Why does subtracting an eigenvalue from the diagonal and solving give the eigenspace?
$begingroup$
Given a matrix $A$, with eigenvalue $lambda$, why does substracting $lambda I$ from $A$, and then solving the matrix give us the eigenspace?
Like, I don't understand the intuition behind subtracting the two, to hence obtain the eigenspace.
I can't find an answer anywhere, a high level explanation would be very much appreciated! Thank you in advance.
linear-algebra matrices eigenvalues-eigenvectors
asked Dec 20 '18 at 10:11
Eoin Ó CoinnighEoin Ó Coinnigh
11
$endgroup$
add a comment |
$begingroup$
Given a matrix $A$, with eigenvalue $lambda$, why does substracting $lambda I$ from $A$, and then solving the matrix give us the eigenspace?
Like, I don't understand the intuition behind subtracting the two, to hence obtain the eigenspace.
I can't find an answer anywhere, a high level explanation would be very much appreciated! Thank you in advance.
linear-algebra matrices eigenvalues-eigenvectors
asked Dec 20 '18 at 10:11
Eoin Ó CoinnighEoin Ó Coinnigh
11
$endgroup$
$begingroup$
Eigenvector by definition is the non-zero $vin V$ such that $Av=lambda v$ where $lambda $ is the corressponding eigenvalue. So to find $v$: $Av=lambda viff (A-lambda I)v=0$, where $0$ is the zero matrix. In other words eigenspace associated to $lambda$ is the $N(A-lambda I)$.
$endgroup$
– Yadati Kiran
Dec 20 '18 at 10:16
$begingroup$
Eigenvectors are nonzero vectors that $A-lambda I$ sends to the zero vector. This can only happen if $lambda$ is chosen so that $A-lambda I$ has zero determinant (otherwise it would be invertible so it would only send the zero vector to the zero vector). That said, finding roots of the characteristic polynomial only gets you the eigenvalues, not the eigenvectors.
$endgroup$
– Ian
Dec 20 '18 at 10:19
$begingroup$
Thanks, that makes perfect sense.
$endgroup$
– Eoin Ó Coinnigh
Dec 20 '18 at 13:23
add a comment |
$begingroup$
Given a matrix $A$, with eigenvalue $lambda$, why does substracting $lambda I$ from $A$, and then solving the matrix give us the eigenspace?
Like, I don't understand the intuition behind subtracting the two, to hence obtain the eigenspace.
I can't find an answer anywhere, a high level explanation would be very much appreciated! Thank you in advance.
linear-algebra matrices eigenvalues-eigenvectors
asked Dec 20 '18 at 10:11
Eoin Ó CoinnighEoin Ó Coinnigh
11
$endgroup$
Given a matrix $A$, with eigenvalue $lambda$, why does substracting $lambda I$ from $A$, and then solving the matrix give us the eigenspace?
Like, I don't understand the intuition behind subtracting the two, to hence obtain the eigenspace.
I can't find an answer anywhere, a high level explanation would be very much appreciated! Thank you in advance.
linear-algebra matrices eigenvalues-eigenvectors
linear-algebra matrices eigenvalues-eigenvectors
asked Dec 20 '18 at 10:11
Eoin Ó CoinnighEoin Ó Coinnigh
11
asked Dec 20 '18 at 10:11
Eoin Ó CoinnighEoin Ó Coinnigh
11
edited Dec 20 '18 at 10:18
Eoin Ó Coinnigh
asked Dec 20 '18 at 10:11
Eoin Ó CoinnighEoin Ó Coinnigh
11
asked Dec 20 '18 at 10:11
Eoin Ó CoinnighEoin Ó Coinnigh
11
asked Dec 20 '18 at 10:11
Eoin Ó CoinnighEoin Ó Coinnigh
11
11
$begingroup$
Eigenvector by definition is the non-zero $vin V$ such that $Av=lambda v$ where $lambda $ is the corressponding eigenvalue. So to find $v$: $Av=lambda viff (A-lambda I)v=0$, where $0$ is the zero matrix. In other words eigenspace associated to $lambda$ is the $N(A-lambda I)$.
$endgroup$
– Yadati Kiran
Dec 20 '18 at 10:16
$begingroup$
Eigenvectors are nonzero vectors that $A-lambda I$ sends to the zero vector. This can only happen if $lambda$ is chosen so that $A-lambda I$ has zero determinant (otherwise it would be invertible so it would only send the zero vector to the zero vector). That said, finding roots of the characteristic polynomial only gets you the eigenvalues, not the eigenvectors.
$endgroup$
– Ian
Dec 20 '18 at 10:19
$begingroup$
Thanks, that makes perfect sense.
$endgroup$
– Eoin Ó Coinnigh
Dec 20 '18 at 13:23
add a comment |
$begingroup$
Eigenvector by definition is the non-zero $vin V$ such that $Av=lambda v$ where $lambda $ is the corressponding eigenvalue. So to find $v$: $Av=lambda viff (A-lambda I)v=0$, where $0$ is the zero matrix. In other words eigenspace associated to $lambda$ is the $N(A-lambda I)$.
$endgroup$
– Yadati Kiran
Dec 20 '18 at 10:16
$begingroup$
Eigenvectors are nonzero vectors that $A-lambda I$ sends to the zero vector. This can only happen if $lambda$ is chosen so that $A-lambda I$ has zero determinant (otherwise it would be invertible so it would only send the zero vector to the zero vector). That said, finding roots of the characteristic polynomial only gets you the eigenvalues, not the eigenvectors.
$endgroup$
– Ian
Dec 20 '18 at 10:19
$begingroup$
Thanks, that makes perfect sense.
$endgroup$
– Eoin Ó Coinnigh
Dec 20 '18 at 13:23
$begingroup$
Eigenvector by definition is the non-zero $vin V$ such that $Av=lambda v$ where $lambda $ is the corressponding eigenvalue. So to find $v$: $Av=lambda viff (A-lambda I)v=0$, where $0$ is the zero matrix. In other words eigenspace associated to $lambda$ is the $N(A-lambda I)$.
$endgroup$
– Yadati Kiran
Dec 20 '18 at 10:16
$begingroup$
Eigenvector by definition is the non-zero $vin V$ such that $Av=lambda v$ where $lambda $ is the corressponding eigenvalue. So to find $v$: $Av=lambda viff (A-lambda I)v=0$, where $0$ is the zero matrix. In other words eigenspace associated to $lambda$ is the $N(A-lambda I)$.
$endgroup$
– Yadati Kiran
Dec 20 '18 at 10:16
$begingroup$
Eigenvectors are nonzero vectors that $A-lambda I$ sends to the zero vector. This can only happen if $lambda$ is chosen so that $A-lambda I$ has zero determinant (otherwise it would be invertible so it would only send the zero vector to the zero vector). That said, finding roots of the characteristic polynomial only gets you the eigenvalues, not the eigenvectors.
$endgroup$
– Ian
Dec 20 '18 at 10:19
$begingroup$
Eigenvectors are nonzero vectors that $A-lambda I$ sends to the zero vector. This can only happen if $lambda$ is chosen so that $A-lambda I$ has zero determinant (otherwise it would be invertible so it would only send the zero vector to the zero vector). That said, finding roots of the characteristic polynomial only gets you the eigenvalues, not the eigenvectors.
$endgroup$
– Ian
Dec 20 '18 at 10:19
$begingroup$
Thanks, that makes perfect sense.
$endgroup$
– Eoin Ó Coinnigh
Dec 20 '18 at 13:23
$begingroup$
Thanks, that makes perfect sense.
$endgroup$
– Eoin Ó Coinnigh
Dec 20 '18 at 13:23
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Let $V$ a vector space. The eigenspace associated to the eigenvalue $lambda $ is the set $E_lambda $ of vector $vin V$ s.t. $$Av=lambda v.$$
In otherword, $$E_lambda ={vin Vmid Av=lambda v}.$$
Therefore $$vin E_lambda iff Av=lambda viff (A-lambda I)v=Av-lambda v=0$$
$$iff vin ker(A-lambda I).$$
We conclude that $$E_lambda =ker(A-lambda I).$$
edited Dec 20 '18 at 20:08
Henning Makholm
241k17308546
answered Dec 20 '18 at 10:17
SurbSurb
38.2k94476
$endgroup$
add a comment |
$begingroup$
Let $V$ be a $n$-dimensional vector space over any field and $A$ any $(n times n)$-Matrix acting on that space. Remember the eigenvalue equation $$Av = lambda v$$ for $v neq 0$. Now, rewrite $v=I v$, where $I$ is the identity matrix, and substract it from the eigenvalue equation, hence
$$(A- lambda I )v = 0.$$ Let us assume for a moment that the matrix $A-lambda I$ is invertible. For that we'd multiply both sides with the inverse and we'd get $v=0$. A contradiction to our requirement that $v neq 0$. Therefore, $A-lambda I$ cannot be invertible and so $det(A-lambda I) = 0$. But the characteristic polyomial is exactly the determinant of $A-lambda I$.
On the other hand, if $lambda$ is a zero of the characteristic polynomial, i.e. $A-lambda I$ is not invertible, $(A-lambda I) v =0$ has a non-trivial solution and its solution is an eigenvector corresponding to the eigenvalue $lambda$.
answered Dec 20 '18 at 10:28
YoungMathYoungMath
195111
$endgroup$
add a comment |
$begingroup$
Notice that
begin{align*}
Av= lambda v,~v ne 0
Longleftrightarrow (A - lambda I)v = 0,~v ne 0 Longleftrightarrow v intext{Nul}(A- lambda I),~v ne 0.
end{align*}
answered Dec 20 '18 at 20:05
Pietro PaparellaPietro Paparella
1,474615
$endgroup$
add a comment |
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3 Answers
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active
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3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $V$ a vector space. The eigenspace associated to the eigenvalue $lambda $ is the set $E_lambda $ of vector $vin V$ s.t. $$Av=lambda v.$$
In otherword, $$E_lambda ={vin Vmid Av=lambda v}.$$
Therefore $$vin E_lambda iff Av=lambda viff (A-lambda I)v=Av-lambda v=0$$
$$iff vin ker(A-lambda I).$$
We conclude that $$E_lambda =ker(A-lambda I).$$
edited Dec 20 '18 at 20:08
Henning Makholm
241k17308546
answered Dec 20 '18 at 10:17
SurbSurb
38.2k94476
$endgroup$
add a comment |
$begingroup$
Let $V$ a vector space. The eigenspace associated to the eigenvalue $lambda $ is the set $E_lambda $ of vector $vin V$ s.t. $$Av=lambda v.$$
In otherword, $$E_lambda ={vin Vmid Av=lambda v}.$$
Therefore $$vin E_lambda iff Av=lambda viff (A-lambda I)v=Av-lambda v=0$$
$$iff vin ker(A-lambda I).$$
We conclude that $$E_lambda =ker(A-lambda I).$$
edited Dec 20 '18 at 20:08
Henning Makholm
241k17308546
answered Dec 20 '18 at 10:17
SurbSurb
38.2k94476
$endgroup$
add a comment |
$begingroup$
Let $V$ a vector space. The eigenspace associated to the eigenvalue $lambda $ is the set $E_lambda $ of vector $vin V$ s.t. $$Av=lambda v.$$
In otherword, $$E_lambda ={vin Vmid Av=lambda v}.$$
Therefore $$vin E_lambda iff Av=lambda viff (A-lambda I)v=Av-lambda v=0$$
$$iff vin ker(A-lambda I).$$
We conclude that $$E_lambda =ker(A-lambda I).$$
edited Dec 20 '18 at 20:08
Henning Makholm
241k17308546
answered Dec 20 '18 at 10:17
SurbSurb
38.2k94476
$endgroup$
Let $V$ a vector space. The eigenspace associated to the eigenvalue $lambda $ is the set $E_lambda $ of vector $vin V$ s.t. $$Av=lambda v.$$
In otherword, $$E_lambda ={vin Vmid Av=lambda v}.$$
Therefore $$vin E_lambda iff Av=lambda viff (A-lambda I)v=Av-lambda v=0$$
$$iff vin ker(A-lambda I).$$
We conclude that $$E_lambda =ker(A-lambda I).$$
edited Dec 20 '18 at 20:08
Henning Makholm
241k17308546
answered Dec 20 '18 at 10:17
SurbSurb
38.2k94476
edited Dec 20 '18 at 20:08
Henning Makholm
241k17308546
edited Dec 20 '18 at 20:08
Henning Makholm
241k17308546
edited Dec 20 '18 at 20:08
Henning Makholm
241k17308546
241k17308546
answered Dec 20 '18 at 10:17
SurbSurb
38.2k94476
answered Dec 20 '18 at 10:17
SurbSurb
38.2k94476
answered Dec 20 '18 at 10:17
SurbSurb
38.2k94476
38.2k94476
add a comment |
add a comment |
$begingroup$
Let $V$ be a $n$-dimensional vector space over any field and $A$ any $(n times n)$-Matrix acting on that space. Remember the eigenvalue equation $$Av = lambda v$$ for $v neq 0$. Now, rewrite $v=I v$, where $I$ is the identity matrix, and substract it from the eigenvalue equation, hence
$$(A- lambda I )v = 0.$$ Let us assume for a moment that the matrix $A-lambda I$ is invertible. For that we'd multiply both sides with the inverse and we'd get $v=0$. A contradiction to our requirement that $v neq 0$. Therefore, $A-lambda I$ cannot be invertible and so $det(A-lambda I) = 0$. But the characteristic polyomial is exactly the determinant of $A-lambda I$.
On the other hand, if $lambda$ is a zero of the characteristic polynomial, i.e. $A-lambda I$ is not invertible, $(A-lambda I) v =0$ has a non-trivial solution and its solution is an eigenvector corresponding to the eigenvalue $lambda$.
answered Dec 20 '18 at 10:28
YoungMathYoungMath
195111
$endgroup$
add a comment |
$begingroup$
Let $V$ be a $n$-dimensional vector space over any field and $A$ any $(n times n)$-Matrix acting on that space. Remember the eigenvalue equation $$Av = lambda v$$ for $v neq 0$. Now, rewrite $v=I v$, where $I$ is the identity matrix, and substract it from the eigenvalue equation, hence
$$(A- lambda I )v = 0.$$ Let us assume for a moment that the matrix $A-lambda I$ is invertible. For that we'd multiply both sides with the inverse and we'd get $v=0$. A contradiction to our requirement that $v neq 0$. Therefore, $A-lambda I$ cannot be invertible and so $det(A-lambda I) = 0$. But the characteristic polyomial is exactly the determinant of $A-lambda I$.
On the other hand, if $lambda$ is a zero of the characteristic polynomial, i.e. $A-lambda I$ is not invertible, $(A-lambda I) v =0$ has a non-trivial solution and its solution is an eigenvector corresponding to the eigenvalue $lambda$.
answered Dec 20 '18 at 10:28
YoungMathYoungMath
195111
$endgroup$
add a comment |
$begingroup$
Let $V$ be a $n$-dimensional vector space over any field and $A$ any $(n times n)$-Matrix acting on that space. Remember the eigenvalue equation $$Av = lambda v$$ for $v neq 0$. Now, rewrite $v=I v$, where $I$ is the identity matrix, and substract it from the eigenvalue equation, hence
$$(A- lambda I )v = 0.$$ Let us assume for a moment that the matrix $A-lambda I$ is invertible. For that we'd multiply both sides with the inverse and we'd get $v=0$. A contradiction to our requirement that $v neq 0$. Therefore, $A-lambda I$ cannot be invertible and so $det(A-lambda I) = 0$. But the characteristic polyomial is exactly the determinant of $A-lambda I$.
On the other hand, if $lambda$ is a zero of the characteristic polynomial, i.e. $A-lambda I$ is not invertible, $(A-lambda I) v =0$ has a non-trivial solution and its solution is an eigenvector corresponding to the eigenvalue $lambda$.
answered Dec 20 '18 at 10:28
YoungMathYoungMath
195111
$endgroup$
Let $V$ be a $n$-dimensional vector space over any field and $A$ any $(n times n)$-Matrix acting on that space. Remember the eigenvalue equation $$Av = lambda v$$ for $v neq 0$. Now, rewrite $v=I v$, where $I$ is the identity matrix, and substract it from the eigenvalue equation, hence
$$(A- lambda I )v = 0.$$ Let us assume for a moment that the matrix $A-lambda I$ is invertible. For that we'd multiply both sides with the inverse and we'd get $v=0$. A contradiction to our requirement that $v neq 0$. Therefore, $A-lambda I$ cannot be invertible and so $det(A-lambda I) = 0$. But the characteristic polyomial is exactly the determinant of $A-lambda I$.
On the other hand, if $lambda$ is a zero of the characteristic polynomial, i.e. $A-lambda I$ is not invertible, $(A-lambda I) v =0$ has a non-trivial solution and its solution is an eigenvector corresponding to the eigenvalue $lambda$.
answered Dec 20 '18 at 10:28
YoungMathYoungMath
195111
answered Dec 20 '18 at 10:28
YoungMathYoungMath
195111
answered Dec 20 '18 at 10:28
YoungMathYoungMath
195111
answered Dec 20 '18 at 10:28
YoungMathYoungMath
195111
195111
add a comment |
add a comment |
$begingroup$
Notice that
begin{align*}
Av= lambda v,~v ne 0
Longleftrightarrow (A - lambda I)v = 0,~v ne 0 Longleftrightarrow v intext{Nul}(A- lambda I),~v ne 0.
end{align*}
answered Dec 20 '18 at 20:05
Pietro PaparellaPietro Paparella
1,474615
$endgroup$
add a comment |
$begingroup$
Notice that
begin{align*}
Av= lambda v,~v ne 0
Longleftrightarrow (A - lambda I)v = 0,~v ne 0 Longleftrightarrow v intext{Nul}(A- lambda I),~v ne 0.
end{align*}
answered Dec 20 '18 at 20:05
Pietro PaparellaPietro Paparella
1,474615
$endgroup$
add a comment |
$begingroup$
Notice that
begin{align*}
Av= lambda v,~v ne 0
Longleftrightarrow (A - lambda I)v = 0,~v ne 0 Longleftrightarrow v intext{Nul}(A- lambda I),~v ne 0.
end{align*}
answered Dec 20 '18 at 20:05
Pietro PaparellaPietro Paparella
1,474615
$endgroup$
Notice that
begin{align*}
Av= lambda v,~v ne 0
Longleftrightarrow (A - lambda I)v = 0,~v ne 0 Longleftrightarrow v intext{Nul}(A- lambda I),~v ne 0.
end{align*}
answered Dec 20 '18 at 20:05
Pietro PaparellaPietro Paparella
1,474615
answered Dec 20 '18 at 20:05
Pietro PaparellaPietro Paparella
1,474615
answered Dec 20 '18 at 20:05
Pietro PaparellaPietro Paparella
1,474615
answered Dec 20 '18 at 20:05
Pietro PaparellaPietro Paparella
1,474615
1,474615
add a comment |
add a comment |
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$begingroup$
Eigenvector by definition is the non-zero $vin V$ such that $Av=lambda v$ where $lambda $ is the corressponding eigenvalue. So to find $v$: $Av=lambda viff (A-lambda I)v=0$, where $0$ is the zero matrix. In other words eigenspace associated to $lambda$ is the $N(A-lambda I)$.
$endgroup$
– Yadati Kiran
Dec 20 '18 at 10:16
$begingroup$
Eigenvectors are nonzero vectors that $A-lambda I$ sends to the zero vector. This can only happen if $lambda$ is chosen so that $A-lambda I$ has zero determinant (otherwise it would be invertible so it would only send the zero vector to the zero vector). That said, finding roots of the characteristic polynomial only gets you the eigenvalues, not the eigenvectors.
$endgroup$
– Ian
Dec 20 '18 at 10:19
$begingroup$
Thanks, that makes perfect sense.
$endgroup$
– Eoin Ó Coinnigh
Dec 20 '18 at 13:23