If we know the rank of a matrix r, can we assume that will have precisely r non-zero eigenvalues?
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I have looked at many answers on the internet regarding the relationship between rank and eigenvalues, and all of them contain complex calculations and descriptions too advanced for me, a beginner student of linear algebra, and do not really answer my question at hand, which is : is it possible to precisely determine the amount of non-zero eigenvalues of a matrix A simply by its rank r? There is a question in my textbook and vaguely alludes to this being the case, but it never actually goes into detail about the relationship between rank and eigenvalues.
linear-algebra eigenvalues-eigenvectors matrix-rank
asked Dec 17 '18 at 0:10
Will BurghardWill Burghard
42
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add a comment |
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I have looked at many answers on the internet regarding the relationship between rank and eigenvalues, and all of them contain complex calculations and descriptions too advanced for me, a beginner student of linear algebra, and do not really answer my question at hand, which is : is it possible to precisely determine the amount of non-zero eigenvalues of a matrix A simply by its rank r? There is a question in my textbook and vaguely alludes to this being the case, but it never actually goes into detail about the relationship between rank and eigenvalues.
linear-algebra eigenvalues-eigenvectors matrix-rank
asked Dec 17 '18 at 0:10
Will BurghardWill Burghard
42
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$begingroup$
Perhaps you may want to look at this. math.stackexchange.com/questions/1349907/…
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– Anurag A
Dec 17 '18 at 0:16
add a comment |
$begingroup$
I have looked at many answers on the internet regarding the relationship between rank and eigenvalues, and all of them contain complex calculations and descriptions too advanced for me, a beginner student of linear algebra, and do not really answer my question at hand, which is : is it possible to precisely determine the amount of non-zero eigenvalues of a matrix A simply by its rank r? There is a question in my textbook and vaguely alludes to this being the case, but it never actually goes into detail about the relationship between rank and eigenvalues.
linear-algebra eigenvalues-eigenvectors matrix-rank
asked Dec 17 '18 at 0:10
Will BurghardWill Burghard
42
$endgroup$
I have looked at many answers on the internet regarding the relationship between rank and eigenvalues, and all of them contain complex calculations and descriptions too advanced for me, a beginner student of linear algebra, and do not really answer my question at hand, which is : is it possible to precisely determine the amount of non-zero eigenvalues of a matrix A simply by its rank r? There is a question in my textbook and vaguely alludes to this being the case, but it never actually goes into detail about the relationship between rank and eigenvalues.
linear-algebra eigenvalues-eigenvectors matrix-rank
linear-algebra eigenvalues-eigenvectors matrix-rank
asked Dec 17 '18 at 0:10
Will BurghardWill Burghard
42
asked Dec 17 '18 at 0:10
Will BurghardWill Burghard
42
asked Dec 17 '18 at 0:10
Will BurghardWill Burghard
42
asked Dec 17 '18 at 0:10
Will BurghardWill Burghard
42
asked Dec 17 '18 at 0:10
Will BurghardWill Burghard
42
42
$begingroup$
Perhaps you may want to look at this. math.stackexchange.com/questions/1349907/…
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– Anurag A
Dec 17 '18 at 0:16
add a comment |
$begingroup$
Perhaps you may want to look at this. math.stackexchange.com/questions/1349907/…
$endgroup$
– Anurag A
Dec 17 '18 at 0:16
$begingroup$
Perhaps you may want to look at this. math.stackexchange.com/questions/1349907/…
$endgroup$
– Anurag A
Dec 17 '18 at 0:16
$begingroup$
Perhaps you may want to look at this. math.stackexchange.com/questions/1349907/…
$endgroup$
– Anurag A
Dec 17 '18 at 0:16
add a comment |
3 Answers
3
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oldest
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Consider the identity matrix in $3$ dimensions. It has rank $3$, but the characteristic polynomial is $(1-lambda)^3$, which only has one root $lambda=1$ of multiplicity $3$.
answered Dec 17 '18 at 0:18
YiFanYiFan
3,7511527
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The matrix will have exactly $r$ non-zero singular values, but not necessarily $r$ non-zero eigenvalues.
Consider the matrix $A = begin{bmatrix}0&1\0&0end{bmatrix}$. We have $text{rank}(A) = 1$, but both eigenvalues of $A$ are $0$.
answered Dec 17 '18 at 0:39
JimmyK4542JimmyK4542
41.1k245106
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add a comment |
$begingroup$
I have looked at many answers on the internet regarding the
relationship between rank and eigenvalues, and all of them contain
complex calculations and descriptions too advanced for me, a beginner
student of linear algebra
Supposing that the matrix is square and admits an eigendecomp then you would have $r$ eigenvalues right but there matrices with rank $r$ that aren't square and have $r$ singular values.
answered Dec 17 '18 at 0:18
Ryan HoweRyan Howe
2,45411323
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add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Consider the identity matrix in $3$ dimensions. It has rank $3$, but the characteristic polynomial is $(1-lambda)^3$, which only has one root $lambda=1$ of multiplicity $3$.
answered Dec 17 '18 at 0:18
YiFanYiFan
3,7511527
$endgroup$
add a comment |
$begingroup$
Consider the identity matrix in $3$ dimensions. It has rank $3$, but the characteristic polynomial is $(1-lambda)^3$, which only has one root $lambda=1$ of multiplicity $3$.
answered Dec 17 '18 at 0:18
YiFanYiFan
3,7511527
$endgroup$
add a comment |
$begingroup$
Consider the identity matrix in $3$ dimensions. It has rank $3$, but the characteristic polynomial is $(1-lambda)^3$, which only has one root $lambda=1$ of multiplicity $3$.
answered Dec 17 '18 at 0:18
YiFanYiFan
3,7511527
$endgroup$
Consider the identity matrix in $3$ dimensions. It has rank $3$, but the characteristic polynomial is $(1-lambda)^3$, which only has one root $lambda=1$ of multiplicity $3$.
answered Dec 17 '18 at 0:18
YiFanYiFan
3,7511527
answered Dec 17 '18 at 0:18
YiFanYiFan
3,7511527
answered Dec 17 '18 at 0:18
YiFanYiFan
3,7511527
answered Dec 17 '18 at 0:18
YiFanYiFan
3,7511527
3,7511527
add a comment |
add a comment |
$begingroup$
The matrix will have exactly $r$ non-zero singular values, but not necessarily $r$ non-zero eigenvalues.
Consider the matrix $A = begin{bmatrix}0&1\0&0end{bmatrix}$. We have $text{rank}(A) = 1$, but both eigenvalues of $A$ are $0$.
answered Dec 17 '18 at 0:39
JimmyK4542JimmyK4542
41.1k245106
$endgroup$
add a comment |
$begingroup$
The matrix will have exactly $r$ non-zero singular values, but not necessarily $r$ non-zero eigenvalues.
Consider the matrix $A = begin{bmatrix}0&1\0&0end{bmatrix}$. We have $text{rank}(A) = 1$, but both eigenvalues of $A$ are $0$.
answered Dec 17 '18 at 0:39
JimmyK4542JimmyK4542
41.1k245106
$endgroup$
add a comment |
$begingroup$
The matrix will have exactly $r$ non-zero singular values, but not necessarily $r$ non-zero eigenvalues.
Consider the matrix $A = begin{bmatrix}0&1\0&0end{bmatrix}$. We have $text{rank}(A) = 1$, but both eigenvalues of $A$ are $0$.
answered Dec 17 '18 at 0:39
JimmyK4542JimmyK4542
41.1k245106
$endgroup$
The matrix will have exactly $r$ non-zero singular values, but not necessarily $r$ non-zero eigenvalues.
Consider the matrix $A = begin{bmatrix}0&1\0&0end{bmatrix}$. We have $text{rank}(A) = 1$, but both eigenvalues of $A$ are $0$.
answered Dec 17 '18 at 0:39
JimmyK4542JimmyK4542
41.1k245106
answered Dec 17 '18 at 0:39
JimmyK4542JimmyK4542
41.1k245106
answered Dec 17 '18 at 0:39
JimmyK4542JimmyK4542
41.1k245106
answered Dec 17 '18 at 0:39
JimmyK4542JimmyK4542
41.1k245106
41.1k245106
add a comment |
add a comment |
$begingroup$
I have looked at many answers on the internet regarding the
relationship between rank and eigenvalues, and all of them contain
complex calculations and descriptions too advanced for me, a beginner
student of linear algebra
Supposing that the matrix is square and admits an eigendecomp then you would have $r$ eigenvalues right but there matrices with rank $r$ that aren't square and have $r$ singular values.
answered Dec 17 '18 at 0:18
Ryan HoweRyan Howe
2,45411323
$endgroup$
add a comment |
$begingroup$
I have looked at many answers on the internet regarding the
relationship between rank and eigenvalues, and all of them contain
complex calculations and descriptions too advanced for me, a beginner
student of linear algebra
Supposing that the matrix is square and admits an eigendecomp then you would have $r$ eigenvalues right but there matrices with rank $r$ that aren't square and have $r$ singular values.
answered Dec 17 '18 at 0:18
Ryan HoweRyan Howe
2,45411323
$endgroup$
add a comment |
$begingroup$
I have looked at many answers on the internet regarding the
relationship between rank and eigenvalues, and all of them contain
complex calculations and descriptions too advanced for me, a beginner
student of linear algebra
Supposing that the matrix is square and admits an eigendecomp then you would have $r$ eigenvalues right but there matrices with rank $r$ that aren't square and have $r$ singular values.
answered Dec 17 '18 at 0:18
Ryan HoweRyan Howe
2,45411323
$endgroup$
I have looked at many answers on the internet regarding the
relationship between rank and eigenvalues, and all of them contain
complex calculations and descriptions too advanced for me, a beginner
student of linear algebra
Supposing that the matrix is square and admits an eigendecomp then you would have $r$ eigenvalues right but there matrices with rank $r$ that aren't square and have $r$ singular values.
answered Dec 17 '18 at 0:18
Ryan HoweRyan Howe
2,45411323
answered Dec 17 '18 at 0:18
Ryan HoweRyan Howe
2,45411323
answered Dec 17 '18 at 0:18
Ryan HoweRyan Howe
2,45411323
answered Dec 17 '18 at 0:18
Ryan HoweRyan Howe
2,45411323
2,45411323
add a comment |
add a comment |
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$begingroup$
Perhaps you may want to look at this. math.stackexchange.com/questions/1349907/…
$endgroup$
– Anurag A
Dec 17 '18 at 0:16