Show the non-zero vectors $v_1,v_2,⋯,v_n$ are linearly independent if $(A−λI)^{j+1}v_j=0$, where...
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Show the non-zero vectors $v_1,v_2,⋯,v_n$ are linearly independent if $(A−λI)^{j+1}v_j=0$, 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 $Sp{v_1,v_2}=Sp{w_1,w_2}$ and {$v_1,w_2$} are linearly independent?linear-Independency is retained over a linear MapShow that if Span${v_1,v_2,v_3}$=$mathbb{R}^3$ 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,
$$v_j = a_1v_1 + a_2v_2 + cdots + a_{j-1}v_{j-1} + a_{j+1}v_{j+1} + cdots + a_nv_n.$$ Applying the operator $(A - lambda I)^{j+1}$ to both sides, we have
$$ 0 = (A - lambda I)^{j+1}(a_1v_1 + a_2v_2 + cdots + a_{j-1}v_{j-1}) + (A - lambda I)^{j+1}(a_{j-1}v_{j-1} + a_{j+1}v_{j+1} + cdots + a_nv_n).$$ The first sum is zero, since the vectors there require the operator $A - lambda I$ applied at most $j$ times to be $0$. Hence, $$0 = (A - lambda I)^{j+1}(a_{j-1}v_{j-1} + a_{j+1}v_{j+1} + cdots + a_nv_n).$$
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 2 days ago
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,
$$v_j = a_1v_1 + a_2v_2 + cdots + a_{j-1}v_{j-1} + a_{j+1}v_{j+1} + cdots + a_nv_n.$$ Applying the operator $(A - lambda I)^{j+1}$ to both sides, we have
$$ 0 = (A - lambda I)^{j+1}(a_1v_1 + a_2v_2 + cdots + a_{j-1}v_{j-1}) + (A - lambda I)^{j+1}(a_{j-1}v_{j-1} + a_{j+1}v_{j+1} + cdots + a_nv_n).$$ The first sum is zero, since the vectors there require the operator $A - lambda I$ applied at most $j$ times to be $0$. Hence, $$0 = (A - lambda I)^{j+1}(a_{j-1}v_{j-1} + a_{j+1}v_{j+1} + cdots + a_nv_n).$$
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 2 days ago
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
2 days ago
1
$begingroup$
Possibly the additional condition $(A−λI)^{j}v_jnot=0$ has to be imposed.
$endgroup$
– Jens Schwaiger
2 days ago
$begingroup$
I imagine that what you mean is that: $$ (A-lambda I)^jv_jne 0quadtext{and}quad (A-lambda I)^{j+1}v_j=0.$$
$endgroup$
– Yiorgos S. Smyrlis
2 days ago
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,
$$v_j = a_1v_1 + a_2v_2 + cdots + a_{j-1}v_{j-1} + a_{j+1}v_{j+1} + cdots + a_nv_n.$$ Applying the operator $(A - lambda I)^{j+1}$ to both sides, we have
$$ 0 = (A - lambda I)^{j+1}(a_1v_1 + a_2v_2 + cdots + a_{j-1}v_{j-1}) + (A - lambda I)^{j+1}(a_{j-1}v_{j-1} + a_{j+1}v_{j+1} + cdots + a_nv_n).$$ The first sum is zero, since the vectors there require the operator $A - lambda I$ applied at most $j$ times to be $0$. Hence, $$0 = (A - lambda I)^{j+1}(a_{j-1}v_{j-1} + a_{j+1}v_{j+1} + cdots + a_nv_n).$$
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 2 days ago
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,
$$v_j = a_1v_1 + a_2v_2 + cdots + a_{j-1}v_{j-1} + a_{j+1}v_{j+1} + cdots + a_nv_n.$$ Applying the operator $(A - lambda I)^{j+1}$ to both sides, we have
$$ 0 = (A - lambda I)^{j+1}(a_1v_1 + a_2v_2 + cdots + a_{j-1}v_{j-1}) + (A - lambda I)^{j+1}(a_{j-1}v_{j-1} + a_{j+1}v_{j+1} + cdots + a_nv_n).$$ The first sum is zero, since the vectors there require the operator $A - lambda I$ applied at most $j$ times to be $0$. Hence, $$0 = (A - lambda I)^{j+1}(a_{j-1}v_{j-1} + a_{j+1}v_{j+1} + cdots + a_nv_n).$$
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 2 days ago
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 2 days ago
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 2 days ago
Yejus
asked 2 days ago
YejusYejus
62
New contributor
Yejus is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked 2 days ago
YejusYejus
62
asked 2 days ago
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.
Yejus is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
$v_j=0$ is an obvious counterexample.
$endgroup$
– Kavi Rama Murthy
2 days ago
1
$begingroup$
Possibly the additional condition $(A−λI)^{j}v_jnot=0$ has to be imposed.
$endgroup$
– Jens Schwaiger
2 days ago
$begingroup$
I imagine that what you mean is that: $$ (A-lambda I)^jv_jne 0quadtext{and}quad (A-lambda I)^{j+1}v_j=0.$$
$endgroup$
– Yiorgos S. Smyrlis
2 days ago
add a comment |
$begingroup$
$v_j=0$ is an obvious counterexample.
$endgroup$
– Kavi Rama Murthy
2 days ago
1
$begingroup$
Possibly the additional condition $(A−λI)^{j}v_jnot=0$ has to be imposed.
$endgroup$
– Jens Schwaiger
2 days ago
$begingroup$
I imagine that what you mean is that: $$ (A-lambda I)^jv_jne 0quadtext{and}quad (A-lambda I)^{j+1}v_j=0.$$
$endgroup$
– Yiorgos S. Smyrlis
2 days ago
$begingroup$
$v_j=0$ is an obvious counterexample.
$endgroup$
– Kavi Rama Murthy
2 days ago
$begingroup$
$v_j=0$ is an obvious counterexample.
$endgroup$
– Kavi Rama Murthy
2 days ago
1
1
$begingroup$
Possibly the additional condition $(A−λI)^{j}v_jnot=0$ has to be imposed.
$endgroup$
– Jens Schwaiger
2 days ago
$begingroup$
Possibly the additional condition $(A−λI)^{j}v_jnot=0$ has to be imposed.
$endgroup$
– Jens Schwaiger
2 days ago
$begingroup$
I imagine that what you mean is that: $$ (A-lambda I)^jv_jne 0quadtext{and}quad (A-lambda I)^{j+1}v_j=0.$$
$endgroup$
– Yiorgos S. Smyrlis
2 days ago
$begingroup$
I imagine that what you mean is that: $$ (A-lambda I)^jv_jne 0quadtext{and}quad (A-lambda I)^{j+1}v_j=0.$$
$endgroup$
– Yiorgos S. Smyrlis
2 days ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I use the assumption that
$$ (A-lambda I)^jv_jne 0quadtext{and}quad (A-lambda I)^{j+1}v_j=0,$$
otherwise it is not true. For example
$$
lambda=0, quad
A=left(begin{array}{cc} 0 & 1\ 0 & 0end{array}right), quad v_j=binom{0}{1}, quad jinmathbb N.
$$
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
$v_k=c_1v_1+cdots+c_{k-1}v_{k-1}$. Then
$$
0ne (A-lambda I)^kv_k
=(A-lambda I)^kbig(c_1v_1+cdots+c_{k-1}v_{k-1}big)=0
$$
answered 2 days ago
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.4k1385164
$endgroup$
add a comment |
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1 Answer
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$begingroup$
I use the assumption that
$$ (A-lambda I)^jv_jne 0quadtext{and}quad (A-lambda I)^{j+1}v_j=0,$$
otherwise it is not true. For example
$$
lambda=0, quad
A=left(begin{array}{cc} 0 & 1\ 0 & 0end{array}right), quad v_j=binom{0}{1}, quad jinmathbb N.
$$
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
$v_k=c_1v_1+cdots+c_{k-1}v_{k-1}$. Then
$$
0ne (A-lambda I)^kv_k
=(A-lambda I)^kbig(c_1v_1+cdots+c_{k-1}v_{k-1}big)=0
$$
answered 2 days ago
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.4k1385164
$endgroup$
add a comment |
$begingroup$
I use the assumption that
$$ (A-lambda I)^jv_jne 0quadtext{and}quad (A-lambda I)^{j+1}v_j=0,$$
otherwise it is not true. For example
$$
lambda=0, quad
A=left(begin{array}{cc} 0 & 1\ 0 & 0end{array}right), quad v_j=binom{0}{1}, quad jinmathbb N.
$$
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
$v_k=c_1v_1+cdots+c_{k-1}v_{k-1}$. Then
$$
0ne (A-lambda I)^kv_k
=(A-lambda I)^kbig(c_1v_1+cdots+c_{k-1}v_{k-1}big)=0
$$
answered 2 days ago
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.4k1385164
$endgroup$
add a comment |
$begingroup$
I use the assumption that
$$ (A-lambda I)^jv_jne 0quadtext{and}quad (A-lambda I)^{j+1}v_j=0,$$
otherwise it is not true. For example
$$
lambda=0, quad
A=left(begin{array}{cc} 0 & 1\ 0 & 0end{array}right), quad v_j=binom{0}{1}, quad jinmathbb N.
$$
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
$v_k=c_1v_1+cdots+c_{k-1}v_{k-1}$. Then
$$
0ne (A-lambda I)^kv_k
=(A-lambda I)^kbig(c_1v_1+cdots+c_{k-1}v_{k-1}big)=0
$$
answered 2 days ago
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.4k1385164
$endgroup$
I use the assumption that
$$ (A-lambda I)^jv_jne 0quadtext{and}quad (A-lambda I)^{j+1}v_j=0,$$
otherwise it is not true. For example
$$
lambda=0, quad
A=left(begin{array}{cc} 0 & 1\ 0 & 0end{array}right), quad v_j=binom{0}{1}, quad jinmathbb N.
$$
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
$v_k=c_1v_1+cdots+c_{k-1}v_{k-1}$. Then
$$
0ne (A-lambda I)^kv_k
=(A-lambda I)^kbig(c_1v_1+cdots+c_{k-1}v_{k-1}big)=0
$$
answered 2 days ago
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.4k1385164
answered 2 days ago
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.4k1385164
answered 2 days ago
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.4k1385164
answered 2 days ago
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.
Yejus is a new contributor. Be nice, and check out our Code of Conduct.
Yejus is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
$v_j=0$ is an obvious counterexample.
$endgroup$
– Kavi Rama Murthy
2 days ago
1
$begingroup$
Possibly the additional condition $(A−λI)^{j}v_jnot=0$ has to be imposed.
$endgroup$
– Jens Schwaiger
2 days ago
$begingroup$
I imagine that what you mean is that: $$ (A-lambda I)^jv_jne 0quadtext{and}quad (A-lambda I)^{j+1}v_j=0.$$
$endgroup$
– Yiorgos S. Smyrlis
2 days ago