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How to prove that eigenvectors from different eigenvalues are linearly independent
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How can I prove that if I have $n$ eigenvectors from different eigenvalues, they are all linearly independent?
linear-algebra eigenvalues-eigenvectors
edited Jan 12 '16 at 16:38
Martin Sleziak
44.7k8117272
asked Mar 27 '11 at 22:00
Corey L.Corey L.
276145
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add a comment |
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How can I prove that if I have $n$ eigenvectors from different eigenvalues, they are all linearly independent?
linear-algebra eigenvalues-eigenvectors
edited Jan 12 '16 at 16:38
Martin Sleziak
44.7k8117272
asked Mar 27 '11 at 22:00
Corey L.Corey L.
276145
$endgroup$
add a comment |
$begingroup$
How can I prove that if I have $n$ eigenvectors from different eigenvalues, they are all linearly independent?
linear-algebra eigenvalues-eigenvectors
edited Jan 12 '16 at 16:38
Martin Sleziak
44.7k8117272
asked Mar 27 '11 at 22:00
Corey L.Corey L.
276145
$endgroup$
How can I prove that if I have $n$ eigenvectors from different eigenvalues, they are all linearly independent?
linear-algebra eigenvalues-eigenvectors
linear-algebra eigenvalues-eigenvectors
edited Jan 12 '16 at 16:38
Martin Sleziak
44.7k8117272
asked Mar 27 '11 at 22:00
Corey L.Corey L.
276145
edited Jan 12 '16 at 16:38
Martin Sleziak
44.7k8117272
asked Mar 27 '11 at 22:00
Corey L.Corey L.
276145
edited Jan 12 '16 at 16:38
Martin Sleziak
44.7k8117272
edited Jan 12 '16 at 16:38
Martin Sleziak
44.7k8117272
edited Jan 12 '16 at 16:38
Martin Sleziak
44.7k8117272
44.7k8117272
asked Mar 27 '11 at 22:00
Corey L.Corey L.
276145
asked Mar 27 '11 at 22:00
Corey L.Corey L.
276145
asked Mar 27 '11 at 22:00
Corey L.Corey L.
276145
276145
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
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I'll do it with two vectors. I'll leave it to you do it in general.
Suppose $mathbf{v}_1$ and $mathbf{v}_2$ correspond to distinct eigenvalues $lambda_1$ and $lambda_2$, respectively.
Take a linear combination that is equal to $0$, $alpha_1mathbf{v}_1+alpha_2mathbf{v}_2 = mathbf{0}$. We need to show that $alpha_1=alpha_2=0$.
Applying $T$ to both sides, we get
$$mathbf{0} = T(mathbf{0}) = T(alpha_1mathbf{v}_1+alpha_2mathbf{v}_2) = alpha_1lambda_1mathbf{v}_1 + alpha_2lambda_2mathbf{v}_2.$$
Now, instead, multiply the original equation by $lambda_1$:
$$mathbf{0} = lambda_1alpha_1mathbf{v}_1 + lambda_1alpha_2mathbf{v}_2.$$
Now take the two equations,
$$begin{align*}
mathbf{0} &= alpha_1lambda_1mathbf{v}_1 + alpha_2lambda_2mathbf{v}_2\
mathbf{0} &= alpha_1lambda_1mathbf{v}_1 + alpha_2lambda_1mathbf{v}_2
end{align*}$$
and taking the difference, we get:
$$mathbf{0} = 0mathbf{v}_1 + alpha_2(lambda_2-lambda_1)mathbf{v}_2 = alpha_2(lambda_2-lambda_1)mathbf{v}_2.$$
Since $lambda_2-lambda_1neq 0$, and since $mathbf{v}_2neqmathbf{0}$ (because $mathbf{v}_2$ is an eigenvector), then $alpha_2=0$. Using this on the original linear combination $mathbf{0} = alpha_1mathbf{v}_1 + alpha_2mathbf{v}_2$, we conclude that $alpha_1=0$ as well (since $mathbf{v}_1neqmathbf{0}$).
So $mathbf{v}_1$ and $mathbf{v}_2$ are linearly independent.
Now try using induction on $n$ for the general case.
edited Dec 2 '15 at 1:00
Community♦
1
answered Mar 27 '11 at 22:06
Arturo MagidinArturo Magidin
261k33585906
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3
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I believe that you wrote $lambda_2$ instead of $lambda_1$ in the row before "Now take"
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– jacob
Mar 14 '14 at 8:08
4
$begingroup$
Is there any intuition behind this? Any pictorial way of thinking?
$endgroup$
– IgNite
May 7 '16 at 8:42
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Maybe this will be of use: from the $0 = alpha_1 v_1 + ... + alpha_n v_n$, if you do $$left|left|lim_{jtoinfty} (1/ lambda_1^j) A^j (alpha_1 v_1 + ... + alpha_n v_n)right|right|$$, from the definition of the eigenvalue (that $Av = lambda v$), we can see the $v_1$ component will grow much faster than the others, so that limit equals $alpha_1$, which equals $0$ if linearly independent.
$endgroup$
– user203509
Apr 29 '17 at 15:54
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@Arturo Very nice.
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– user412674
May 27 '17 at 20:24
add a comment |
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Alternative:
Let $j$ be the maximal $j$ such that $v_1,dots,v_j$ are independent. Then there exists $c_i$, $1leq ileq j$ so that $sum_{i=1}^j c_iv_i=v_{j+1}$. But by applying $T$ we also have that
$$sum_{i=1}^j c_ilambda_iv_i=lambda_{j+1}v_{j+1}=lambda_{j+1}sum_{i=1}^j c_i v_i$$ Hence $$sum_{i=1}^j left(lambda_i-lambda_{j+1}right) c_iv_i=0$$ which is a contradiction since $lambda_ineq lambda_{j+1}$ for $1leq ileq j$.
Hope that helps,
answered Mar 28 '11 at 0:34
Eric NaslundEric Naslund
60.2k10138240
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4
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P.S. The argument uses the well-ordering principle on the naturals (by looking at the least $j$ such that $v_1,ldots,v_{j+1}$ is dependent). Well-ordering for the naturals is equivalent to induction.
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– Arturo Magidin
Mar 28 '11 at 1:20
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@Arturo: Thanks! Should be fixed now. (Didn't realize I was using well ordering in that way)
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– Eric Naslund
Mar 28 '11 at 1:30
1
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No problem; I'll delete the other two comments since it's been fixed. In any case, many people prefer arguments along these lines to an explicit induction, even if they are logically equivalent (you can think of the argument you give as a proof by contradiction of the inductive step, with the base being taken for granted [or as trivial, since $v_1$ is nonzero]; cast that way, it may be clearer why the two arguments are closely connected).
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– Arturo Magidin
Mar 28 '11 at 1:34
1
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@Eric Naslund This proofs seems very similar to the one given in Axler...
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– user38268
Aug 31 '11 at 7:29
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@D B Lim: What is Axler?
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– Eric Naslund
Aug 31 '11 at 14:42
|
show 1 more comment
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Hey I think there's a slick way to do this without induction. Suppose that $T$ is a linear transformation of a vector space $V$ and that $v_1,ldots,v_n in V$ are eigenvectors of $T$ with corresponding eigenvalues $lambda_1,ldots,lambda_n in F$ ($F$ the field of scalars). We want to show that, if $sum_{i=1}^n c_i v_i = 0$, where the coefficients $c_i$ are in $F$, then necessarily each $c_i$ is zero.
For simplicity, I will just explain why $c_1 = 0$. Consider the polynomial $p_1(x) in F[x]$ given as $p_1(x) = (x-lambda_2) cdots (x-lambda_n)$. Note that the $x-lambda_1$ term is "missing" here. Now, since each $v_i$ is an eigenvector of $T$, we have
begin{align*}
p_1(T) v_i = p_1(lambda_i) v_i &&
text{ where} && p_1(lambda_i) = begin{cases}
0 & text{ if } i neq 1 \
p_1(lambda_1) neq 0 & text{ if } i = 1
end{cases}.
end{align*}
Thus, applying $p_1(T)$ to the sum $sum_{i=1}^n c_i v_i = 0$, we get
$$ p_1(lambda_1) c_1 v_1 = 0 $$
which implies $c_1 = 0$, since $p_1(lambda_1) neq 0$ and $v_1 neq 0$.
answered Nov 2 '15 at 1:27
Mike FMike F
12.4k23481
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For eigenvectors $vec{v^1},vec{v^2},dots,vec{v^n}$ with different eigenvalues $lambda_1neqlambda_2neq dots neqlambda_n$ of a $ ntimes n$ matrix $A$.
Given the $ ntimes n$ matrix $P$ of the eigenvectors (with eigenvectors as the columns).
$$P=Big[vec{v^1},vec{v^2},dots,vec{v^n}Big]$$
Given the $ ntimes n$ matrix $Lambda$ of the eigenvalues on the diagonal (zeros elsewhere):
$$Lambda = begin{bmatrix}
lambda_1 & 0 & dots & 0 \
0 & lambda_2 & dots & 0 \
vdots & vdots & ddots & vdots \
0 & 0 & dots & lambda_n
end{bmatrix}
$$
Let $vec{c}=(c_1,c_2,dots,c_n)^T$
We need to show that only $c_1=c_2=...=c_n=0$ can satisfy the following:
$$c_1vec{v^1}+c_2vec{v^2}+...= vec{0^{}}$$
Applying the matrix to this equation gives:
$$c_1lambda_1vec{v^1}+c_2lambda_2vec{v^2}+...+c_nlambda_nvec{v^n}= vec{0^{}}$$
We can write this equation in the form of vectors and matrices:
$$PLambda vec{c^{}}=vec{0^{}}$$
But with since $A$ can be diagonalised to $Lambda$, we know $PLambda=AP$
$$implies APvec{c^{}}=vec{0^{}}$$
since $APneq 0$, we have $vec{c}=0$.
answered Mar 13 '16 at 1:58
Zeeshan AhmadZeeshan Ahmad
1239
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Let me add a comment to the last. In order to have $c=0$ the only solution, $AP$ must be invertible but since $A$ is already invertible because $lambda_i neq lambda_j forall ineq j$, the only demand is $P$ to be invertible $Rightarrow$ independent eigenvectors.
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– Thoth
Sep 13 '16 at 16:36
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
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votes
$begingroup$
I'll do it with two vectors. I'll leave it to you do it in general.
Suppose $mathbf{v}_1$ and $mathbf{v}_2$ correspond to distinct eigenvalues $lambda_1$ and $lambda_2$, respectively.
Take a linear combination that is equal to $0$, $alpha_1mathbf{v}_1+alpha_2mathbf{v}_2 = mathbf{0}$. We need to show that $alpha_1=alpha_2=0$.
Applying $T$ to both sides, we get
$$mathbf{0} = T(mathbf{0}) = T(alpha_1mathbf{v}_1+alpha_2mathbf{v}_2) = alpha_1lambda_1mathbf{v}_1 + alpha_2lambda_2mathbf{v}_2.$$
Now, instead, multiply the original equation by $lambda_1$:
$$mathbf{0} = lambda_1alpha_1mathbf{v}_1 + lambda_1alpha_2mathbf{v}_2.$$
Now take the two equations,
$$begin{align*}
mathbf{0} &= alpha_1lambda_1mathbf{v}_1 + alpha_2lambda_2mathbf{v}_2\
mathbf{0} &= alpha_1lambda_1mathbf{v}_1 + alpha_2lambda_1mathbf{v}_2
end{align*}$$
and taking the difference, we get:
$$mathbf{0} = 0mathbf{v}_1 + alpha_2(lambda_2-lambda_1)mathbf{v}_2 = alpha_2(lambda_2-lambda_1)mathbf{v}_2.$$
Since $lambda_2-lambda_1neq 0$, and since $mathbf{v}_2neqmathbf{0}$ (because $mathbf{v}_2$ is an eigenvector), then $alpha_2=0$. Using this on the original linear combination $mathbf{0} = alpha_1mathbf{v}_1 + alpha_2mathbf{v}_2$, we conclude that $alpha_1=0$ as well (since $mathbf{v}_1neqmathbf{0}$).
So $mathbf{v}_1$ and $mathbf{v}_2$ are linearly independent.
Now try using induction on $n$ for the general case.
edited Dec 2 '15 at 1:00
Community♦
1
answered Mar 27 '11 at 22:06
Arturo MagidinArturo Magidin
261k33585906
$endgroup$
3
$begingroup$
I believe that you wrote $lambda_2$ instead of $lambda_1$ in the row before "Now take"
$endgroup$
– jacob
Mar 14 '14 at 8:08
4
$begingroup$
Is there any intuition behind this? Any pictorial way of thinking?
$endgroup$
– IgNite
May 7 '16 at 8:42
$begingroup$
Maybe this will be of use: from the $0 = alpha_1 v_1 + ... + alpha_n v_n$, if you do $$left|left|lim_{jtoinfty} (1/ lambda_1^j) A^j (alpha_1 v_1 + ... + alpha_n v_n)right|right|$$, from the definition of the eigenvalue (that $Av = lambda v$), we can see the $v_1$ component will grow much faster than the others, so that limit equals $alpha_1$, which equals $0$ if linearly independent.
$endgroup$
– user203509
Apr 29 '17 at 15:54
$begingroup$
@Arturo Very nice.
$endgroup$
– user412674
May 27 '17 at 20:24
add a comment |
$begingroup$
I'll do it with two vectors. I'll leave it to you do it in general.
Suppose $mathbf{v}_1$ and $mathbf{v}_2$ correspond to distinct eigenvalues $lambda_1$ and $lambda_2$, respectively.
Take a linear combination that is equal to $0$, $alpha_1mathbf{v}_1+alpha_2mathbf{v}_2 = mathbf{0}$. We need to show that $alpha_1=alpha_2=0$.
Applying $T$ to both sides, we get
$$mathbf{0} = T(mathbf{0}) = T(alpha_1mathbf{v}_1+alpha_2mathbf{v}_2) = alpha_1lambda_1mathbf{v}_1 + alpha_2lambda_2mathbf{v}_2.$$
Now, instead, multiply the original equation by $lambda_1$:
$$mathbf{0} = lambda_1alpha_1mathbf{v}_1 + lambda_1alpha_2mathbf{v}_2.$$
Now take the two equations,
$$begin{align*}
mathbf{0} &= alpha_1lambda_1mathbf{v}_1 + alpha_2lambda_2mathbf{v}_2\
mathbf{0} &= alpha_1lambda_1mathbf{v}_1 + alpha_2lambda_1mathbf{v}_2
end{align*}$$
and taking the difference, we get:
$$mathbf{0} = 0mathbf{v}_1 + alpha_2(lambda_2-lambda_1)mathbf{v}_2 = alpha_2(lambda_2-lambda_1)mathbf{v}_2.$$
Since $lambda_2-lambda_1neq 0$, and since $mathbf{v}_2neqmathbf{0}$ (because $mathbf{v}_2$ is an eigenvector), then $alpha_2=0$. Using this on the original linear combination $mathbf{0} = alpha_1mathbf{v}_1 + alpha_2mathbf{v}_2$, we conclude that $alpha_1=0$ as well (since $mathbf{v}_1neqmathbf{0}$).
So $mathbf{v}_1$ and $mathbf{v}_2$ are linearly independent.
Now try using induction on $n$ for the general case.
edited Dec 2 '15 at 1:00
Community♦
1
answered Mar 27 '11 at 22:06
Arturo MagidinArturo Magidin
261k33585906
$endgroup$
3
$begingroup$
I believe that you wrote $lambda_2$ instead of $lambda_1$ in the row before "Now take"
$endgroup$
– jacob
Mar 14 '14 at 8:08
4
$begingroup$
Is there any intuition behind this? Any pictorial way of thinking?
$endgroup$
– IgNite
May 7 '16 at 8:42
$begingroup$
Maybe this will be of use: from the $0 = alpha_1 v_1 + ... + alpha_n v_n$, if you do $$left|left|lim_{jtoinfty} (1/ lambda_1^j) A^j (alpha_1 v_1 + ... + alpha_n v_n)right|right|$$, from the definition of the eigenvalue (that $Av = lambda v$), we can see the $v_1$ component will grow much faster than the others, so that limit equals $alpha_1$, which equals $0$ if linearly independent.
$endgroup$
– user203509
Apr 29 '17 at 15:54
$begingroup$
@Arturo Very nice.
$endgroup$
– user412674
May 27 '17 at 20:24
add a comment |
$begingroup$
I'll do it with two vectors. I'll leave it to you do it in general.
Suppose $mathbf{v}_1$ and $mathbf{v}_2$ correspond to distinct eigenvalues $lambda_1$ and $lambda_2$, respectively.
Take a linear combination that is equal to $0$, $alpha_1mathbf{v}_1+alpha_2mathbf{v}_2 = mathbf{0}$. We need to show that $alpha_1=alpha_2=0$.
Applying $T$ to both sides, we get
$$mathbf{0} = T(mathbf{0}) = T(alpha_1mathbf{v}_1+alpha_2mathbf{v}_2) = alpha_1lambda_1mathbf{v}_1 + alpha_2lambda_2mathbf{v}_2.$$
Now, instead, multiply the original equation by $lambda_1$:
$$mathbf{0} = lambda_1alpha_1mathbf{v}_1 + lambda_1alpha_2mathbf{v}_2.$$
Now take the two equations,
$$begin{align*}
mathbf{0} &= alpha_1lambda_1mathbf{v}_1 + alpha_2lambda_2mathbf{v}_2\
mathbf{0} &= alpha_1lambda_1mathbf{v}_1 + alpha_2lambda_1mathbf{v}_2
end{align*}$$
and taking the difference, we get:
$$mathbf{0} = 0mathbf{v}_1 + alpha_2(lambda_2-lambda_1)mathbf{v}_2 = alpha_2(lambda_2-lambda_1)mathbf{v}_2.$$
Since $lambda_2-lambda_1neq 0$, and since $mathbf{v}_2neqmathbf{0}$ (because $mathbf{v}_2$ is an eigenvector), then $alpha_2=0$. Using this on the original linear combination $mathbf{0} = alpha_1mathbf{v}_1 + alpha_2mathbf{v}_2$, we conclude that $alpha_1=0$ as well (since $mathbf{v}_1neqmathbf{0}$).
So $mathbf{v}_1$ and $mathbf{v}_2$ are linearly independent.
Now try using induction on $n$ for the general case.
edited Dec 2 '15 at 1:00
Community♦
1
answered Mar 27 '11 at 22:06
Arturo MagidinArturo Magidin
261k33585906
$endgroup$
I'll do it with two vectors. I'll leave it to you do it in general.
Suppose $mathbf{v}_1$ and $mathbf{v}_2$ correspond to distinct eigenvalues $lambda_1$ and $lambda_2$, respectively.
Take a linear combination that is equal to $0$, $alpha_1mathbf{v}_1+alpha_2mathbf{v}_2 = mathbf{0}$. We need to show that $alpha_1=alpha_2=0$.
Applying $T$ to both sides, we get
$$mathbf{0} = T(mathbf{0}) = T(alpha_1mathbf{v}_1+alpha_2mathbf{v}_2) = alpha_1lambda_1mathbf{v}_1 + alpha_2lambda_2mathbf{v}_2.$$
Now, instead, multiply the original equation by $lambda_1$:
$$mathbf{0} = lambda_1alpha_1mathbf{v}_1 + lambda_1alpha_2mathbf{v}_2.$$
Now take the two equations,
$$begin{align*}
mathbf{0} &= alpha_1lambda_1mathbf{v}_1 + alpha_2lambda_2mathbf{v}_2\
mathbf{0} &= alpha_1lambda_1mathbf{v}_1 + alpha_2lambda_1mathbf{v}_2
end{align*}$$
and taking the difference, we get:
$$mathbf{0} = 0mathbf{v}_1 + alpha_2(lambda_2-lambda_1)mathbf{v}_2 = alpha_2(lambda_2-lambda_1)mathbf{v}_2.$$
Since $lambda_2-lambda_1neq 0$, and since $mathbf{v}_2neqmathbf{0}$ (because $mathbf{v}_2$ is an eigenvector), then $alpha_2=0$. Using this on the original linear combination $mathbf{0} = alpha_1mathbf{v}_1 + alpha_2mathbf{v}_2$, we conclude that $alpha_1=0$ as well (since $mathbf{v}_1neqmathbf{0}$).
So $mathbf{v}_1$ and $mathbf{v}_2$ are linearly independent.
Now try using induction on $n$ for the general case.
edited Dec 2 '15 at 1:00
Community♦
1
answered Mar 27 '11 at 22:06
Arturo MagidinArturo Magidin
261k33585906
edited Dec 2 '15 at 1:00
Community♦
1
edited Dec 2 '15 at 1:00
Community♦
1
edited Dec 2 '15 at 1:00
Community♦
1
1
answered Mar 27 '11 at 22:06
Arturo MagidinArturo Magidin
261k33585906
answered Mar 27 '11 at 22:06
Arturo MagidinArturo Magidin
261k33585906
answered Mar 27 '11 at 22:06
Arturo MagidinArturo Magidin
261k33585906
261k33585906
3
$begingroup$
I believe that you wrote $lambda_2$ instead of $lambda_1$ in the row before "Now take"
$endgroup$
– jacob
Mar 14 '14 at 8:08
4
$begingroup$
Is there any intuition behind this? Any pictorial way of thinking?
$endgroup$
– IgNite
May 7 '16 at 8:42
$begingroup$
Maybe this will be of use: from the $0 = alpha_1 v_1 + ... + alpha_n v_n$, if you do $$left|left|lim_{jtoinfty} (1/ lambda_1^j) A^j (alpha_1 v_1 + ... + alpha_n v_n)right|right|$$, from the definition of the eigenvalue (that $Av = lambda v$), we can see the $v_1$ component will grow much faster than the others, so that limit equals $alpha_1$, which equals $0$ if linearly independent.
$endgroup$
– user203509
Apr 29 '17 at 15:54
$begingroup$
@Arturo Very nice.
$endgroup$
– user412674
May 27 '17 at 20:24
add a comment |
3
$begingroup$
I believe that you wrote $lambda_2$ instead of $lambda_1$ in the row before "Now take"
$endgroup$
– jacob
Mar 14 '14 at 8:08
4
$begingroup$
Is there any intuition behind this? Any pictorial way of thinking?
$endgroup$
– IgNite
May 7 '16 at 8:42
$begingroup$
Maybe this will be of use: from the $0 = alpha_1 v_1 + ... + alpha_n v_n$, if you do $$left|left|lim_{jtoinfty} (1/ lambda_1^j) A^j (alpha_1 v_1 + ... + alpha_n v_n)right|right|$$, from the definition of the eigenvalue (that $Av = lambda v$), we can see the $v_1$ component will grow much faster than the others, so that limit equals $alpha_1$, which equals $0$ if linearly independent.
$endgroup$
– user203509
Apr 29 '17 at 15:54
$begingroup$
@Arturo Very nice.
$endgroup$
– user412674
May 27 '17 at 20:24
3
3
$begingroup$
I believe that you wrote $lambda_2$ instead of $lambda_1$ in the row before "Now take"
$endgroup$
– jacob
Mar 14 '14 at 8:08
$begingroup$
I believe that you wrote $lambda_2$ instead of $lambda_1$ in the row before "Now take"
$endgroup$
– jacob
Mar 14 '14 at 8:08
4
4
$begingroup$
Is there any intuition behind this? Any pictorial way of thinking?
$endgroup$
– IgNite
May 7 '16 at 8:42
$begingroup$
Is there any intuition behind this? Any pictorial way of thinking?
$endgroup$
– IgNite
May 7 '16 at 8:42
$begingroup$
Maybe this will be of use: from the $0 = alpha_1 v_1 + ... + alpha_n v_n$, if you do $$left|left|lim_{jtoinfty} (1/ lambda_1^j) A^j (alpha_1 v_1 + ... + alpha_n v_n)right|right|$$, from the definition of the eigenvalue (that $Av = lambda v$), we can see the $v_1$ component will grow much faster than the others, so that limit equals $alpha_1$, which equals $0$ if linearly independent.
$endgroup$
– user203509
Apr 29 '17 at 15:54
$begingroup$
Maybe this will be of use: from the $0 = alpha_1 v_1 + ... + alpha_n v_n$, if you do $$left|left|lim_{jtoinfty} (1/ lambda_1^j) A^j (alpha_1 v_1 + ... + alpha_n v_n)right|right|$$, from the definition of the eigenvalue (that $Av = lambda v$), we can see the $v_1$ component will grow much faster than the others, so that limit equals $alpha_1$, which equals $0$ if linearly independent.
$endgroup$
– user203509
Apr 29 '17 at 15:54
$begingroup$
@Arturo Very nice.
$endgroup$
– user412674
May 27 '17 at 20:24
$begingroup$
@Arturo Very nice.
$endgroup$
– user412674
May 27 '17 at 20:24
add a comment |
$begingroup$
Alternative:
Let $j$ be the maximal $j$ such that $v_1,dots,v_j$ are independent. Then there exists $c_i$, $1leq ileq j$ so that $sum_{i=1}^j c_iv_i=v_{j+1}$. But by applying $T$ we also have that
$$sum_{i=1}^j c_ilambda_iv_i=lambda_{j+1}v_{j+1}=lambda_{j+1}sum_{i=1}^j c_i v_i$$ Hence $$sum_{i=1}^j left(lambda_i-lambda_{j+1}right) c_iv_i=0$$ which is a contradiction since $lambda_ineq lambda_{j+1}$ for $1leq ileq j$.
Hope that helps,
answered Mar 28 '11 at 0:34
Eric NaslundEric Naslund
60.2k10138240
$endgroup$
4
$begingroup$
P.S. The argument uses the well-ordering principle on the naturals (by looking at the least $j$ such that $v_1,ldots,v_{j+1}$ is dependent). Well-ordering for the naturals is equivalent to induction.
$endgroup$
– Arturo Magidin
Mar 28 '11 at 1:20
$begingroup$
@Arturo: Thanks! Should be fixed now. (Didn't realize I was using well ordering in that way)
$endgroup$
– Eric Naslund
Mar 28 '11 at 1:30
1
$begingroup$
No problem; I'll delete the other two comments since it's been fixed. In any case, many people prefer arguments along these lines to an explicit induction, even if they are logically equivalent (you can think of the argument you give as a proof by contradiction of the inductive step, with the base being taken for granted [or as trivial, since $v_1$ is nonzero]; cast that way, it may be clearer why the two arguments are closely connected).
$endgroup$
– Arturo Magidin
Mar 28 '11 at 1:34
1
$begingroup$
@Eric Naslund This proofs seems very similar to the one given in Axler...
$endgroup$
– user38268
Aug 31 '11 at 7:29
$begingroup$
@D B Lim: What is Axler?
$endgroup$
– Eric Naslund
Aug 31 '11 at 14:42
|
show 1 more comment
$begingroup$
Alternative:
Let $j$ be the maximal $j$ such that $v_1,dots,v_j$ are independent. Then there exists $c_i$, $1leq ileq j$ so that $sum_{i=1}^j c_iv_i=v_{j+1}$. But by applying $T$ we also have that
$$sum_{i=1}^j c_ilambda_iv_i=lambda_{j+1}v_{j+1}=lambda_{j+1}sum_{i=1}^j c_i v_i$$ Hence $$sum_{i=1}^j left(lambda_i-lambda_{j+1}right) c_iv_i=0$$ which is a contradiction since $lambda_ineq lambda_{j+1}$ for $1leq ileq j$.
Hope that helps,
answered Mar 28 '11 at 0:34
Eric NaslundEric Naslund
60.2k10138240
$endgroup$
4
$begingroup$
P.S. The argument uses the well-ordering principle on the naturals (by looking at the least $j$ such that $v_1,ldots,v_{j+1}$ is dependent). Well-ordering for the naturals is equivalent to induction.
$endgroup$
– Arturo Magidin
Mar 28 '11 at 1:20
$begingroup$
@Arturo: Thanks! Should be fixed now. (Didn't realize I was using well ordering in that way)
$endgroup$
– Eric Naslund
Mar 28 '11 at 1:30
1
$begingroup$
No problem; I'll delete the other two comments since it's been fixed. In any case, many people prefer arguments along these lines to an explicit induction, even if they are logically equivalent (you can think of the argument you give as a proof by contradiction of the inductive step, with the base being taken for granted [or as trivial, since $v_1$ is nonzero]; cast that way, it may be clearer why the two arguments are closely connected).
$endgroup$
– Arturo Magidin
Mar 28 '11 at 1:34
1
$begingroup$
@Eric Naslund This proofs seems very similar to the one given in Axler...
$endgroup$
– user38268
Aug 31 '11 at 7:29
$begingroup$
@D B Lim: What is Axler?
$endgroup$
– Eric Naslund
Aug 31 '11 at 14:42
|
show 1 more comment
$begingroup$
Alternative:
Let $j$ be the maximal $j$ such that $v_1,dots,v_j$ are independent. Then there exists $c_i$, $1leq ileq j$ so that $sum_{i=1}^j c_iv_i=v_{j+1}$. But by applying $T$ we also have that
$$sum_{i=1}^j c_ilambda_iv_i=lambda_{j+1}v_{j+1}=lambda_{j+1}sum_{i=1}^j c_i v_i$$ Hence $$sum_{i=1}^j left(lambda_i-lambda_{j+1}right) c_iv_i=0$$ which is a contradiction since $lambda_ineq lambda_{j+1}$ for $1leq ileq j$.
Hope that helps,
answered Mar 28 '11 at 0:34
Eric NaslundEric Naslund
60.2k10138240
$endgroup$
Alternative:
Let $j$ be the maximal $j$ such that $v_1,dots,v_j$ are independent. Then there exists $c_i$, $1leq ileq j$ so that $sum_{i=1}^j c_iv_i=v_{j+1}$. But by applying $T$ we also have that
$$sum_{i=1}^j c_ilambda_iv_i=lambda_{j+1}v_{j+1}=lambda_{j+1}sum_{i=1}^j c_i v_i$$ Hence $$sum_{i=1}^j left(lambda_i-lambda_{j+1}right) c_iv_i=0$$ which is a contradiction since $lambda_ineq lambda_{j+1}$ for $1leq ileq j$.
Hope that helps,
answered Mar 28 '11 at 0:34
Eric NaslundEric Naslund
60.2k10138240
edited Aug 22 '17 at 14:27
answered Mar 28 '11 at 0:34
Eric NaslundEric Naslund
60.2k10138240
answered Mar 28 '11 at 0:34
Eric NaslundEric Naslund
60.2k10138240
answered Mar 28 '11 at 0:34
Eric NaslundEric Naslund
60.2k10138240
60.2k10138240
4
$begingroup$
P.S. The argument uses the well-ordering principle on the naturals (by looking at the least $j$ such that $v_1,ldots,v_{j+1}$ is dependent). Well-ordering for the naturals is equivalent to induction.
$endgroup$
– Arturo Magidin
Mar 28 '11 at 1:20
$begingroup$
@Arturo: Thanks! Should be fixed now. (Didn't realize I was using well ordering in that way)
$endgroup$
– Eric Naslund
Mar 28 '11 at 1:30
1
$begingroup$
No problem; I'll delete the other two comments since it's been fixed. In any case, many people prefer arguments along these lines to an explicit induction, even if they are logically equivalent (you can think of the argument you give as a proof by contradiction of the inductive step, with the base being taken for granted [or as trivial, since $v_1$ is nonzero]; cast that way, it may be clearer why the two arguments are closely connected).
$endgroup$
– Arturo Magidin
Mar 28 '11 at 1:34
1
$begingroup$
@Eric Naslund This proofs seems very similar to the one given in Axler...
$endgroup$
– user38268
Aug 31 '11 at 7:29
$begingroup$
@D B Lim: What is Axler?
$endgroup$
– Eric Naslund
Aug 31 '11 at 14:42
|
show 1 more comment
4
$begingroup$
P.S. The argument uses the well-ordering principle on the naturals (by looking at the least $j$ such that $v_1,ldots,v_{j+1}$ is dependent). Well-ordering for the naturals is equivalent to induction.
$endgroup$
– Arturo Magidin
Mar 28 '11 at 1:20
$begingroup$
@Arturo: Thanks! Should be fixed now. (Didn't realize I was using well ordering in that way)
$endgroup$
– Eric Naslund
Mar 28 '11 at 1:30
1
$begingroup$
No problem; I'll delete the other two comments since it's been fixed. In any case, many people prefer arguments along these lines to an explicit induction, even if they are logically equivalent (you can think of the argument you give as a proof by contradiction of the inductive step, with the base being taken for granted [or as trivial, since $v_1$ is nonzero]; cast that way, it may be clearer why the two arguments are closely connected).
$endgroup$
– Arturo Magidin
Mar 28 '11 at 1:34
1
$begingroup$
@Eric Naslund This proofs seems very similar to the one given in Axler...
$endgroup$
– user38268
Aug 31 '11 at 7:29
$begingroup$
@D B Lim: What is Axler?
$endgroup$
– Eric Naslund
Aug 31 '11 at 14:42
4
4
$begingroup$
P.S. The argument uses the well-ordering principle on the naturals (by looking at the least $j$ such that $v_1,ldots,v_{j+1}$ is dependent). Well-ordering for the naturals is equivalent to induction.
$endgroup$
– Arturo Magidin
Mar 28 '11 at 1:20
$begingroup$
P.S. The argument uses the well-ordering principle on the naturals (by looking at the least $j$ such that $v_1,ldots,v_{j+1}$ is dependent). Well-ordering for the naturals is equivalent to induction.
$endgroup$
– Arturo Magidin
Mar 28 '11 at 1:20
$begingroup$
@Arturo: Thanks! Should be fixed now. (Didn't realize I was using well ordering in that way)
$endgroup$
– Eric Naslund
Mar 28 '11 at 1:30
$begingroup$
@Arturo: Thanks! Should be fixed now. (Didn't realize I was using well ordering in that way)
$endgroup$
– Eric Naslund
Mar 28 '11 at 1:30
1
1
$begingroup$
No problem; I'll delete the other two comments since it's been fixed. In any case, many people prefer arguments along these lines to an explicit induction, even if they are logically equivalent (you can think of the argument you give as a proof by contradiction of the inductive step, with the base being taken for granted [or as trivial, since $v_1$ is nonzero]; cast that way, it may be clearer why the two arguments are closely connected).
$endgroup$
– Arturo Magidin
Mar 28 '11 at 1:34
$begingroup$
No problem; I'll delete the other two comments since it's been fixed. In any case, many people prefer arguments along these lines to an explicit induction, even if they are logically equivalent (you can think of the argument you give as a proof by contradiction of the inductive step, with the base being taken for granted [or as trivial, since $v_1$ is nonzero]; cast that way, it may be clearer why the two arguments are closely connected).
$endgroup$
– Arturo Magidin
Mar 28 '11 at 1:34
1
1
$begingroup$
@Eric Naslund This proofs seems very similar to the one given in Axler...
$endgroup$
– user38268
Aug 31 '11 at 7:29
$begingroup$
@Eric Naslund This proofs seems very similar to the one given in Axler...
$endgroup$
– user38268
Aug 31 '11 at 7:29
$begingroup$
@D B Lim: What is Axler?
$endgroup$
– Eric Naslund
Aug 31 '11 at 14:42
$begingroup$
@D B Lim: What is Axler?
$endgroup$
– Eric Naslund
Aug 31 '11 at 14:42
|
show 1 more comment
$begingroup$
Hey I think there's a slick way to do this without induction. Suppose that $T$ is a linear transformation of a vector space $V$ and that $v_1,ldots,v_n in V$ are eigenvectors of $T$ with corresponding eigenvalues $lambda_1,ldots,lambda_n in F$ ($F$ the field of scalars). We want to show that, if $sum_{i=1}^n c_i v_i = 0$, where the coefficients $c_i$ are in $F$, then necessarily each $c_i$ is zero.
For simplicity, I will just explain why $c_1 = 0$. Consider the polynomial $p_1(x) in F[x]$ given as $p_1(x) = (x-lambda_2) cdots (x-lambda_n)$. Note that the $x-lambda_1$ term is "missing" here. Now, since each $v_i$ is an eigenvector of $T$, we have
begin{align*}
p_1(T) v_i = p_1(lambda_i) v_i &&
text{ where} && p_1(lambda_i) = begin{cases}
0 & text{ if } i neq 1 \
p_1(lambda_1) neq 0 & text{ if } i = 1
end{cases}.
end{align*}
Thus, applying $p_1(T)$ to the sum $sum_{i=1}^n c_i v_i = 0$, we get
$$ p_1(lambda_1) c_1 v_1 = 0 $$
which implies $c_1 = 0$, since $p_1(lambda_1) neq 0$ and $v_1 neq 0$.
answered Nov 2 '15 at 1:27
Mike FMike F
12.4k23481
$endgroup$
add a comment |
$begingroup$
Hey I think there's a slick way to do this without induction. Suppose that $T$ is a linear transformation of a vector space $V$ and that $v_1,ldots,v_n in V$ are eigenvectors of $T$ with corresponding eigenvalues $lambda_1,ldots,lambda_n in F$ ($F$ the field of scalars). We want to show that, if $sum_{i=1}^n c_i v_i = 0$, where the coefficients $c_i$ are in $F$, then necessarily each $c_i$ is zero.
For simplicity, I will just explain why $c_1 = 0$. Consider the polynomial $p_1(x) in F[x]$ given as $p_1(x) = (x-lambda_2) cdots (x-lambda_n)$. Note that the $x-lambda_1$ term is "missing" here. Now, since each $v_i$ is an eigenvector of $T$, we have
begin{align*}
p_1(T) v_i = p_1(lambda_i) v_i &&
text{ where} && p_1(lambda_i) = begin{cases}
0 & text{ if } i neq 1 \
p_1(lambda_1) neq 0 & text{ if } i = 1
end{cases}.
end{align*}
Thus, applying $p_1(T)$ to the sum $sum_{i=1}^n c_i v_i = 0$, we get
$$ p_1(lambda_1) c_1 v_1 = 0 $$
which implies $c_1 = 0$, since $p_1(lambda_1) neq 0$ and $v_1 neq 0$.
answered Nov 2 '15 at 1:27
Mike FMike F
12.4k23481
$endgroup$
add a comment |
$begingroup$
Hey I think there's a slick way to do this without induction. Suppose that $T$ is a linear transformation of a vector space $V$ and that $v_1,ldots,v_n in V$ are eigenvectors of $T$ with corresponding eigenvalues $lambda_1,ldots,lambda_n in F$ ($F$ the field of scalars). We want to show that, if $sum_{i=1}^n c_i v_i = 0$, where the coefficients $c_i$ are in $F$, then necessarily each $c_i$ is zero.
For simplicity, I will just explain why $c_1 = 0$. Consider the polynomial $p_1(x) in F[x]$ given as $p_1(x) = (x-lambda_2) cdots (x-lambda_n)$. Note that the $x-lambda_1$ term is "missing" here. Now, since each $v_i$ is an eigenvector of $T$, we have
begin{align*}
p_1(T) v_i = p_1(lambda_i) v_i &&
text{ where} && p_1(lambda_i) = begin{cases}
0 & text{ if } i neq 1 \
p_1(lambda_1) neq 0 & text{ if } i = 1
end{cases}.
end{align*}
Thus, applying $p_1(T)$ to the sum $sum_{i=1}^n c_i v_i = 0$, we get
$$ p_1(lambda_1) c_1 v_1 = 0 $$
which implies $c_1 = 0$, since $p_1(lambda_1) neq 0$ and $v_1 neq 0$.
answered Nov 2 '15 at 1:27
Mike FMike F
12.4k23481
$endgroup$
Hey I think there's a slick way to do this without induction. Suppose that $T$ is a linear transformation of a vector space $V$ and that $v_1,ldots,v_n in V$ are eigenvectors of $T$ with corresponding eigenvalues $lambda_1,ldots,lambda_n in F$ ($F$ the field of scalars). We want to show that, if $sum_{i=1}^n c_i v_i = 0$, where the coefficients $c_i$ are in $F$, then necessarily each $c_i$ is zero.
For simplicity, I will just explain why $c_1 = 0$. Consider the polynomial $p_1(x) in F[x]$ given as $p_1(x) = (x-lambda_2) cdots (x-lambda_n)$. Note that the $x-lambda_1$ term is "missing" here. Now, since each $v_i$ is an eigenvector of $T$, we have
begin{align*}
p_1(T) v_i = p_1(lambda_i) v_i &&
text{ where} && p_1(lambda_i) = begin{cases}
0 & text{ if } i neq 1 \
p_1(lambda_1) neq 0 & text{ if } i = 1
end{cases}.
end{align*}
Thus, applying $p_1(T)$ to the sum $sum_{i=1}^n c_i v_i = 0$, we get
$$ p_1(lambda_1) c_1 v_1 = 0 $$
which implies $c_1 = 0$, since $p_1(lambda_1) neq 0$ and $v_1 neq 0$.
answered Nov 2 '15 at 1:27
Mike FMike F
12.4k23481
edited Nov 2 '15 at 2:34
answered Nov 2 '15 at 1:27
Mike FMike F
12.4k23481
answered Nov 2 '15 at 1:27
Mike FMike F
12.4k23481
answered Nov 2 '15 at 1:27
Mike FMike F
12.4k23481
12.4k23481
add a comment |
add a comment |
$begingroup$
For eigenvectors $vec{v^1},vec{v^2},dots,vec{v^n}$ with different eigenvalues $lambda_1neqlambda_2neq dots neqlambda_n$ of a $ ntimes n$ matrix $A$.
Given the $ ntimes n$ matrix $P$ of the eigenvectors (with eigenvectors as the columns).
$$P=Big[vec{v^1},vec{v^2},dots,vec{v^n}Big]$$
Given the $ ntimes n$ matrix $Lambda$ of the eigenvalues on the diagonal (zeros elsewhere):
$$Lambda = begin{bmatrix}
lambda_1 & 0 & dots & 0 \
0 & lambda_2 & dots & 0 \
vdots & vdots & ddots & vdots \
0 & 0 & dots & lambda_n
end{bmatrix}
$$
Let $vec{c}=(c_1,c_2,dots,c_n)^T$
We need to show that only $c_1=c_2=...=c_n=0$ can satisfy the following:
$$c_1vec{v^1}+c_2vec{v^2}+...= vec{0^{}}$$
Applying the matrix to this equation gives:
$$c_1lambda_1vec{v^1}+c_2lambda_2vec{v^2}+...+c_nlambda_nvec{v^n}= vec{0^{}}$$
We can write this equation in the form of vectors and matrices:
$$PLambda vec{c^{}}=vec{0^{}}$$
But with since $A$ can be diagonalised to $Lambda$, we know $PLambda=AP$
$$implies APvec{c^{}}=vec{0^{}}$$
since $APneq 0$, we have $vec{c}=0$.
answered Mar 13 '16 at 1:58
Zeeshan AhmadZeeshan Ahmad
1239
$endgroup$
$begingroup$
Let me add a comment to the last. In order to have $c=0$ the only solution, $AP$ must be invertible but since $A$ is already invertible because $lambda_i neq lambda_j forall ineq j$, the only demand is $P$ to be invertible $Rightarrow$ independent eigenvectors.
$endgroup$
– Thoth
Sep 13 '16 at 16:36
add a comment |
$begingroup$
For eigenvectors $vec{v^1},vec{v^2},dots,vec{v^n}$ with different eigenvalues $lambda_1neqlambda_2neq dots neqlambda_n$ of a $ ntimes n$ matrix $A$.
Given the $ ntimes n$ matrix $P$ of the eigenvectors (with eigenvectors as the columns).
$$P=Big[vec{v^1},vec{v^2},dots,vec{v^n}Big]$$
Given the $ ntimes n$ matrix $Lambda$ of the eigenvalues on the diagonal (zeros elsewhere):
$$Lambda = begin{bmatrix}
lambda_1 & 0 & dots & 0 \
0 & lambda_2 & dots & 0 \
vdots & vdots & ddots & vdots \
0 & 0 & dots & lambda_n
end{bmatrix}
$$
Let $vec{c}=(c_1,c_2,dots,c_n)^T$
We need to show that only $c_1=c_2=...=c_n=0$ can satisfy the following:
$$c_1vec{v^1}+c_2vec{v^2}+...= vec{0^{}}$$
Applying the matrix to this equation gives:
$$c_1lambda_1vec{v^1}+c_2lambda_2vec{v^2}+...+c_nlambda_nvec{v^n}= vec{0^{}}$$
We can write this equation in the form of vectors and matrices:
$$PLambda vec{c^{}}=vec{0^{}}$$
But with since $A$ can be diagonalised to $Lambda$, we know $PLambda=AP$
$$implies APvec{c^{}}=vec{0^{}}$$
since $APneq 0$, we have $vec{c}=0$.
answered Mar 13 '16 at 1:58
Zeeshan AhmadZeeshan Ahmad
1239
$endgroup$
$begingroup$
Let me add a comment to the last. In order to have $c=0$ the only solution, $AP$ must be invertible but since $A$ is already invertible because $lambda_i neq lambda_j forall ineq j$, the only demand is $P$ to be invertible $Rightarrow$ independent eigenvectors.
$endgroup$
– Thoth
Sep 13 '16 at 16:36
add a comment |
$begingroup$
For eigenvectors $vec{v^1},vec{v^2},dots,vec{v^n}$ with different eigenvalues $lambda_1neqlambda_2neq dots neqlambda_n$ of a $ ntimes n$ matrix $A$.
Given the $ ntimes n$ matrix $P$ of the eigenvectors (with eigenvectors as the columns).
$$P=Big[vec{v^1},vec{v^2},dots,vec{v^n}Big]$$
Given the $ ntimes n$ matrix $Lambda$ of the eigenvalues on the diagonal (zeros elsewhere):
$$Lambda = begin{bmatrix}
lambda_1 & 0 & dots & 0 \
0 & lambda_2 & dots & 0 \
vdots & vdots & ddots & vdots \
0 & 0 & dots & lambda_n
end{bmatrix}
$$
Let $vec{c}=(c_1,c_2,dots,c_n)^T$
We need to show that only $c_1=c_2=...=c_n=0$ can satisfy the following:
$$c_1vec{v^1}+c_2vec{v^2}+...= vec{0^{}}$$
Applying the matrix to this equation gives:
$$c_1lambda_1vec{v^1}+c_2lambda_2vec{v^2}+...+c_nlambda_nvec{v^n}= vec{0^{}}$$
We can write this equation in the form of vectors and matrices:
$$PLambda vec{c^{}}=vec{0^{}}$$
But with since $A$ can be diagonalised to $Lambda$, we know $PLambda=AP$
$$implies APvec{c^{}}=vec{0^{}}$$
since $APneq 0$, we have $vec{c}=0$.
answered Mar 13 '16 at 1:58
Zeeshan AhmadZeeshan Ahmad
1239
$endgroup$
For eigenvectors $vec{v^1},vec{v^2},dots,vec{v^n}$ with different eigenvalues $lambda_1neqlambda_2neq dots neqlambda_n$ of a $ ntimes n$ matrix $A$.
Given the $ ntimes n$ matrix $P$ of the eigenvectors (with eigenvectors as the columns).
$$P=Big[vec{v^1},vec{v^2},dots,vec{v^n}Big]$$
Given the $ ntimes n$ matrix $Lambda$ of the eigenvalues on the diagonal (zeros elsewhere):
$$Lambda = begin{bmatrix}
lambda_1 & 0 & dots & 0 \
0 & lambda_2 & dots & 0 \
vdots & vdots & ddots & vdots \
0 & 0 & dots & lambda_n
end{bmatrix}
$$
Let $vec{c}=(c_1,c_2,dots,c_n)^T$
We need to show that only $c_1=c_2=...=c_n=0$ can satisfy the following:
$$c_1vec{v^1}+c_2vec{v^2}+...= vec{0^{}}$$
Applying the matrix to this equation gives:
$$c_1lambda_1vec{v^1}+c_2lambda_2vec{v^2}+...+c_nlambda_nvec{v^n}= vec{0^{}}$$
We can write this equation in the form of vectors and matrices:
$$PLambda vec{c^{}}=vec{0^{}}$$
But with since $A$ can be diagonalised to $Lambda$, we know $PLambda=AP$
$$implies APvec{c^{}}=vec{0^{}}$$
since $APneq 0$, we have $vec{c}=0$.
answered Mar 13 '16 at 1:58
Zeeshan AhmadZeeshan Ahmad
1239
answered Mar 13 '16 at 1:58
Zeeshan AhmadZeeshan Ahmad
1239
answered Mar 13 '16 at 1:58
Zeeshan AhmadZeeshan Ahmad
1239
answered Mar 13 '16 at 1:58
Zeeshan AhmadZeeshan Ahmad
1239
1239
$begingroup$
Let me add a comment to the last. In order to have $c=0$ the only solution, $AP$ must be invertible but since $A$ is already invertible because $lambda_i neq lambda_j forall ineq j$, the only demand is $P$ to be invertible $Rightarrow$ independent eigenvectors.
$endgroup$
– Thoth
Sep 13 '16 at 16:36
add a comment |
$begingroup$
Let me add a comment to the last. In order to have $c=0$ the only solution, $AP$ must be invertible but since $A$ is already invertible because $lambda_i neq lambda_j forall ineq j$, the only demand is $P$ to be invertible $Rightarrow$ independent eigenvectors.
$endgroup$
– Thoth
Sep 13 '16 at 16:36
$begingroup$
Let me add a comment to the last. In order to have $c=0$ the only solution, $AP$ must be invertible but since $A$ is already invertible because $lambda_i neq lambda_j forall ineq j$, the only demand is $P$ to be invertible $Rightarrow$ independent eigenvectors.
$endgroup$
– Thoth
Sep 13 '16 at 16:36
$begingroup$
Let me add a comment to the last. In order to have $c=0$ the only solution, $AP$ must be invertible but since $A$ is already invertible because $lambda_i neq lambda_j forall ineq j$, the only demand is $P$ to be invertible $Rightarrow$ independent eigenvectors.
$endgroup$
– Thoth
Sep 13 '16 at 16:36
add a comment |
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