'm still stuck), It will help me a lot. Thanks!
**All I know about similar matrices is that there exists $P$ such that $A=P^{-1}BP$, so the solution should be quite basic, using that fact only.
linear-algebra matrices
linear-algebra matrices
asked Dec 28 '18 at 20:56
OmerOmer
535110
asked Dec 28 '18 at 20:56
OmerOmer
535110
edited Dec 28 '18 at 21:09
Omer
asked Dec 28 '18 at 20:56
OmerOmer
535110
asked Dec 28 '18 at 20:56
OmerOmer
535110
asked Dec 28 '18 at 20:56
OmerOmer
535110
535110
1
$begingroup$
Hint: Prove that similar matrices has the same characteristic polynomial and the fact a matrix satisfies its own characteristic polynomial.
$endgroup$
– dezdichado
Dec 28 '18 at 21:01
$begingroup$
@dezdichado Hi, I haven't learnt about characterisitc polynomial yet, all I know about similar matrices is that there exists P such that A=P^-1*B*P, and that's it. They said in the question that it is enough to show this for n=2 and than explain how that also implies it for all values of n. sorry for not mentioning that, I'll edit my post now.
$endgroup$
– Omer
Dec 28 '18 at 21:07
add a comment |
1
$begingroup$
Hint: Prove that similar matrices has the same characteristic polynomial and the fact a matrix satisfies its own characteristic polynomial.
$endgroup$
– dezdichado
Dec 28 '18 at 21:01
$begingroup$
@dezdichado Hi, I haven't learnt about characterisitc polynomial yet, all I know about similar matrices is that there exists P such that A=P^-1*B*P, and that's it. They said in the question that it is enough to show this for n=2 and than explain how that also implies it for all values of n. sorry for not mentioning that, I'll edit my post now.
$endgroup$
– Omer
Dec 28 '18 at 21:07
1
1
$begingroup$
Hint: Prove that similar matrices has the same characteristic polynomial and the fact a matrix satisfies its own characteristic polynomial.
$endgroup$
– dezdichado
Dec 28 '18 at 21:01
$begingroup$
Hint: Prove that similar matrices has the same characteristic polynomial and the fact a matrix satisfies its own characteristic polynomial.
$endgroup$
– dezdichado
Dec 28 '18 at 21:01
$begingroup$
@dezdichado Hi, I haven't learnt about characterisitc polynomial yet, all I know about similar matrices is that there exists P such that A=P^-1*B*P, and that's it. They said in the question that it is enough to show this for n=2 and than explain how that also implies it for all values of n. sorry for not mentioning that, I'll edit my post now.
$endgroup$
– Omer
Dec 28 '18 at 21:07
$begingroup$
@dezdichado Hi, I haven't learnt about characterisitc polynomial yet, all I know about similar matrices is that there exists P such that A=P^-1*B*P, and that's it. They said in the question that it is enough to show this for n=2 and than explain how that also implies it for all values of n. sorry for not mentioning that, I'll edit my post now.
$endgroup$
– Omer
Dec 28 '18 at 21:07
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
Hint:
First check that $;A-alpha I=P^{-1}DP-alpha I=P^{-1}(D-alpha I)P$.
Second, check that $;(D-alpha_1I)(D-alpha_2I)dots(D-alpha_nI)=0$
answered Dec 28 '18 at 21:22
BernardBernard
124k741117
$endgroup$
add a comment |
$begingroup$
Prove that if $p(x)=c_0+c_1x+dots+c^kx^k$ is a polynomial and
$$
p(B)=c_0I_n+c_1B+dots+c_kB^k=0
$$
(briefly, $B$ satisfies $p$) then also $p(A)=0$, whenever $A$ is similar to $B$.
Next prove that $D=operatorname{diag}(a_1,dots,a_n)$ (the diagonal matrix with the given elements on the diagonal) satisfies $p(x)=(x-a_1)(x-a_2)dots(x-a_n)$.
answered Dec 28 '18 at 21:11
egregegreg
186k1486208
$endgroup$
add a comment |
$begingroup$
Hint: PT
$$
A= P^{-1}begin{bmatrix}
a_1 & & \
&a_2 & \
& & ddots \
end{bmatrix}Pimplies
A-a_{1}I= P^{-1}begin{bmatrix}
0 & & \
&a_{2}-a_{1} & \
& & ddots
end{bmatrix}P
$$
then multiply all the $A-a_{i}I$.
answered Dec 28 '18 at 21:14
zimbra314zimbra314
640312
$endgroup$
add a comment |
$begingroup$
Let $P$ be as in the OP and let $e_k$; $k=1,2,ldots, n$ denote the vector such that the $k$-th component is 1 and every other component is 0.
Fact 1: Then on the one hand, the $P^{-1}e_k$s span $mathbb{F}^n$, so if there is a matrix $B$ such that $B(P^{-1}e_k)$ is 0 for all $k$, then $B$ must be the 0 matrix.
[Make sure you see why Fact 1 is true.]
On the other hand, $A(P^{-1} e_k) = P^{-1}DP (P^{-1}e_k) = P^{-1}D(PP^{-1})e_k$ $ = P^{-1}De_k = P^{-1} a_ke_k = a_kP^{-1}e_k$. Thus $(A-a_kI)(P^{-1}e_k) = 0$ for each such $k$.
This implies $[prod_{l=1}^n (A-a_lI)](P^{-1}e_k) = 0$ for each such $k$. This and Fact 1 implies $[prod_l (A-a_lI)]$ is the 0 matrix.
answered Dec 28 '18 at 21:16
MikeMike
4,641512
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint:
First check that $;A-alpha I=P^{-1}DP-alpha I=P^{-1}(D-alpha I)P$.
Second, check that $;(D-alpha_1I)(D-alpha_2I)dots(D-alpha_nI)=0$
answered Dec 28 '18 at 21:22
BernardBernard
124k741117
$endgroup$
add a comment |
$begingroup$
Hint:
First check that $;A-alpha I=P^{-1}DP-alpha I=P^{-1}(D-alpha I)P$.
Second, check that $;(D-alpha_1I)(D-alpha_2I)dots(D-alpha_nI)=0$
answered Dec 28 '18 at 21:22
BernardBernard
124k741117
$endgroup$
add a comment |
$begingroup$
Hint:
First check that $;A-alpha I=P^{-1}DP-alpha I=P^{-1}(D-alpha I)P$.
Second, check that $;(D-alpha_1I)(D-alpha_2I)dots(D-alpha_nI)=0$
answered Dec 28 '18 at 21:22
BernardBernard
124k741117
$endgroup$
Hint:
First check that $;A-alpha I=P^{-1}DP-alpha I=P^{-1}(D-alpha I)P$.
Second, check that $;(D-alpha_1I)(D-alpha_2I)dots(D-alpha_nI)=0$
answered Dec 28 '18 at 21:22
BernardBernard
124k741117
answered Dec 28 '18 at 21:22
BernardBernard
124k741117
answered Dec 28 '18 at 21:22
BernardBernard
124k741117
answered Dec 28 '18 at 21:22
BernardBernard
124k741117
124k741117
add a comment |
add a comment |
$begingroup$
Prove that if $p(x)=c_0+c_1x+dots+c^kx^k$ is a polynomial and
$$
p(B)=c_0I_n+c_1B+dots+c_kB^k=0
$$
(briefly, $B$ satisfies $p$) then also $p(A)=0$, whenever $A$ is similar to $B$.
Next prove that $D=operatorname{diag}(a_1,dots,a_n)$ (the diagonal matrix with the given elements on the diagonal) satisfies $p(x)=(x-a_1)(x-a_2)dots(x-a_n)$.
answered Dec 28 '18 at 21:11
egregegreg
186k1486208
$endgroup$
add a comment |
$begingroup$
Prove that if $p(x)=c_0+c_1x+dots+c^kx^k$ is a polynomial and
$$
p(B)=c_0I_n+c_1B+dots+c_kB^k=0
$$
(briefly, $B$ satisfies $p$) then also $p(A)=0$, whenever $A$ is similar to $B$.
Next prove that $D=operatorname{diag}(a_1,dots,a_n)$ (the diagonal matrix with the given elements on the diagonal) satisfies $p(x)=(x-a_1)(x-a_2)dots(x-a_n)$.
answered Dec 28 '18 at 21:11
egregegreg
186k1486208
$endgroup$
add a comment |
$begingroup$
Prove that if $p(x)=c_0+c_1x+dots+c^kx^k$ is a polynomial and
$$
p(B)=c_0I_n+c_1B+dots+c_kB^k=0
$$
(briefly, $B$ satisfies $p$) then also $p(A)=0$, whenever $A$ is similar to $B$.
Next prove that $D=operatorname{diag}(a_1,dots,a_n)$ (the diagonal matrix with the given elements on the diagonal) satisfies $p(x)=(x-a_1)(x-a_2)dots(x-a_n)$.
answered Dec 28 '18 at 21:11
egregegreg
186k1486208
$endgroup$
Prove that if $p(x)=c_0+c_1x+dots+c^kx^k$ is a polynomial and
$$
p(B)=c_0I_n+c_1B+dots+c_kB^k=0
$$
(briefly, $B$ satisfies $p$) then also $p(A)=0$, whenever $A$ is similar to $B$.
Next prove that $D=operatorname{diag}(a_1,dots,a_n)$ (the diagonal matrix with the given elements on the diagonal) satisfies $p(x)=(x-a_1)(x-a_2)dots(x-a_n)$.
answered Dec 28 '18 at 21:11
egregegreg
186k1486208
answered Dec 28 '18 at 21:11
egregegreg
186k1486208
answered Dec 28 '18 at 21:11
egregegreg
186k1486208
answered Dec 28 '18 at 21:11
egregegreg
186k1486208
186k1486208
add a comment |
add a comment |
$begingroup$
Hint: PT
$$
A= P^{-1}begin{bmatrix}
a_1 & & \
&a_2 & \
& & ddots \
end{bmatrix}Pimplies
A-a_{1}I= P^{-1}begin{bmatrix}
0 & & \
&a_{2}-a_{1} & \
& & ddots
end{bmatrix}P
$$
then multiply all the $A-a_{i}I$.
answered Dec 28 '18 at 21:14
zimbra314zimbra314
640312
$endgroup$
add a comment |
$begingroup$
Hint: PT
$$
A= P^{-1}begin{bmatrix}
a_1 & & \
&a_2 & \
& & ddots \
end{bmatrix}Pimplies
A-a_{1}I= P^{-1}begin{bmatrix}
0 & & \
&a_{2}-a_{1} & \
& & ddots
end{bmatrix}P
$$
then multiply all the $A-a_{i}I$.
answered Dec 28 '18 at 21:14
zimbra314zimbra314
640312
$endgroup$
add a comment |
$begingroup$
Hint: PT
$$
A= P^{-1}begin{bmatrix}
a_1 & & \
&a_2 & \
& & ddots \
end{bmatrix}Pimplies
A-a_{1}I= P^{-1}begin{bmatrix}
0 & & \
&a_{2}-a_{1} & \
& & ddots
end{bmatrix}P
$$
then multiply all the $A-a_{i}I$.
answered Dec 28 '18 at 21:14
zimbra314zimbra314
640312
$endgroup$
Hint: PT
$$
A= P^{-1}begin{bmatrix}
a_1 & & \
&a_2 & \
& & ddots \
end{bmatrix}Pimplies
A-a_{1}I= P^{-1}begin{bmatrix}
0 & & \
&a_{2}-a_{1} & \
& & ddots
end{bmatrix}P
$$
then multiply all the $A-a_{i}I$.
answered Dec 28 '18 at 21:14
zimbra314zimbra314
640312
answered Dec 28 '18 at 21:14
zimbra314zimbra314
640312
answered Dec 28 '18 at 21:14
zimbra314zimbra314
640312
answered Dec 28 '18 at 21:14
zimbra314zimbra314
640312
640312
add a comment |
add a comment |
$begingroup$
Let $P$ be as in the OP and let $e_k$; $k=1,2,ldots, n$ denote the vector such that the $k$-th component is 1 and every other component is 0.
Fact 1: Then on the one hand, the $P^{-1}e_k$s span $mathbb{F}^n$, so if there is a matrix $B$ such that $B(P^{-1}e_k)$ is 0 for all $k$, then $B$ must be the 0 matrix.
[Make sure you see why Fact 1 is true.]
On the other hand, $A(P^{-1} e_k) = P^{-1}DP (P^{-1}e_k) = P^{-1}D(PP^{-1})e_k$ $ = P^{-1}De_k = P^{-1} a_ke_k = a_kP^{-1}e_k$. Thus $(A-a_kI)(P^{-1}e_k) = 0$ for each such $k$.
This implies $[prod_{l=1}^n (A-a_lI)](P^{-1}e_k) = 0$ for each such $k$. This and Fact 1 implies $[prod_l (A-a_lI)]$ is the 0 matrix.
answered Dec 28 '18 at 21:16
MikeMike
4,641512
$endgroup$
add a comment |
$begingroup$
Let $P$ be as in the OP and let $e_k$; $k=1,2,ldots, n$ denote the vector such that the $k$-th component is 1 and every other component is 0.
Fact 1: Then on the one hand, the $P^{-1}e_k$s span $mathbb{F}^n$, so if there is a matrix $B$ such that $B(P^{-1}e_k)$ is 0 for all $k$, then $B$ must be the 0 matrix.
[Make sure you see why Fact 1 is true.]
On the other hand, $A(P^{-1} e_k) = P^{-1}DP (P^{-1}e_k) = P^{-1}D(PP^{-1})e_k$ $ = P^{-1}De_k = P^{-1} a_ke_k = a_kP^{-1}e_k$. Thus $(A-a_kI)(P^{-1}e_k) = 0$ for each such $k$.
This implies $[prod_{l=1}^n (A-a_lI)](P^{-1}e_k) = 0$ for each such $k$. This and Fact 1 implies $[prod_l (A-a_lI)]$ is the 0 matrix.
answered Dec 28 '18 at 21:16
MikeMike
4,641512
$endgroup$
add a comment |
$begingroup$
Let $P$ be as in the OP and let $e_k$; $k=1,2,ldots, n$ denote the vector such that the $k$-th component is 1 and every other component is 0.
Fact 1: Then on the one hand, the $P^{-1}e_k$s span $mathbb{F}^n$, so if there is a matrix $B$ such that $B(P^{-1}e_k)$ is 0 for all $k$, then $B$ must be the 0 matrix.
[Make sure you see why Fact 1 is true.]
On the other hand, $A(P^{-1} e_k) = P^{-1}DP (P^{-1}e_k) = P^{-1}D(PP^{-1})e_k$ $ = P^{-1}De_k = P^{-1} a_ke_k = a_kP^{-1}e_k$. Thus $(A-a_kI)(P^{-1}e_k) = 0$ for each such $k$.
This implies $[prod_{l=1}^n (A-a_lI)](P^{-1}e_k) = 0$ for each such $k$. This and Fact 1 implies $[prod_l (A-a_lI)]$ is the 0 matrix.
answered Dec 28 '18 at 21:16
MikeMike
4,641512
$endgroup$
Let $P$ be as in the OP and let $e_k$; $k=1,2,ldots, n$ denote the vector such that the $k$-th component is 1 and every other component is 0.
Fact 1: Then on the one hand, the $P^{-1}e_k$s span $mathbb{F}^n$, so if there is a matrix $B$ such that $B(P^{-1}e_k)$ is 0 for all $k$, then $B$ must be the 0 matrix.
[Make sure you see why Fact 1 is true.]
On the other hand, $A(P^{-1} e_k) = P^{-1}DP (P^{-1}e_k) = P^{-1}D(PP^{-1})e_k$ $ = P^{-1}De_k = P^{-1} a_ke_k = a_kP^{-1}e_k$. Thus $(A-a_kI)(P^{-1}e_k) = 0$ for each such $k$.
This implies $[prod_{l=1}^n (A-a_lI)](P^{-1}e_k) = 0$ for each such $k$. This and Fact 1 implies $[prod_l (A-a_lI)]$ is the 0 matrix.
answered Dec 28 '18 at 21:16
MikeMike
4,641512
edited Dec 28 '18 at 21:21
answered Dec 28 '18 at 21:16
MikeMike
4,641512
answered Dec 28 '18 at 21:16
MikeMike
4,641512
answered Dec 28 '18 at 21:16
MikeMike
4,641512
4,641512
add a comment |
add a comment |
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$begingroup$
Hint: Prove that similar matrices has the same characteristic polynomial and the fact a matrix satisfies its own characteristic polynomial.
$endgroup$
– dezdichado
Dec 28 '18 at 21:01
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@dezdichado Hi, I haven't learnt about characterisitc polynomial yet, all I know about similar matrices is that there exists P such that A=P^-1*B*P, and that's it. They said in the question that it is enough to show this for n=2 and than explain how that also implies it for all values of n. sorry for not mentioning that, I'll edit my post now.
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– Omer
Dec 28 '18 at 21:07