Show the non-zero vectors $v_1,v_2,⋯,v_n$ are linearly independent if $(A−λI)^j+1v_j=0$, where $j=0,1,⋯,n$.A question about linear independenceUnique linear combination problemIs $span(v_1, . . . ,v_m)$ a linearly dependent or linearly independent set of vectors? Also, what will happen if we take span of span?Showing Linear dependencyHow to prove that $v_1,v_2$ are linearly independent if $Spv_1,v_2=Spw_1,w_2$ and $v_1,w_2$ are linearly independent?linear-Independency is retained over a linear MapShow that if Span$v_1,v_2,v_3$=$mathbbR^3$ then $v_1,v_2,v_3$ is linearly independent.Let $v_1,v_2,ldots,v_n$ a linear independent set of vectors, and let $w inlangle v_1,v_2,ldots,v_nrangle$, prove linear independenceThe length of every linearly independent list of vectors is less than or equal to the length of every spanning list of vectors.Proof of $j=1$ where $v_j in span(v_1, …,v_j-1)$
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Show the non-zero vectors $v_1,v_2,⋯,v_n$ are linearly independent if $(A−λI)^j+1v_j=0$, where $j=0,1,⋯,n$.
A question about linear independenceUnique linear combination problemIs $span(v_1, . . . ,v_m)$ a linearly dependent or linearly independent set of vectors? Also, what will happen if we take span of span?Showing Linear dependencyHow to prove that $v_1,v_2$ are linearly independent if $Spv_1,v_2=Spw_1,w_2$ and $v_1,w_2$ are linearly independent?linear-Independency is retained over a linear MapShow that if Span$v_1,v_2,v_3$=$mathbbR^3$ then $v_1,v_2,v_3$ is linearly independent.Let $v_1,v_2,ldots,v_n$ a linear independent set of vectors, and let $w inlangle v_1,v_2,ldots,v_nrangle$, prove linear independenceThe length of every linearly independent list of vectors is less than or equal to the length of every spanning list of vectors.Proof of $j=1$ where $v_j in span(v_1, …,v_j-1)$
$begingroup$
My attempt at a solution uses the linear dependence lemma. Assume on the contrary that the list of vectors is linearly dependent. Then it is possible to find some $v_j$ that can be written as a linear combination of the remaining vectors: that is,
$$v_j = a_1v_1 + a_2v_2 + cdots + a_j-1v_j-1 + a_j+1v_j+1 + cdots + a_nv_n.$$ Applying the operator $(A - lambda I)^j+1$ to both sides, we have
$$ 0 = (A - lambda I)^j+1(a_1v_1 + a_2v_2 + cdots + a_j-1v_j-1) + (A - lambda I)^j+1(a_j-1v_j-1 + a_j+1v_j+1 + cdots + a_nv_n).$$ The first sum is zero, since the vectors there require the operator $A - lambda I$ applied at most $j$ times to be $0$. Hence, $$0 = (A - lambda I)^j+1(a_j-1v_j-1 + a_j+1v_j+1 + cdots + a_nv_n).$$
And this is where I'm stuck. How do I use this to show a contradiction and conclude they are linearly independent?
linear-algebra vector-spaces vectors
asked yesterday
YejusYejus
62
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Yejus is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$
add a comment |
$begingroup$
My attempt at a solution uses the linear dependence lemma. Assume on the contrary that the list of vectors is linearly dependent. Then it is possible to find some $v_j$ that can be written as a linear combination of the remaining vectors: that is,
$$v_j = a_1v_1 + a_2v_2 + cdots + a_j-1v_j-1 + a_j+1v_j+1 + cdots + a_nv_n.$$ Applying the operator $(A - lambda I)^j+1$ to both sides, we have
$$ 0 = (A - lambda I)^j+1(a_1v_1 + a_2v_2 + cdots + a_j-1v_j-1) + (A - lambda I)^j+1(a_j-1v_j-1 + a_j+1v_j+1 + cdots + a_nv_n).$$ The first sum is zero, since the vectors there require the operator $A - lambda I$ applied at most $j$ times to be $0$. Hence, $$0 = (A - lambda I)^j+1(a_j-1v_j-1 + a_j+1v_j+1 + cdots + a_nv_n).$$
And this is where I'm stuck. How do I use this to show a contradiction and conclude they are linearly independent?
linear-algebra vector-spaces vectors
asked yesterday
YejusYejus
62
New contributor
Yejus is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
$v_j=0$ is an obvious counterexample.
$endgroup$
– Kavi Rama Murthy
yesterday
1
$begingroup$
Possibly the additional condition $(A−λI)^jv_jnot=0$ has to be imposed.
$endgroup$
– Jens Schwaiger
yesterday
$begingroup$
I imagine that what you mean is that: $$ (A-lambda I)^jv_jne 0quadtextandquad (A-lambda I)^j+1v_j=0.$$
$endgroup$
– Yiorgos S. Smyrlis
yesterday
add a comment |
$begingroup$
My attempt at a solution uses the linear dependence lemma. Assume on the contrary that the list of vectors is linearly dependent. Then it is possible to find some $v_j$ that can be written as a linear combination of the remaining vectors: that is,
$$v_j = a_1v_1 + a_2v_2 + cdots + a_j-1v_j-1 + a_j+1v_j+1 + cdots + a_nv_n.$$ Applying the operator $(A - lambda I)^j+1$ to both sides, we have
$$ 0 = (A - lambda I)^j+1(a_1v_1 + a_2v_2 + cdots + a_j-1v_j-1) + (A - lambda I)^j+1(a_j-1v_j-1 + a_j+1v_j+1 + cdots + a_nv_n).$$ The first sum is zero, since the vectors there require the operator $A - lambda I$ applied at most $j$ times to be $0$. Hence, $$0 = (A - lambda I)^j+1(a_j-1v_j-1 + a_j+1v_j+1 + cdots + a_nv_n).$$
And this is where I'm stuck. How do I use this to show a contradiction and conclude they are linearly independent?
linear-algebra vector-spaces vectors
asked yesterday
YejusYejus
62
New contributor
Yejus is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
My attempt at a solution uses the linear dependence lemma. Assume on the contrary that the list of vectors is linearly dependent. Then it is possible to find some $v_j$ that can be written as a linear combination of the remaining vectors: that is,
$$v_j = a_1v_1 + a_2v_2 + cdots + a_j-1v_j-1 + a_j+1v_j+1 + cdots + a_nv_n.$$ Applying the operator $(A - lambda I)^j+1$ to both sides, we have
$$ 0 = (A - lambda I)^j+1(a_1v_1 + a_2v_2 + cdots + a_j-1v_j-1) + (A - lambda I)^j+1(a_j-1v_j-1 + a_j+1v_j+1 + cdots + a_nv_n).$$ The first sum is zero, since the vectors there require the operator $A - lambda I$ applied at most $j$ times to be $0$. Hence, $$0 = (A - lambda I)^j+1(a_j-1v_j-1 + a_j+1v_j+1 + cdots + a_nv_n).$$
And this is where I'm stuck. How do I use this to show a contradiction and conclude they are linearly independent?
linear-algebra vector-spaces vectors
linear-algebra vector-spaces vectors
asked yesterday
YejusYejus
62
New contributor
Yejus is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
YejusYejus
62
New contributor
Yejus is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
Yejus
asked yesterday
YejusYejus
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New contributor
Yejus is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked yesterday
YejusYejus
62
asked yesterday
YejusYejus
62
62
New contributor
Yejus is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor
Yejus is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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$begingroup$
$v_j=0$ is an obvious counterexample.
$endgroup$
– Kavi Rama Murthy
yesterday
1
$begingroup$
Possibly the additional condition $(A−λI)^jv_jnot=0$ has to be imposed.
$endgroup$
– Jens Schwaiger
yesterday
$begingroup$
I imagine that what you mean is that: $$ (A-lambda I)^jv_jne 0quadtextandquad (A-lambda I)^j+1v_j=0.$$
$endgroup$
– Yiorgos S. Smyrlis
yesterday
add a comment |
$begingroup$
$v_j=0$ is an obvious counterexample.
$endgroup$
– Kavi Rama Murthy
yesterday
1
$begingroup$
Possibly the additional condition $(A−λI)^jv_jnot=0$ has to be imposed.
$endgroup$
– Jens Schwaiger
yesterday
$begingroup$
I imagine that what you mean is that: $$ (A-lambda I)^jv_jne 0quadtextandquad (A-lambda I)^j+1v_j=0.$$
$endgroup$
– Yiorgos S. Smyrlis
yesterday
$begingroup$
$v_j=0$ is an obvious counterexample.
$endgroup$
– Kavi Rama Murthy
yesterday
$begingroup$
$v_j=0$ is an obvious counterexample.
$endgroup$
– Kavi Rama Murthy
yesterday
1
1
$begingroup$
Possibly the additional condition $(A−λI)^jv_jnot=0$ has to be imposed.
$endgroup$
– Jens Schwaiger
yesterday
$begingroup$
Possibly the additional condition $(A−λI)^jv_jnot=0$ has to be imposed.
$endgroup$
– Jens Schwaiger
yesterday
$begingroup$
I imagine that what you mean is that: $$ (A-lambda I)^jv_jne 0quadtextandquad (A-lambda I)^j+1v_j=0.$$
$endgroup$
– Yiorgos S. Smyrlis
yesterday
$begingroup$
I imagine that what you mean is that: $$ (A-lambda I)^jv_jne 0quadtextandquad (A-lambda I)^j+1v_j=0.$$
$endgroup$
– Yiorgos S. Smyrlis
yesterday
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I use the assumption that
$$ (A-lambda I)^jv_jne 0quadtextandquad (A-lambda I)^j+1v_j=0,$$
otherwise it is not true. For example
$$
lambda=0, quad
A=left(beginarraycc 0 & 1\ 0 & 0endarrayright), quad v_j=binom01, quad jinmathbb N.
$$
In order to prove that $v_1,ldots,v_n$ are linearly independent, it suffices to show that, for all $k$, the vector $v_k$ can not be expressed as a linear combination of $v_1,ldots,v_k-1$.
Clearly, for $k=1$, this holds, since $(A-lambda)v_1ne 0$, and hence $v_1ne 0$.
Assume that
$v_k=c_1v_1+cdots+c_k-1v_k-1$. Then
$$
0ne (A-lambda I)^kv_k
=(A-lambda I)^kbig(c_1v_1+cdots+c_k-1v_k-1big)=0
$$
answered yesterday
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.4k1385164
$endgroup$
add a comment |
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$begingroup$
I use the assumption that
$$ (A-lambda I)^jv_jne 0quadtextandquad (A-lambda I)^j+1v_j=0,$$
otherwise it is not true. For example
$$
lambda=0, quad
A=left(beginarraycc 0 & 1\ 0 & 0endarrayright), quad v_j=binom01, quad jinmathbb N.
$$
In order to prove that $v_1,ldots,v_n$ are linearly independent, it suffices to show that, for all $k$, the vector $v_k$ can not be expressed as a linear combination of $v_1,ldots,v_k-1$.
Clearly, for $k=1$, this holds, since $(A-lambda)v_1ne 0$, and hence $v_1ne 0$.
Assume that
$v_k=c_1v_1+cdots+c_k-1v_k-1$. Then
$$
0ne (A-lambda I)^kv_k
=(A-lambda I)^kbig(c_1v_1+cdots+c_k-1v_k-1big)=0
$$
answered yesterday
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.4k1385164
$endgroup$
add a comment |
$begingroup$
I use the assumption that
$$ (A-lambda I)^jv_jne 0quadtextandquad (A-lambda I)^j+1v_j=0,$$
otherwise it is not true. For example
$$
lambda=0, quad
A=left(beginarraycc 0 & 1\ 0 & 0endarrayright), quad v_j=binom01, quad jinmathbb N.
$$
In order to prove that $v_1,ldots,v_n$ are linearly independent, it suffices to show that, for all $k$, the vector $v_k$ can not be expressed as a linear combination of $v_1,ldots,v_k-1$.
Clearly, for $k=1$, this holds, since $(A-lambda)v_1ne 0$, and hence $v_1ne 0$.
Assume that
$v_k=c_1v_1+cdots+c_k-1v_k-1$. Then
$$
0ne (A-lambda I)^kv_k
=(A-lambda I)^kbig(c_1v_1+cdots+c_k-1v_k-1big)=0
$$
answered yesterday
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.4k1385164
$endgroup$
add a comment |
$begingroup$
I use the assumption that
$$ (A-lambda I)^jv_jne 0quadtextandquad (A-lambda I)^j+1v_j=0,$$
otherwise it is not true. For example
$$
lambda=0, quad
A=left(beginarraycc 0 & 1\ 0 & 0endarrayright), quad v_j=binom01, quad jinmathbb N.
$$
In order to prove that $v_1,ldots,v_n$ are linearly independent, it suffices to show that, for all $k$, the vector $v_k$ can not be expressed as a linear combination of $v_1,ldots,v_k-1$.
Clearly, for $k=1$, this holds, since $(A-lambda)v_1ne 0$, and hence $v_1ne 0$.
Assume that
$v_k=c_1v_1+cdots+c_k-1v_k-1$. Then
$$
0ne (A-lambda I)^kv_k
=(A-lambda I)^kbig(c_1v_1+cdots+c_k-1v_k-1big)=0
$$
answered yesterday
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.4k1385164
$endgroup$
I use the assumption that
$$ (A-lambda I)^jv_jne 0quadtextandquad (A-lambda I)^j+1v_j=0,$$
otherwise it is not true. For example
$$
lambda=0, quad
A=left(beginarraycc 0 & 1\ 0 & 0endarrayright), quad v_j=binom01, quad jinmathbb N.
$$
In order to prove that $v_1,ldots,v_n$ are linearly independent, it suffices to show that, for all $k$, the vector $v_k$ can not be expressed as a linear combination of $v_1,ldots,v_k-1$.
Clearly, for $k=1$, this holds, since $(A-lambda)v_1ne 0$, and hence $v_1ne 0$.
Assume that
$v_k=c_1v_1+cdots+c_k-1v_k-1$. Then
$$
0ne (A-lambda I)^kv_k
=(A-lambda I)^kbig(c_1v_1+cdots+c_k-1v_k-1big)=0
$$
answered yesterday
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.4k1385164
answered yesterday
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.4k1385164
answered yesterday
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.4k1385164
answered yesterday
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.4k1385164
63.4k1385164
add a comment |
add a comment |
Yejus is a new contributor. Be nice, and check out our Code of Conduct.
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8mt,xl5fh0eLvRZE1 OKHTQQ9 L,AJJ,2wQ ZpRocI
$begingroup$
$v_j=0$ is an obvious counterexample.
$endgroup$
– Kavi Rama Murthy
yesterday
1
$begingroup$
Possibly the additional condition $(A−λI)^jv_jnot=0$ has to be imposed.
$endgroup$
– Jens Schwaiger
yesterday
$begingroup$
I imagine that what you mean is that: $$ (A-lambda I)^jv_jne 0quadtextandquad (A-lambda I)^j+1v_j=0.$$
$endgroup$
– Yiorgos S. Smyrlis
yesterday