Finding standard matrix and proving questions
$begingroup$
Let $A$ be an $ntimes n$ matrix. For each $w∈Bbb R$, we define a linear transformation $T_w:Bbb R^ntoBbb R^n$ such that $T_w(u) = Au - wu$ for $u∈Bbb R^n$.
a) Write down the standard matrix for $T_w$.
b) For any $w$, $l∈Bbb R$, show that
$(A-wI)(A-lI) = (A-lI)(A-wI)$.
c) Suppose $A$ is diagonalizable and the eigenvalues of $A$ are $w_1$, $w_2$, ... , $w_k$. If $v$ is an eigenvector of A, say $Av=w_iv$ for some $i$, show that $(A-w_1I)(A-w_2I)...(A-w_kI)v=0$.
What I have done so far:
By letting $A$ =
$$ left[
begin{array}{cccc}
a_{11}&a_{12}&cdots&a_{1n}\
a_{21}&a_{22}&cdots&a_{2n}\
vdots&vdots&ddots&vdots\
a_{n1}&a_{n2}&cdots&a_{nn}
end{array}
right] $$, standard matrix = $(A-wI)$. However, I am not quite sure how to proceed from here to parts b and c. Do I prove part b by applying matrix multiplication manually, or is there another way to do it?
linear-algebra matrices eigenvalues-eigenvectors
edited Nov 26 '18 at 8:39
Tianlalu
3,08621038
asked Nov 26 '18 at 8:07
Cheryl Cheryl
755
$endgroup$
add a comment |
$begingroup$
Let $A$ be an $ntimes n$ matrix. For each $w∈Bbb R$, we define a linear transformation $T_w:Bbb R^ntoBbb R^n$ such that $T_w(u) = Au - wu$ for $u∈Bbb R^n$.
a) Write down the standard matrix for $T_w$.
b) For any $w$, $l∈Bbb R$, show that
$(A-wI)(A-lI) = (A-lI)(A-wI)$.
c) Suppose $A$ is diagonalizable and the eigenvalues of $A$ are $w_1$, $w_2$, ... , $w_k$. If $v$ is an eigenvector of A, say $Av=w_iv$ for some $i$, show that $(A-w_1I)(A-w_2I)...(A-w_kI)v=0$.
What I have done so far:
By letting $A$ =
$$ left[
begin{array}{cccc}
a_{11}&a_{12}&cdots&a_{1n}\
a_{21}&a_{22}&cdots&a_{2n}\
vdots&vdots&ddots&vdots\
a_{n1}&a_{n2}&cdots&a_{nn}
end{array}
right] $$, standard matrix = $(A-wI)$. However, I am not quite sure how to proceed from here to parts b and c. Do I prove part b by applying matrix multiplication manually, or is there another way to do it?
linear-algebra matrices eigenvalues-eigenvectors
edited Nov 26 '18 at 8:39
Tianlalu
3,08621038
asked Nov 26 '18 at 8:07
Cheryl Cheryl
755
$endgroup$
add a comment |
$begingroup$
Let $A$ be an $ntimes n$ matrix. For each $w∈Bbb R$, we define a linear transformation $T_w:Bbb R^ntoBbb R^n$ such that $T_w(u) = Au - wu$ for $u∈Bbb R^n$.
a) Write down the standard matrix for $T_w$.
b) For any $w$, $l∈Bbb R$, show that
$(A-wI)(A-lI) = (A-lI)(A-wI)$.
c) Suppose $A$ is diagonalizable and the eigenvalues of $A$ are $w_1$, $w_2$, ... , $w_k$. If $v$ is an eigenvector of A, say $Av=w_iv$ for some $i$, show that $(A-w_1I)(A-w_2I)...(A-w_kI)v=0$.
What I have done so far:
By letting $A$ =
$$ left[
begin{array}{cccc}
a_{11}&a_{12}&cdots&a_{1n}\
a_{21}&a_{22}&cdots&a_{2n}\
vdots&vdots&ddots&vdots\
a_{n1}&a_{n2}&cdots&a_{nn}
end{array}
right] $$, standard matrix = $(A-wI)$. However, I am not quite sure how to proceed from here to parts b and c. Do I prove part b by applying matrix multiplication manually, or is there another way to do it?
linear-algebra matrices eigenvalues-eigenvectors
edited Nov 26 '18 at 8:39
Tianlalu
3,08621038
asked Nov 26 '18 at 8:07
Cheryl Cheryl
755
$endgroup$
Let $A$ be an $ntimes n$ matrix. For each $w∈Bbb R$, we define a linear transformation $T_w:Bbb R^ntoBbb R^n$ such that $T_w(u) = Au - wu$ for $u∈Bbb R^n$.
a) Write down the standard matrix for $T_w$.
b) For any $w$, $l∈Bbb R$, show that
$(A-wI)(A-lI) = (A-lI)(A-wI)$.
c) Suppose $A$ is diagonalizable and the eigenvalues of $A$ are $w_1$, $w_2$, ... , $w_k$. If $v$ is an eigenvector of A, say $Av=w_iv$ for some $i$, show that $(A-w_1I)(A-w_2I)...(A-w_kI)v=0$.
What I have done so far:
By letting $A$ =
$$ left[
begin{array}{cccc}
a_{11}&a_{12}&cdots&a_{1n}\
a_{21}&a_{22}&cdots&a_{2n}\
vdots&vdots&ddots&vdots\
a_{n1}&a_{n2}&cdots&a_{nn}
end{array}
right] $$, standard matrix = $(A-wI)$. However, I am not quite sure how to proceed from here to parts b and c. Do I prove part b by applying matrix multiplication manually, or is there another way to do it?
linear-algebra matrices eigenvalues-eigenvectors
linear-algebra matrices eigenvalues-eigenvectors
edited Nov 26 '18 at 8:39
Tianlalu
3,08621038
asked Nov 26 '18 at 8:07
Cheryl Cheryl
755
edited Nov 26 '18 at 8:39
Tianlalu
3,08621038
asked Nov 26 '18 at 8:07
Cheryl Cheryl
755
edited Nov 26 '18 at 8:39
Tianlalu
3,08621038
edited Nov 26 '18 at 8:39
Tianlalu
3,08621038
edited Nov 26 '18 at 8:39
Tianlalu
3,08621038
3,08621038
asked Nov 26 '18 at 8:07
Cheryl Cheryl
755
asked Nov 26 '18 at 8:07
Cheryl Cheryl
755
asked Nov 26 '18 at 8:07
Cheryl Cheryl
755
755
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Your answer to part a is correct. For part b note that $Aw=wA$ and $Al=lA$; expanding both sides of the given equation then quickly shows that they are indeed equal.
answered Nov 26 '18 at 8:37
ServaesServaes
23.1k33893
$endgroup$
add a comment |
$begingroup$
For b)
What you want to do is use the fact that matrix multiplication is linear.
$$(A−wI)(A−lI)= A cdot A -l A cdot I -w I cdot A + w l I cdot I $$
Now what you should use is that the identity matrix does not do much, we can always multiply any matrix by it, certainly it commutes $A cdot I = I cdot A$.
I hope this is enough of a hint and you can now rearrange terms so you can rewrite the left hand side to become the right hand side.
For C) In question $b)$ we learnt that you can interchange order, so what if we bring the term corresponding to $(Av=w_i v)$ to the back, we can certainly do so because of associativity ($(ABC) v=(AB)Cv=AB(Cv)$), what I mean is:
$$ ((A-w_1I)(A-w_2I)dots(A-w_iI) dots(A-w_kI))v$$ $$=left( (A-w_1I)(A-w_2I) dots(A-w_kI)(A-w_iI) right)v$$
Here we used this rule from $b$ repeatedly until the right term was at the front. Now just first multiply out the last term by associativity, do you notice something?
answered Nov 26 '18 at 8:59
Wesley StrikWesley Strik
1,665423
$endgroup$
add a comment |
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2 Answers
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active
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2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Your answer to part a is correct. For part b note that $Aw=wA$ and $Al=lA$; expanding both sides of the given equation then quickly shows that they are indeed equal.
answered Nov 26 '18 at 8:37
ServaesServaes
23.1k33893
$endgroup$
add a comment |
$begingroup$
Your answer to part a is correct. For part b note that $Aw=wA$ and $Al=lA$; expanding both sides of the given equation then quickly shows that they are indeed equal.
answered Nov 26 '18 at 8:37
ServaesServaes
23.1k33893
$endgroup$
add a comment |
$begingroup$
Your answer to part a is correct. For part b note that $Aw=wA$ and $Al=lA$; expanding both sides of the given equation then quickly shows that they are indeed equal.
answered Nov 26 '18 at 8:37
ServaesServaes
23.1k33893
$endgroup$
Your answer to part a is correct. For part b note that $Aw=wA$ and $Al=lA$; expanding both sides of the given equation then quickly shows that they are indeed equal.
answered Nov 26 '18 at 8:37
ServaesServaes
23.1k33893
answered Nov 26 '18 at 8:37
ServaesServaes
23.1k33893
answered Nov 26 '18 at 8:37
ServaesServaes
23.1k33893
answered Nov 26 '18 at 8:37
ServaesServaes
23.1k33893
23.1k33893
add a comment |
add a comment |
$begingroup$
For b)
What you want to do is use the fact that matrix multiplication is linear.
$$(A−wI)(A−lI)= A cdot A -l A cdot I -w I cdot A + w l I cdot I $$
Now what you should use is that the identity matrix does not do much, we can always multiply any matrix by it, certainly it commutes $A cdot I = I cdot A$.
I hope this is enough of a hint and you can now rearrange terms so you can rewrite the left hand side to become the right hand side.
For C) In question $b)$ we learnt that you can interchange order, so what if we bring the term corresponding to $(Av=w_i v)$ to the back, we can certainly do so because of associativity ($(ABC) v=(AB)Cv=AB(Cv)$), what I mean is:
$$ ((A-w_1I)(A-w_2I)dots(A-w_iI) dots(A-w_kI))v$$ $$=left( (A-w_1I)(A-w_2I) dots(A-w_kI)(A-w_iI) right)v$$
Here we used this rule from $b$ repeatedly until the right term was at the front. Now just first multiply out the last term by associativity, do you notice something?
answered Nov 26 '18 at 8:59
Wesley StrikWesley Strik
1,665423
$endgroup$
add a comment |
$begingroup$
For b)
What you want to do is use the fact that matrix multiplication is linear.
$$(A−wI)(A−lI)= A cdot A -l A cdot I -w I cdot A + w l I cdot I $$
Now what you should use is that the identity matrix does not do much, we can always multiply any matrix by it, certainly it commutes $A cdot I = I cdot A$.
I hope this is enough of a hint and you can now rearrange terms so you can rewrite the left hand side to become the right hand side.
For C) In question $b)$ we learnt that you can interchange order, so what if we bring the term corresponding to $(Av=w_i v)$ to the back, we can certainly do so because of associativity ($(ABC) v=(AB)Cv=AB(Cv)$), what I mean is:
$$ ((A-w_1I)(A-w_2I)dots(A-w_iI) dots(A-w_kI))v$$ $$=left( (A-w_1I)(A-w_2I) dots(A-w_kI)(A-w_iI) right)v$$
Here we used this rule from $b$ repeatedly until the right term was at the front. Now just first multiply out the last term by associativity, do you notice something?
answered Nov 26 '18 at 8:59
Wesley StrikWesley Strik
1,665423
$endgroup$
add a comment |
$begingroup$
For b)
What you want to do is use the fact that matrix multiplication is linear.
$$(A−wI)(A−lI)= A cdot A -l A cdot I -w I cdot A + w l I cdot I $$
Now what you should use is that the identity matrix does not do much, we can always multiply any matrix by it, certainly it commutes $A cdot I = I cdot A$.
I hope this is enough of a hint and you can now rearrange terms so you can rewrite the left hand side to become the right hand side.
For C) In question $b)$ we learnt that you can interchange order, so what if we bring the term corresponding to $(Av=w_i v)$ to the back, we can certainly do so because of associativity ($(ABC) v=(AB)Cv=AB(Cv)$), what I mean is:
$$ ((A-w_1I)(A-w_2I)dots(A-w_iI) dots(A-w_kI))v$$ $$=left( (A-w_1I)(A-w_2I) dots(A-w_kI)(A-w_iI) right)v$$
Here we used this rule from $b$ repeatedly until the right term was at the front. Now just first multiply out the last term by associativity, do you notice something?
answered Nov 26 '18 at 8:59
Wesley StrikWesley Strik
1,665423
$endgroup$
For b)
What you want to do is use the fact that matrix multiplication is linear.
$$(A−wI)(A−lI)= A cdot A -l A cdot I -w I cdot A + w l I cdot I $$
Now what you should use is that the identity matrix does not do much, we can always multiply any matrix by it, certainly it commutes $A cdot I = I cdot A$.
I hope this is enough of a hint and you can now rearrange terms so you can rewrite the left hand side to become the right hand side.
For C) In question $b)$ we learnt that you can interchange order, so what if we bring the term corresponding to $(Av=w_i v)$ to the back, we can certainly do so because of associativity ($(ABC) v=(AB)Cv=AB(Cv)$), what I mean is:
$$ ((A-w_1I)(A-w_2I)dots(A-w_iI) dots(A-w_kI))v$$ $$=left( (A-w_1I)(A-w_2I) dots(A-w_kI)(A-w_iI) right)v$$
Here we used this rule from $b$ repeatedly until the right term was at the front. Now just first multiply out the last term by associativity, do you notice something?
answered Nov 26 '18 at 8:59
Wesley StrikWesley Strik
1,665423
edited Nov 26 '18 at 9:16
answered Nov 26 '18 at 8:59
Wesley StrikWesley Strik
1,665423
answered Nov 26 '18 at 8:59
Wesley StrikWesley Strik
1,665423
answered Nov 26 '18 at 8:59
Wesley StrikWesley Strik
1,665423
1,665423
add a comment |
add a comment |
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