Real n-by-n Matrices…
Clash Royale CLAN TAG#URR8PPP
$begingroup$
Let $M_n(mathbbR)$ denote the vector space of real $ntimes n$ matrics, and let $A in M_n(mathbbR)$.
Part (a) of this question says: Suppose $B in M_n(mathbbR)$ such that $AB = I_n$ (the $n times n$ identity matrix. If $C in M_n(mathbbR)$ such that $CA = 0$, then prove $C = 0$.
I have already proven part (a).
Part (b) asks: Assume there exists a least positive integer $m$ such that $t_0I + t_1A + dots + t_mA^m = 0$ for some $t_0, dots, t_m in mathbbR$ with $t_m neq 0$. Also, suppose that $AB = I_n$ for some $B in M_n(mathbbR)$.Prove that $t_0 neq 0$. (Hint: Use the result from part (a)).
My idea is to use induction on $m$. That is, suppose
$$t_0I = 0.$$
But this implies that $t_0 = 0$ since $I$ is the identity. But we could see this as
beginalign*
t_0I &= 0
\
t_0AB &= 0
\
t_0CAB &= 0
\
t_0C &= 0.
endalign*
But I don't think this is the right approach. So I have a few questions: (i) Is this the correct approach? (ii) How do I use the result from part (a) properly? (iii) What's a good resource for these types of questions? The book that I am using is "Linear Algebra Done Right by Sheldon Axler".
I am studying for my linear algebra comp in a few weeks so any help is appreciated!
linear-algebra matrices vector-spaces
asked Jan 1 at 20:25
Taylor McMillanTaylor McMillan
695
$endgroup$
add a comment |
$begingroup$
Let $M_n(mathbbR)$ denote the vector space of real $ntimes n$ matrics, and let $A in M_n(mathbbR)$.
Part (a) of this question says: Suppose $B in M_n(mathbbR)$ such that $AB = I_n$ (the $n times n$ identity matrix. If $C in M_n(mathbbR)$ such that $CA = 0$, then prove $C = 0$.
I have already proven part (a).
Part (b) asks: Assume there exists a least positive integer $m$ such that $t_0I + t_1A + dots + t_mA^m = 0$ for some $t_0, dots, t_m in mathbbR$ with $t_m neq 0$. Also, suppose that $AB = I_n$ for some $B in M_n(mathbbR)$.Prove that $t_0 neq 0$. (Hint: Use the result from part (a)).
My idea is to use induction on $m$. That is, suppose
$$t_0I = 0.$$
But this implies that $t_0 = 0$ since $I$ is the identity. But we could see this as
beginalign*
t_0I &= 0
\
t_0AB &= 0
\
t_0CAB &= 0
\
t_0C &= 0.
endalign*
But I don't think this is the right approach. So I have a few questions: (i) Is this the correct approach? (ii) How do I use the result from part (a) properly? (iii) What's a good resource for these types of questions? The book that I am using is "Linear Algebra Done Right by Sheldon Axler".
I am studying for my linear algebra comp in a few weeks so any help is appreciated!
linear-algebra matrices vector-spaces
asked Jan 1 at 20:25
Taylor McMillanTaylor McMillan
695
$endgroup$
$begingroup$
Once you have got $t_0=0$ then anything you multiply it with gonna be zero. Since we assume $t_mne 0$ then $mge 1$.
$endgroup$
– A.Γ.
Jan 1 at 20:44
add a comment |
$begingroup$
Let $M_n(mathbbR)$ denote the vector space of real $ntimes n$ matrics, and let $A in M_n(mathbbR)$.
Part (a) of this question says: Suppose $B in M_n(mathbbR)$ such that $AB = I_n$ (the $n times n$ identity matrix. If $C in M_n(mathbbR)$ such that $CA = 0$, then prove $C = 0$.
I have already proven part (a).
Part (b) asks: Assume there exists a least positive integer $m$ such that $t_0I + t_1A + dots + t_mA^m = 0$ for some $t_0, dots, t_m in mathbbR$ with $t_m neq 0$. Also, suppose that $AB = I_n$ for some $B in M_n(mathbbR)$.Prove that $t_0 neq 0$. (Hint: Use the result from part (a)).
My idea is to use induction on $m$. That is, suppose
$$t_0I = 0.$$
But this implies that $t_0 = 0$ since $I$ is the identity. But we could see this as
beginalign*
t_0I &= 0
\
t_0AB &= 0
\
t_0CAB &= 0
\
t_0C &= 0.
endalign*
But I don't think this is the right approach. So I have a few questions: (i) Is this the correct approach? (ii) How do I use the result from part (a) properly? (iii) What's a good resource for these types of questions? The book that I am using is "Linear Algebra Done Right by Sheldon Axler".
I am studying for my linear algebra comp in a few weeks so any help is appreciated!
linear-algebra matrices vector-spaces
asked Jan 1 at 20:25
Taylor McMillanTaylor McMillan
695
$endgroup$
Let $M_n(mathbbR)$ denote the vector space of real $ntimes n$ matrics, and let $A in M_n(mathbbR)$.
Part (a) of this question says: Suppose $B in M_n(mathbbR)$ such that $AB = I_n$ (the $n times n$ identity matrix. If $C in M_n(mathbbR)$ such that $CA = 0$, then prove $C = 0$.
I have already proven part (a).
Part (b) asks: Assume there exists a least positive integer $m$ such that $t_0I + t_1A + dots + t_mA^m = 0$ for some $t_0, dots, t_m in mathbbR$ with $t_m neq 0$. Also, suppose that $AB = I_n$ for some $B in M_n(mathbbR)$.Prove that $t_0 neq 0$. (Hint: Use the result from part (a)).
My idea is to use induction on $m$. That is, suppose
$$t_0I = 0.$$
But this implies that $t_0 = 0$ since $I$ is the identity. But we could see this as
beginalign*
t_0I &= 0
\
t_0AB &= 0
\
t_0CAB &= 0
\
t_0C &= 0.
endalign*
But I don't think this is the right approach. So I have a few questions: (i) Is this the correct approach? (ii) How do I use the result from part (a) properly? (iii) What's a good resource for these types of questions? The book that I am using is "Linear Algebra Done Right by Sheldon Axler".
I am studying for my linear algebra comp in a few weeks so any help is appreciated!
linear-algebra matrices vector-spaces
linear-algebra matrices vector-spaces
asked Jan 1 at 20:25
Taylor McMillanTaylor McMillan
695
asked Jan 1 at 20:25
Taylor McMillanTaylor McMillan
695
asked Jan 1 at 20:25
Taylor McMillanTaylor McMillan
695
asked Jan 1 at 20:25
Taylor McMillanTaylor McMillan
695
asked Jan 1 at 20:25
Taylor McMillanTaylor McMillan
695
695
$begingroup$
Once you have got $t_0=0$ then anything you multiply it with gonna be zero. Since we assume $t_mne 0$ then $mge 1$.
$endgroup$
– A.Γ.
Jan 1 at 20:44
add a comment |
$begingroup$
Once you have got $t_0=0$ then anything you multiply it with gonna be zero. Since we assume $t_mne 0$ then $mge 1$.
$endgroup$
– A.Γ.
Jan 1 at 20:44
$begingroup$
Once you have got $t_0=0$ then anything you multiply it with gonna be zero. Since we assume $t_mne 0$ then $mge 1$.
$endgroup$
– A.Γ.
Jan 1 at 20:44
$begingroup$
Once you have got $t_0=0$ then anything you multiply it with gonna be zero. Since we assume $t_mne 0$ then $mge 1$.
$endgroup$
– A.Γ.
Jan 1 at 20:44
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Suppose exists $B$ such that $AB=I$ and that $m$ is the smallest integer such that, exist coefficients $t_i$ such that,
$$
t_0 I + t_1 A + cdots + t_m A^m = 0
$$
If $t_0=0$ you have,
$$
t_1 A + cdots + t_m A^m = 0
$$
Now, see if you can use the fact that $AB = I$ to find new coefficients $t_i'$ satisfying
$$
t_0'I + t_1'A + cdots t_n ' A^n = 0
$$
where $n<m$.
Since by assumption $m$ is the smallest integer such that this type of combination exists, then you have a contradiction.
I liked that book as a first course in linear algebra. You could also see if past exam problems are available to study from. To me, studying for exams is somewhat dependent on the format of exam itself so its hard to give general advice.
answered Jan 1 at 20:46
tchtch
639210
$endgroup$
$begingroup$
Thank you for this. My goto for using contradiction is typically when the problem screams "this is obviously true". Didn't really think that this result was intuitive so I didn't take this route. When in doubt try it out, I guess.
$endgroup$
– Taylor McMillan
Jan 1 at 21:05
$begingroup$
That is what I am doing. My university has several past exams posted I am doing all of those problems.
$endgroup$
– Taylor McMillan
Jan 1 at 21:07
add a comment |
$begingroup$
Hint: assume $t_0=0$ then
$$
t_1A+ldots+t_mA^m=0.
$$
Multiply by $B$ and get the contradiction with the minimality of $m$.
answered Jan 1 at 20:46
A.Γ.A.Γ.
22.7k32656
$endgroup$
add a comment |
$begingroup$
Here is another hint/solution:
- assume it to equal $0$,
- multiply your equation by $A$, and factor by $A$,
- using part (a), get a contradiction.
edited Jan 1 at 22:14
amWhy
192k28225439
answered Jan 1 at 21:25
Cauchy is my masterCauchy is my master
92
$endgroup$
$begingroup$
Bullet lists need a blank line before them in order to appear—I've added one. Welcome to the site!
$endgroup$
– timtfj
Jan 1 at 22:18
add a comment |
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Suppose exists $B$ such that $AB=I$ and that $m$ is the smallest integer such that, exist coefficients $t_i$ such that,
$$
t_0 I + t_1 A + cdots + t_m A^m = 0
$$
If $t_0=0$ you have,
$$
t_1 A + cdots + t_m A^m = 0
$$
Now, see if you can use the fact that $AB = I$ to find new coefficients $t_i'$ satisfying
$$
t_0'I + t_1'A + cdots t_n ' A^n = 0
$$
where $n<m$.
Since by assumption $m$ is the smallest integer such that this type of combination exists, then you have a contradiction.
I liked that book as a first course in linear algebra. You could also see if past exam problems are available to study from. To me, studying for exams is somewhat dependent on the format of exam itself so its hard to give general advice.
answered Jan 1 at 20:46
tchtch
639210
$endgroup$
$begingroup$
Thank you for this. My goto for using contradiction is typically when the problem screams "this is obviously true". Didn't really think that this result was intuitive so I didn't take this route. When in doubt try it out, I guess.
$endgroup$
– Taylor McMillan
Jan 1 at 21:05
$begingroup$
That is what I am doing. My university has several past exams posted I am doing all of those problems.
$endgroup$
– Taylor McMillan
Jan 1 at 21:07
add a comment |
$begingroup$
Suppose exists $B$ such that $AB=I$ and that $m$ is the smallest integer such that, exist coefficients $t_i$ such that,
$$
t_0 I + t_1 A + cdots + t_m A^m = 0
$$
If $t_0=0$ you have,
$$
t_1 A + cdots + t_m A^m = 0
$$
Now, see if you can use the fact that $AB = I$ to find new coefficients $t_i'$ satisfying
$$
t_0'I + t_1'A + cdots t_n ' A^n = 0
$$
where $n<m$.
Since by assumption $m$ is the smallest integer such that this type of combination exists, then you have a contradiction.
I liked that book as a first course in linear algebra. You could also see if past exam problems are available to study from. To me, studying for exams is somewhat dependent on the format of exam itself so its hard to give general advice.
answered Jan 1 at 20:46
tchtch
639210
$endgroup$
$begingroup$
Thank you for this. My goto for using contradiction is typically when the problem screams "this is obviously true". Didn't really think that this result was intuitive so I didn't take this route. When in doubt try it out, I guess.
$endgroup$
– Taylor McMillan
Jan 1 at 21:05
$begingroup$
That is what I am doing. My university has several past exams posted I am doing all of those problems.
$endgroup$
– Taylor McMillan
Jan 1 at 21:07
add a comment |
$begingroup$
Suppose exists $B$ such that $AB=I$ and that $m$ is the smallest integer such that, exist coefficients $t_i$ such that,
$$
t_0 I + t_1 A + cdots + t_m A^m = 0
$$
If $t_0=0$ you have,
$$
t_1 A + cdots + t_m A^m = 0
$$
Now, see if you can use the fact that $AB = I$ to find new coefficients $t_i'$ satisfying
$$
t_0'I + t_1'A + cdots t_n ' A^n = 0
$$
where $n<m$.
Since by assumption $m$ is the smallest integer such that this type of combination exists, then you have a contradiction.
I liked that book as a first course in linear algebra. You could also see if past exam problems are available to study from. To me, studying for exams is somewhat dependent on the format of exam itself so its hard to give general advice.
answered Jan 1 at 20:46
tchtch
639210
$endgroup$
Suppose exists $B$ such that $AB=I$ and that $m$ is the smallest integer such that, exist coefficients $t_i$ such that,
$$
t_0 I + t_1 A + cdots + t_m A^m = 0
$$
If $t_0=0$ you have,
$$
t_1 A + cdots + t_m A^m = 0
$$
Now, see if you can use the fact that $AB = I$ to find new coefficients $t_i'$ satisfying
$$
t_0'I + t_1'A + cdots t_n ' A^n = 0
$$
where $n<m$.
Since by assumption $m$ is the smallest integer such that this type of combination exists, then you have a contradiction.
I liked that book as a first course in linear algebra. You could also see if past exam problems are available to study from. To me, studying for exams is somewhat dependent on the format of exam itself so its hard to give general advice.
answered Jan 1 at 20:46
tchtch
639210
answered Jan 1 at 20:46
tchtch
639210
answered Jan 1 at 20:46
tchtch
639210
answered Jan 1 at 20:46
tchtch
639210
639210
$begingroup$
Thank you for this. My goto for using contradiction is typically when the problem screams "this is obviously true". Didn't really think that this result was intuitive so I didn't take this route. When in doubt try it out, I guess.
$endgroup$
– Taylor McMillan
Jan 1 at 21:05
$begingroup$
That is what I am doing. My university has several past exams posted I am doing all of those problems.
$endgroup$
– Taylor McMillan
Jan 1 at 21:07
add a comment |
$begingroup$
Thank you for this. My goto for using contradiction is typically when the problem screams "this is obviously true". Didn't really think that this result was intuitive so I didn't take this route. When in doubt try it out, I guess.
$endgroup$
– Taylor McMillan
Jan 1 at 21:05
$begingroup$
That is what I am doing. My university has several past exams posted I am doing all of those problems.
$endgroup$
– Taylor McMillan
Jan 1 at 21:07
$begingroup$
Thank you for this. My goto for using contradiction is typically when the problem screams "this is obviously true". Didn't really think that this result was intuitive so I didn't take this route. When in doubt try it out, I guess.
$endgroup$
– Taylor McMillan
Jan 1 at 21:05
$begingroup$
Thank you for this. My goto for using contradiction is typically when the problem screams "this is obviously true". Didn't really think that this result was intuitive so I didn't take this route. When in doubt try it out, I guess.
$endgroup$
– Taylor McMillan
Jan 1 at 21:05
$begingroup$
That is what I am doing. My university has several past exams posted I am doing all of those problems.
$endgroup$
– Taylor McMillan
Jan 1 at 21:07
$begingroup$
That is what I am doing. My university has several past exams posted I am doing all of those problems.
$endgroup$
– Taylor McMillan
Jan 1 at 21:07
add a comment |
$begingroup$
Hint: assume $t_0=0$ then
$$
t_1A+ldots+t_mA^m=0.
$$
Multiply by $B$ and get the contradiction with the minimality of $m$.
answered Jan 1 at 20:46
A.Γ.A.Γ.
22.7k32656
$endgroup$
add a comment |
$begingroup$
Hint: assume $t_0=0$ then
$$
t_1A+ldots+t_mA^m=0.
$$
Multiply by $B$ and get the contradiction with the minimality of $m$.
answered Jan 1 at 20:46
A.Γ.A.Γ.
22.7k32656
$endgroup$
add a comment |
$begingroup$
Hint: assume $t_0=0$ then
$$
t_1A+ldots+t_mA^m=0.
$$
Multiply by $B$ and get the contradiction with the minimality of $m$.
answered Jan 1 at 20:46
A.Γ.A.Γ.
22.7k32656
$endgroup$
Hint: assume $t_0=0$ then
$$
t_1A+ldots+t_mA^m=0.
$$
Multiply by $B$ and get the contradiction with the minimality of $m$.
answered Jan 1 at 20:46
A.Γ.A.Γ.
22.7k32656
answered Jan 1 at 20:46
A.Γ.A.Γ.
22.7k32656
answered Jan 1 at 20:46
A.Γ.A.Γ.
22.7k32656
answered Jan 1 at 20:46
A.Γ.A.Γ.
22.7k32656
22.7k32656
add a comment |
add a comment |
$begingroup$
Here is another hint/solution:
- assume it to equal $0$,
- multiply your equation by $A$, and factor by $A$,
- using part (a), get a contradiction.
edited Jan 1 at 22:14
amWhy
192k28225439
answered Jan 1 at 21:25
Cauchy is my masterCauchy is my master
92
$endgroup$
$begingroup$
Bullet lists need a blank line before them in order to appear—I've added one. Welcome to the site!
$endgroup$
– timtfj
Jan 1 at 22:18
add a comment |
$begingroup$
Here is another hint/solution:
- assume it to equal $0$,
- multiply your equation by $A$, and factor by $A$,
- using part (a), get a contradiction.
edited Jan 1 at 22:14
amWhy
192k28225439
answered Jan 1 at 21:25
Cauchy is my masterCauchy is my master
92
$endgroup$
$begingroup$
Bullet lists need a blank line before them in order to appear—I've added one. Welcome to the site!
$endgroup$
– timtfj
Jan 1 at 22:18
add a comment |
$begingroup$
Here is another hint/solution:
- assume it to equal $0$,
- multiply your equation by $A$, and factor by $A$,
- using part (a), get a contradiction.
edited Jan 1 at 22:14
amWhy
192k28225439
answered Jan 1 at 21:25
Cauchy is my masterCauchy is my master
92
$endgroup$
Here is another hint/solution:
- assume it to equal $0$,
- multiply your equation by $A$, and factor by $A$,
- using part (a), get a contradiction.
edited Jan 1 at 22:14
amWhy
192k28225439
answered Jan 1 at 21:25
Cauchy is my masterCauchy is my master
92
edited Jan 1 at 22:14
amWhy
192k28225439
edited Jan 1 at 22:14
amWhy
192k28225439
edited Jan 1 at 22:14
amWhy
192k28225439
192k28225439
answered Jan 1 at 21:25
Cauchy is my masterCauchy is my master
92
answered Jan 1 at 21:25
Cauchy is my masterCauchy is my master
92
answered Jan 1 at 21:25
Cauchy is my masterCauchy is my master
92
92
$begingroup$
Bullet lists need a blank line before them in order to appear—I've added one. Welcome to the site!
$endgroup$
– timtfj
Jan 1 at 22:18
add a comment |
$begingroup$
Bullet lists need a blank line before them in order to appear—I've added one. Welcome to the site!
$endgroup$
– timtfj
Jan 1 at 22:18
$begingroup$
Bullet lists need a blank line before them in order to appear—I've added one. Welcome to the site!
$endgroup$
– timtfj
Jan 1 at 22:18
$begingroup$
Bullet lists need a blank line before them in order to appear—I've added one. Welcome to the site!
$endgroup$
– timtfj
Jan 1 at 22:18
add a comment |
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$begingroup$
Once you have got $t_0=0$ then anything you multiply it with gonna be zero. Since we assume $t_mne 0$ then $mge 1$.
$endgroup$
– A.Γ.
Jan 1 at 20:44