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Suppose A is a square matrix that satisfies the matrix equation $A^5 = A$. Find, with justification, all possible eigenvalues of A.
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From what I understand, because any eigenvalue $lambda$ of A will solve the equation
$$
lambda^5=lambda\
0=(A^5−A)v=(lambda^5−λ)v
$$
Is this a sufficient answer? It seems lame to just say that any possible eigenvalue of A is the answer..
linear-algebra
edited Jul 27 at 0:10
gt6989b
30.2k22148
asked Jul 27 at 0:00
shroomyshroomy
91
add a comment |Â
up vote
0
down vote
favorite
From what I understand, because any eigenvalue $lambda$ of A will solve the equation
$$
lambda^5=lambda\
0=(A^5−A)v=(lambda^5−λ)v
$$
Is this a sufficient answer? It seems lame to just say that any possible eigenvalue of A is the answer..
linear-algebra
edited Jul 27 at 0:10
gt6989b
30.2k22148
asked Jul 27 at 0:00
shroomyshroomy
91
2
You've left out important context. What kind of matrix is $A$? See this introduction to posting mathematical expressions.
– hardmath
Jul 27 at 0:10
All thats given in the question is that it is a square matrix. No other information is provided.
– shroomyshroomy
Jul 27 at 0:14
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
From what I understand, because any eigenvalue $lambda$ of A will solve the equation
$$
lambda^5=lambda\
0=(A^5−A)v=(lambda^5−λ)v
$$
Is this a sufficient answer? It seems lame to just say that any possible eigenvalue of A is the answer..
linear-algebra
edited Jul 27 at 0:10
gt6989b
30.2k22148
asked Jul 27 at 0:00
shroomyshroomy
91
From what I understand, because any eigenvalue $lambda$ of A will solve the equation
$$
lambda^5=lambda\
0=(A^5−A)v=(lambda^5−λ)v
$$
Is this a sufficient answer? It seems lame to just say that any possible eigenvalue of A is the answer..
linear-algebra
edited Jul 27 at 0:10
gt6989b
30.2k22148
asked Jul 27 at 0:00
shroomyshroomy
91
edited Jul 27 at 0:10
gt6989b
30.2k22148
edited Jul 27 at 0:10
gt6989b
30.2k22148
edited Jul 27 at 0:10
gt6989b
30.2k22148
30.2k22148
asked Jul 27 at 0:00
shroomyshroomy
91
asked Jul 27 at 0:00
shroomyshroomy
91
asked Jul 27 at 0:00
shroomyshroomy
91
91
2
You've left out important context. What kind of matrix is $A$? See this introduction to posting mathematical expressions.
– hardmath
Jul 27 at 0:10
All thats given in the question is that it is a square matrix. No other information is provided.
– shroomyshroomy
Jul 27 at 0:14
add a comment |Â
2
You've left out important context. What kind of matrix is $A$? See this introduction to posting mathematical expressions.
– hardmath
Jul 27 at 0:10
All thats given in the question is that it is a square matrix. No other information is provided.
– shroomyshroomy
Jul 27 at 0:14
2
2
You've left out important context. What kind of matrix is $A$? See this introduction to posting mathematical expressions.
– hardmath
Jul 27 at 0:10
You've left out important context. What kind of matrix is $A$? See this introduction to posting mathematical expressions.
– hardmath
Jul 27 at 0:10
All thats given in the question is that it is a square matrix. No other information is provided.
– shroomyshroomy
Jul 27 at 0:14
All thats given in the question is that it is a square matrix. No other information is provided.
– shroomyshroomy
Jul 27 at 0:14
add a comment |Â
1 Answer
1
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1
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Certainly, assuming $v$ is an eigenvector, you have $v ne vec0$, thus you must have
$$
0 = lambda^5 - lambda = lambdaleft(lambda^4 - 1right)
= lambdaleft(lambda^2 - 1right)left(lambda^2 + 1right),
$$
which easily factorizes further... Can you finish this?
answered Jul 27 at 0:13
gt6989b
30.2k22148
so setting λ(λ^2−1)(λ^2+1)=0 we could solve for lambda = 0, or +/- 1?
– shroomyshroomy
Jul 27 at 0:21
@shroomyshroomy: there are two more roots than that. They are complex, but that is no problem. I am sure you are expected to find them as well.
– Ross Millikan
Jul 27 at 0:24
thats where you lose me.. I can factor to ) but im unsure what other complex roots one could find λ(λ−1)(λ+1)(λ^2+1)=0
– shroomyshroomy
Jul 27 at 0:29
1
Is there a way to verify with just the information given? This practice problem is from an old exam that weights it as 4/100 exam total points for reference. Most questions worth ~10-15 marks each. Seems like it should be a quick little answer type q
– shroomyshroomy
Jul 27 at 0:32
1
For a $4-$point question, I'm sure you would just be expected to answer $0, pm 1, pm i$
– saulspatz
Jul 27 at 1:26
 |Â
show 1 more comment
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Certainly, assuming $v$ is an eigenvector, you have $v ne vec0$, thus you must have
$$
0 = lambda^5 - lambda = lambdaleft(lambda^4 - 1right)
= lambdaleft(lambda^2 - 1right)left(lambda^2 + 1right),
$$
which easily factorizes further... Can you finish this?
answered Jul 27 at 0:13
gt6989b
30.2k22148
so setting λ(λ^2−1)(λ^2+1)=0 we could solve for lambda = 0, or +/- 1?
– shroomyshroomy
Jul 27 at 0:21
@shroomyshroomy: there are two more roots than that. They are complex, but that is no problem. I am sure you are expected to find them as well.
– Ross Millikan
Jul 27 at 0:24
thats where you lose me.. I can factor to ) but im unsure what other complex roots one could find λ(λ−1)(λ+1)(λ^2+1)=0
– shroomyshroomy
Jul 27 at 0:29
1
Is there a way to verify with just the information given? This practice problem is from an old exam that weights it as 4/100 exam total points for reference. Most questions worth ~10-15 marks each. Seems like it should be a quick little answer type q
– shroomyshroomy
Jul 27 at 0:32
1
For a $4-$point question, I'm sure you would just be expected to answer $0, pm 1, pm i$
– saulspatz
Jul 27 at 1:26
 |Â
show 1 more comment
up vote
1
down vote
Certainly, assuming $v$ is an eigenvector, you have $v ne vec0$, thus you must have
$$
0 = lambda^5 - lambda = lambdaleft(lambda^4 - 1right)
= lambdaleft(lambda^2 - 1right)left(lambda^2 + 1right),
$$
which easily factorizes further... Can you finish this?
answered Jul 27 at 0:13
gt6989b
30.2k22148
so setting λ(λ^2−1)(λ^2+1)=0 we could solve for lambda = 0, or +/- 1?
– shroomyshroomy
Jul 27 at 0:21
@shroomyshroomy: there are two more roots than that. They are complex, but that is no problem. I am sure you are expected to find them as well.
– Ross Millikan
Jul 27 at 0:24
thats where you lose me.. I can factor to ) but im unsure what other complex roots one could find λ(λ−1)(λ+1)(λ^2+1)=0
– shroomyshroomy
Jul 27 at 0:29
1
Is there a way to verify with just the information given? This practice problem is from an old exam that weights it as 4/100 exam total points for reference. Most questions worth ~10-15 marks each. Seems like it should be a quick little answer type q
– shroomyshroomy
Jul 27 at 0:32
1
For a $4-$point question, I'm sure you would just be expected to answer $0, pm 1, pm i$
– saulspatz
Jul 27 at 1:26
 |Â
show 1 more comment
up vote
1
down vote
up vote
1
down vote
Certainly, assuming $v$ is an eigenvector, you have $v ne vec0$, thus you must have
$$
0 = lambda^5 - lambda = lambdaleft(lambda^4 - 1right)
= lambdaleft(lambda^2 - 1right)left(lambda^2 + 1right),
$$
which easily factorizes further... Can you finish this?
answered Jul 27 at 0:13
gt6989b
30.2k22148
Certainly, assuming $v$ is an eigenvector, you have $v ne vec0$, thus you must have
$$
0 = lambda^5 - lambda = lambdaleft(lambda^4 - 1right)
= lambdaleft(lambda^2 - 1right)left(lambda^2 + 1right),
$$
which easily factorizes further... Can you finish this?
answered Jul 27 at 0:13
gt6989b
30.2k22148
answered Jul 27 at 0:13
gt6989b
30.2k22148
answered Jul 27 at 0:13
gt6989b
30.2k22148
answered Jul 27 at 0:13
gt6989b
30.2k22148
30.2k22148
so setting λ(λ^2−1)(λ^2+1)=0 we could solve for lambda = 0, or +/- 1?
– shroomyshroomy
Jul 27 at 0:21
@shroomyshroomy: there are two more roots than that. They are complex, but that is no problem. I am sure you are expected to find them as well.
– Ross Millikan
Jul 27 at 0:24
thats where you lose me.. I can factor to ) but im unsure what other complex roots one could find λ(λ−1)(λ+1)(λ^2+1)=0
– shroomyshroomy
Jul 27 at 0:29
1
Is there a way to verify with just the information given? This practice problem is from an old exam that weights it as 4/100 exam total points for reference. Most questions worth ~10-15 marks each. Seems like it should be a quick little answer type q
– shroomyshroomy
Jul 27 at 0:32
1
For a $4-$point question, I'm sure you would just be expected to answer $0, pm 1, pm i$
– saulspatz
Jul 27 at 1:26
 |Â
show 1 more comment
so setting λ(λ^2−1)(λ^2+1)=0 we could solve for lambda = 0, or +/- 1?
– shroomyshroomy
Jul 27 at 0:21
@shroomyshroomy: there are two more roots than that. They are complex, but that is no problem. I am sure you are expected to find them as well.
– Ross Millikan
Jul 27 at 0:24
thats where you lose me.. I can factor to ) but im unsure what other complex roots one could find λ(λ−1)(λ+1)(λ^2+1)=0
– shroomyshroomy
Jul 27 at 0:29
1
Is there a way to verify with just the information given? This practice problem is from an old exam that weights it as 4/100 exam total points for reference. Most questions worth ~10-15 marks each. Seems like it should be a quick little answer type q
– shroomyshroomy
Jul 27 at 0:32
1
For a $4-$point question, I'm sure you would just be expected to answer $0, pm 1, pm i$
– saulspatz
Jul 27 at 1:26
so setting λ(λ^2−1)(λ^2+1)=0 we could solve for lambda = 0, or +/- 1?
– shroomyshroomy
Jul 27 at 0:21
so setting λ(λ^2−1)(λ^2+1)=0 we could solve for lambda = 0, or +/- 1?
– shroomyshroomy
Jul 27 at 0:21
@shroomyshroomy: there are two more roots than that. They are complex, but that is no problem. I am sure you are expected to find them as well.
– Ross Millikan
Jul 27 at 0:24
@shroomyshroomy: there are two more roots than that. They are complex, but that is no problem. I am sure you are expected to find them as well.
– Ross Millikan
Jul 27 at 0:24
thats where you lose me.. I can factor to ) but im unsure what other complex roots one could find λ(λ−1)(λ+1)(λ^2+1)=0
– shroomyshroomy
Jul 27 at 0:29
thats where you lose me.. I can factor to ) but im unsure what other complex roots one could find λ(λ−1)(λ+1)(λ^2+1)=0
– shroomyshroomy
Jul 27 at 0:29
1
1
Is there a way to verify with just the information given? This practice problem is from an old exam that weights it as 4/100 exam total points for reference. Most questions worth ~10-15 marks each. Seems like it should be a quick little answer type q
– shroomyshroomy
Jul 27 at 0:32
Is there a way to verify with just the information given? This practice problem is from an old exam that weights it as 4/100 exam total points for reference. Most questions worth ~10-15 marks each. Seems like it should be a quick little answer type q
– shroomyshroomy
Jul 27 at 0:32
1
1
For a $4-$point question, I'm sure you would just be expected to answer $0, pm 1, pm i$
– saulspatz
Jul 27 at 1:26
For a $4-$point question, I'm sure you would just be expected to answer $0, pm 1, pm i$
– saulspatz
Jul 27 at 1:26
 |Â
show 1 more comment
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2
You've left out important context. What kind of matrix is $A$? See this introduction to posting mathematical expressions.
– hardmath
Jul 27 at 0:10
All thats given in the question is that it is a square matrix. No other information is provided.
– shroomyshroomy
Jul 27 at 0:14