Linear independence proof of sublist from a list of dependent vectors
$begingroup$
Let $lambda$ be an eigenvalue of $A$, such that no eigenvector of $A$ associated with $lambda$ has a zero entry. Then prove that every list of $n-1$ columns of $A-lambda I$ is linearly independent.
This problem is originally from the book Matrix Analysis (2ed) by Roger A. Horn. It's 1.4.P12 on page 82. I know how to prove the reverse direction, but I have no idea for above direction. Any suggestions?
linear-algebra eigenvalues-eigenvectors vectors independence
edited Jan 21 at 9:48
egreg
182k1486204
asked Jan 21 at 8:34
Ton MreTon Mre
82
$endgroup$
add a comment |
$begingroup$
Let $lambda$ be an eigenvalue of $A$, such that no eigenvector of $A$ associated with $lambda$ has a zero entry. Then prove that every list of $n-1$ columns of $A-lambda I$ is linearly independent.
This problem is originally from the book Matrix Analysis (2ed) by Roger A. Horn. It's 1.4.P12 on page 82. I know how to prove the reverse direction, but I have no idea for above direction. Any suggestions?
linear-algebra eigenvalues-eigenvectors vectors independence
edited Jan 21 at 9:48
egreg
182k1486204
asked Jan 21 at 8:34
Ton MreTon Mre
82
$endgroup$
add a comment |
$begingroup$
Let $lambda$ be an eigenvalue of $A$, such that no eigenvector of $A$ associated with $lambda$ has a zero entry. Then prove that every list of $n-1$ columns of $A-lambda I$ is linearly independent.
This problem is originally from the book Matrix Analysis (2ed) by Roger A. Horn. It's 1.4.P12 on page 82. I know how to prove the reverse direction, but I have no idea for above direction. Any suggestions?
linear-algebra eigenvalues-eigenvectors vectors independence
edited Jan 21 at 9:48
egreg
182k1486204
asked Jan 21 at 8:34
Ton MreTon Mre
82
$endgroup$
Let $lambda$ be an eigenvalue of $A$, such that no eigenvector of $A$ associated with $lambda$ has a zero entry. Then prove that every list of $n-1$ columns of $A-lambda I$ is linearly independent.
This problem is originally from the book Matrix Analysis (2ed) by Roger A. Horn. It's 1.4.P12 on page 82. I know how to prove the reverse direction, but I have no idea for above direction. Any suggestions?
linear-algebra eigenvalues-eigenvectors vectors independence
linear-algebra eigenvalues-eigenvectors vectors independence
edited Jan 21 at 9:48
egreg
182k1486204
asked Jan 21 at 8:34
Ton MreTon Mre
82
edited Jan 21 at 9:48
egreg
182k1486204
asked Jan 21 at 8:34
Ton MreTon Mre
82
edited Jan 21 at 9:48
egreg
182k1486204
edited Jan 21 at 9:48
egreg
182k1486204
edited Jan 21 at 9:48
egreg
182k1486204
182k1486204
asked Jan 21 at 8:34
Ton MreTon Mre
82
asked Jan 21 at 8:34
Ton MreTon Mre
82
asked Jan 21 at 8:34
Ton MreTon Mre
82
82
add a comment |
add a comment |
1 Answer
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The null space of $A_{ntimes n}-lambda I$ is the eigenspace of $lambda$. Since no eigenvector corresponding to $lambda$ contains a $0$ entry, the basis of the eigenspace must contain a single eigenvector of $A$. This means the nullity of $A-lambda I$ is $1$, and by the rank-nullity theorem, its rank $n-1$.
Now recall that $(A-lambda I)v$ is a linear combination of the columns of $A-lambda I$. This means their exists a tuple $(v_1,v_2,...,v_n)$ with $v_ine0$ that satisfies $v_1C_1+v_2C_2+cdotcdotcdot+v_nC_n=0$. Since the rank of these vectors is $n-1$, there is one linearly dependent column $C_j$, and since $v_jne0$ for all $j$, that $C_j$ could be any column
answered Jan 21 at 8:51
Shubham JohriShubham Johri
5,186717
$endgroup$
$begingroup$
Simple but effective.I've fidued it out.Tks a lot!
$endgroup$
– Ton Mre
Jan 21 at 13:11
$begingroup$
@TonMre Recall that $(A-lambda I)v$ is a linear combination of the columns of $A-lambda I$. This means their exists a tuple $(v_1,v_2,...,v_n)$ with $v_ine0$ that satisfies $v_1C_1+v_2C_2+cdotcdotcdot+v_nC_n=0$. Since the rank of these vectors is $n-1$, there is one linearly dependent column $C_j$, and since $v_jne0$ for all $j$, that $C_j$ could be any column
$endgroup$
– Shubham Johri
Jan 21 at 13:19
add a comment |
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1 Answer
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$begingroup$
The null space of $A_{ntimes n}-lambda I$ is the eigenspace of $lambda$. Since no eigenvector corresponding to $lambda$ contains a $0$ entry, the basis of the eigenspace must contain a single eigenvector of $A$. This means the nullity of $A-lambda I$ is $1$, and by the rank-nullity theorem, its rank $n-1$.
Now recall that $(A-lambda I)v$ is a linear combination of the columns of $A-lambda I$. This means their exists a tuple $(v_1,v_2,...,v_n)$ with $v_ine0$ that satisfies $v_1C_1+v_2C_2+cdotcdotcdot+v_nC_n=0$. Since the rank of these vectors is $n-1$, there is one linearly dependent column $C_j$, and since $v_jne0$ for all $j$, that $C_j$ could be any column
answered Jan 21 at 8:51
Shubham JohriShubham Johri
5,186717
$endgroup$
$begingroup$
Simple but effective.I've fidued it out.Tks a lot!
$endgroup$
– Ton Mre
Jan 21 at 13:11
$begingroup$
@TonMre Recall that $(A-lambda I)v$ is a linear combination of the columns of $A-lambda I$. This means their exists a tuple $(v_1,v_2,...,v_n)$ with $v_ine0$ that satisfies $v_1C_1+v_2C_2+cdotcdotcdot+v_nC_n=0$. Since the rank of these vectors is $n-1$, there is one linearly dependent column $C_j$, and since $v_jne0$ for all $j$, that $C_j$ could be any column
$endgroup$
– Shubham Johri
Jan 21 at 13:19
add a comment |
$begingroup$
The null space of $A_{ntimes n}-lambda I$ is the eigenspace of $lambda$. Since no eigenvector corresponding to $lambda$ contains a $0$ entry, the basis of the eigenspace must contain a single eigenvector of $A$. This means the nullity of $A-lambda I$ is $1$, and by the rank-nullity theorem, its rank $n-1$.
Now recall that $(A-lambda I)v$ is a linear combination of the columns of $A-lambda I$. This means their exists a tuple $(v_1,v_2,...,v_n)$ with $v_ine0$ that satisfies $v_1C_1+v_2C_2+cdotcdotcdot+v_nC_n=0$. Since the rank of these vectors is $n-1$, there is one linearly dependent column $C_j$, and since $v_jne0$ for all $j$, that $C_j$ could be any column
answered Jan 21 at 8:51
Shubham JohriShubham Johri
5,186717
$endgroup$
$begingroup$
Simple but effective.I've fidued it out.Tks a lot!
$endgroup$
– Ton Mre
Jan 21 at 13:11
$begingroup$
@TonMre Recall that $(A-lambda I)v$ is a linear combination of the columns of $A-lambda I$. This means their exists a tuple $(v_1,v_2,...,v_n)$ with $v_ine0$ that satisfies $v_1C_1+v_2C_2+cdotcdotcdot+v_nC_n=0$. Since the rank of these vectors is $n-1$, there is one linearly dependent column $C_j$, and since $v_jne0$ for all $j$, that $C_j$ could be any column
$endgroup$
– Shubham Johri
Jan 21 at 13:19
add a comment |
$begingroup$
The null space of $A_{ntimes n}-lambda I$ is the eigenspace of $lambda$. Since no eigenvector corresponding to $lambda$ contains a $0$ entry, the basis of the eigenspace must contain a single eigenvector of $A$. This means the nullity of $A-lambda I$ is $1$, and by the rank-nullity theorem, its rank $n-1$.
Now recall that $(A-lambda I)v$ is a linear combination of the columns of $A-lambda I$. This means their exists a tuple $(v_1,v_2,...,v_n)$ with $v_ine0$ that satisfies $v_1C_1+v_2C_2+cdotcdotcdot+v_nC_n=0$. Since the rank of these vectors is $n-1$, there is one linearly dependent column $C_j$, and since $v_jne0$ for all $j$, that $C_j$ could be any column
answered Jan 21 at 8:51
Shubham JohriShubham Johri
5,186717
$endgroup$
The null space of $A_{ntimes n}-lambda I$ is the eigenspace of $lambda$. Since no eigenvector corresponding to $lambda$ contains a $0$ entry, the basis of the eigenspace must contain a single eigenvector of $A$. This means the nullity of $A-lambda I$ is $1$, and by the rank-nullity theorem, its rank $n-1$.
Now recall that $(A-lambda I)v$ is a linear combination of the columns of $A-lambda I$. This means their exists a tuple $(v_1,v_2,...,v_n)$ with $v_ine0$ that satisfies $v_1C_1+v_2C_2+cdotcdotcdot+v_nC_n=0$. Since the rank of these vectors is $n-1$, there is one linearly dependent column $C_j$, and since $v_jne0$ for all $j$, that $C_j$ could be any column
answered Jan 21 at 8:51
Shubham JohriShubham Johri
5,186717
edited Jan 21 at 13:21
answered Jan 21 at 8:51
Shubham JohriShubham Johri
5,186717
answered Jan 21 at 8:51
Shubham JohriShubham Johri
5,186717
answered Jan 21 at 8:51
Shubham JohriShubham Johri
5,186717
5,186717
$begingroup$
Simple but effective.I've fidued it out.Tks a lot!
$endgroup$
– Ton Mre
Jan 21 at 13:11
$begingroup$
@TonMre Recall that $(A-lambda I)v$ is a linear combination of the columns of $A-lambda I$. This means their exists a tuple $(v_1,v_2,...,v_n)$ with $v_ine0$ that satisfies $v_1C_1+v_2C_2+cdotcdotcdot+v_nC_n=0$. Since the rank of these vectors is $n-1$, there is one linearly dependent column $C_j$, and since $v_jne0$ for all $j$, that $C_j$ could be any column
$endgroup$
– Shubham Johri
Jan 21 at 13:19
add a comment |
$begingroup$
Simple but effective.I've fidued it out.Tks a lot!
$endgroup$
– Ton Mre
Jan 21 at 13:11
$begingroup$
@TonMre Recall that $(A-lambda I)v$ is a linear combination of the columns of $A-lambda I$. This means their exists a tuple $(v_1,v_2,...,v_n)$ with $v_ine0$ that satisfies $v_1C_1+v_2C_2+cdotcdotcdot+v_nC_n=0$. Since the rank of these vectors is $n-1$, there is one linearly dependent column $C_j$, and since $v_jne0$ for all $j$, that $C_j$ could be any column
$endgroup$
– Shubham Johri
Jan 21 at 13:19
$begingroup$
Simple but effective.I've fidued it out.Tks a lot!
$endgroup$
– Ton Mre
Jan 21 at 13:11
$begingroup$
Simple but effective.I've fidued it out.Tks a lot!
$endgroup$
– Ton Mre
Jan 21 at 13:11
$begingroup$
@TonMre Recall that $(A-lambda I)v$ is a linear combination of the columns of $A-lambda I$. This means their exists a tuple $(v_1,v_2,...,v_n)$ with $v_ine0$ that satisfies $v_1C_1+v_2C_2+cdotcdotcdot+v_nC_n=0$. Since the rank of these vectors is $n-1$, there is one linearly dependent column $C_j$, and since $v_jne0$ for all $j$, that $C_j$ could be any column
$endgroup$
– Shubham Johri
Jan 21 at 13:19
$begingroup$
@TonMre Recall that $(A-lambda I)v$ is a linear combination of the columns of $A-lambda I$. This means their exists a tuple $(v_1,v_2,...,v_n)$ with $v_ine0$ that satisfies $v_1C_1+v_2C_2+cdotcdotcdot+v_nC_n=0$. Since the rank of these vectors is $n-1$, there is one linearly dependent column $C_j$, and since $v_jne0$ for all $j$, that $C_j$ could be any column
$endgroup$
– Shubham Johri
Jan 21 at 13:19
add a comment |
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