Matrix $C in M_n(mathbbR)$ $lambda_1=lambda_2=0$ rankMatrix rank from rank of sub matricessimilar matrices, real eigenvalues, matrix rank,Finding the rank of an non-invertible matrixRank Deficient matrixRank of a real matrixCan a matrix whose rows are linear combinations of rows of a full-rank matrix be guaranteed to be full rank?Rank of a matrix plus a numberRank of a Hermitian matrix in terms of Eigen values?Rank of matrix relatedWhy is the adjugate matrix the null matrix?
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Matrix $C in M_n(mathbbR)$ $lambda_1=lambda_2=0$ rank
Matrix rank from rank of sub matricessimilar matrices, real eigenvalues, matrix rank,Finding the rank of an non-invertible matrixRank Deficient matrixRank of a real matrixCan a matrix whose rows are linear combinations of rows of a full-rank matrix be guaranteed to be full rank?Rank of a matrix plus a numberRank of a Hermitian matrix in terms of Eigen values?Rank of matrix relatedWhy is the adjugate matrix the null matrix?
$begingroup$
If a have a matrix $C in M_n(mathbbR)$ with $C^2=O_n$ and $n=2k+1$ that has at least to eigenvalues equal to $0$. Can I say from this that the $rank Cleq fracn-12?$
matrix-rank
asked Mar 17 at 12:17
mathlearningmathlearning
1887
$endgroup$
add a comment |
$begingroup$
If a have a matrix $C in M_n(mathbbR)$ with $C^2=O_n$ and $n=2k+1$ that has at least to eigenvalues equal to $0$. Can I say from this that the $rank Cleq fracn-12?$
matrix-rank
asked Mar 17 at 12:17
mathlearningmathlearning
1887
$endgroup$
2
$begingroup$
You need only assume $C^2 = O$, then in fact all eigenvalues are $0$, since $C$ is nilpotent.
$endgroup$
– Minus One-Twelfth
Mar 17 at 12:46
1
$begingroup$
Since $C^2=0$, then $operatornameIm(C)subseteqoperatornameKer(C)$. Then $n=operatornamedimoperatornameKer(C)+operatornamedimoperatornameIm(C)geq 2operatornamedimoperatornameIm(C)$. Since $n$ is odd and $operatornamedimoperatornameIm(C)$ a natural number, then $fracn-12geq operatornamedimoperatornameIm(C)$.
$endgroup$
– user647486
Mar 17 at 12:56
add a comment |
$begingroup$
If a have a matrix $C in M_n(mathbbR)$ with $C^2=O_n$ and $n=2k+1$ that has at least to eigenvalues equal to $0$. Can I say from this that the $rank Cleq fracn-12?$
matrix-rank
asked Mar 17 at 12:17
mathlearningmathlearning
1887
$endgroup$
If a have a matrix $C in M_n(mathbbR)$ with $C^2=O_n$ and $n=2k+1$ that has at least to eigenvalues equal to $0$. Can I say from this that the $rank Cleq fracn-12?$
matrix-rank
matrix-rank
asked Mar 17 at 12:17
mathlearningmathlearning
1887
asked Mar 17 at 12:17
mathlearningmathlearning
1887
asked Mar 17 at 12:17
mathlearningmathlearning
1887
asked Mar 17 at 12:17
mathlearningmathlearning
1887
asked Mar 17 at 12:17
mathlearningmathlearning
1887
1887
2
$begingroup$
You need only assume $C^2 = O$, then in fact all eigenvalues are $0$, since $C$ is nilpotent.
$endgroup$
– Minus One-Twelfth
Mar 17 at 12:46
1
$begingroup$
Since $C^2=0$, then $operatornameIm(C)subseteqoperatornameKer(C)$. Then $n=operatornamedimoperatornameKer(C)+operatornamedimoperatornameIm(C)geq 2operatornamedimoperatornameIm(C)$. Since $n$ is odd and $operatornamedimoperatornameIm(C)$ a natural number, then $fracn-12geq operatornamedimoperatornameIm(C)$.
$endgroup$
– user647486
Mar 17 at 12:56
add a comment |
2
$begingroup$
You need only assume $C^2 = O$, then in fact all eigenvalues are $0$, since $C$ is nilpotent.
$endgroup$
– Minus One-Twelfth
Mar 17 at 12:46
1
$begingroup$
Since $C^2=0$, then $operatornameIm(C)subseteqoperatornameKer(C)$. Then $n=operatornamedimoperatornameKer(C)+operatornamedimoperatornameIm(C)geq 2operatornamedimoperatornameIm(C)$. Since $n$ is odd and $operatornamedimoperatornameIm(C)$ a natural number, then $fracn-12geq operatornamedimoperatornameIm(C)$.
$endgroup$
– user647486
Mar 17 at 12:56
2
2
$begingroup$
You need only assume $C^2 = O$, then in fact all eigenvalues are $0$, since $C$ is nilpotent.
$endgroup$
– Minus One-Twelfth
Mar 17 at 12:46
$begingroup$
You need only assume $C^2 = O$, then in fact all eigenvalues are $0$, since $C$ is nilpotent.
$endgroup$
– Minus One-Twelfth
Mar 17 at 12:46
1
1
$begingroup$
Since $C^2=0$, then $operatornameIm(C)subseteqoperatornameKer(C)$. Then $n=operatornamedimoperatornameKer(C)+operatornamedimoperatornameIm(C)geq 2operatornamedimoperatornameIm(C)$. Since $n$ is odd and $operatornamedimoperatornameIm(C)$ a natural number, then $fracn-12geq operatornamedimoperatornameIm(C)$.
$endgroup$
– user647486
Mar 17 at 12:56
$begingroup$
Since $C^2=0$, then $operatornameIm(C)subseteqoperatornameKer(C)$. Then $n=operatornamedimoperatornameKer(C)+operatornamedimoperatornameIm(C)geq 2operatornamedimoperatornameIm(C)$. Since $n$ is odd and $operatornamedimoperatornameIm(C)$ a natural number, then $fracn-12geq operatornamedimoperatornameIm(C)$.
$endgroup$
– user647486
Mar 17 at 12:56
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$newcommandrankoperatornameranknewcommandnullityoperatornamenullitynewcommandcmathbfcnewcommandmathbf0$Yes. Hints: You want to show that $rank(C)le k$. Let the columns of $C$ be $c_1,ldots,c_n$. Then since $C^2 = O$, we have $Cc_j = $ for all $j=1,ldots, n$. Now suppose by way of contradiction that $rank(C) > k$. Then since (by the rank-nullity theorem) $$nullity(C) = 2k+1-rank(C),$$ we have $$nullity(C) < 2k+1-k = k+1 Rightarrow nullity(C) le k.$$ But the columns of $C$ are all from the null space of $C$ and $rank(C) > k$. In view of the above, can you explain why this is impossible?
answered Mar 17 at 12:44
Minus One-TwelfthMinus One-Twelfth
2,823413
$endgroup$
add a comment |
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$begingroup$
$newcommandrankoperatornameranknewcommandnullityoperatornamenullitynewcommandcmathbfcnewcommandmathbf0$Yes. Hints: You want to show that $rank(C)le k$. Let the columns of $C$ be $c_1,ldots,c_n$. Then since $C^2 = O$, we have $Cc_j = $ for all $j=1,ldots, n$. Now suppose by way of contradiction that $rank(C) > k$. Then since (by the rank-nullity theorem) $$nullity(C) = 2k+1-rank(C),$$ we have $$nullity(C) < 2k+1-k = k+1 Rightarrow nullity(C) le k.$$ But the columns of $C$ are all from the null space of $C$ and $rank(C) > k$. In view of the above, can you explain why this is impossible?
answered Mar 17 at 12:44
Minus One-TwelfthMinus One-Twelfth
2,823413
$endgroup$
add a comment |
$begingroup$
$newcommandrankoperatornameranknewcommandnullityoperatornamenullitynewcommandcmathbfcnewcommandmathbf0$Yes. Hints: You want to show that $rank(C)le k$. Let the columns of $C$ be $c_1,ldots,c_n$. Then since $C^2 = O$, we have $Cc_j = $ for all $j=1,ldots, n$. Now suppose by way of contradiction that $rank(C) > k$. Then since (by the rank-nullity theorem) $$nullity(C) = 2k+1-rank(C),$$ we have $$nullity(C) < 2k+1-k = k+1 Rightarrow nullity(C) le k.$$ But the columns of $C$ are all from the null space of $C$ and $rank(C) > k$. In view of the above, can you explain why this is impossible?
answered Mar 17 at 12:44
Minus One-TwelfthMinus One-Twelfth
2,823413
$endgroup$
add a comment |
$begingroup$
$newcommandrankoperatornameranknewcommandnullityoperatornamenullitynewcommandcmathbfcnewcommandmathbf0$Yes. Hints: You want to show that $rank(C)le k$. Let the columns of $C$ be $c_1,ldots,c_n$. Then since $C^2 = O$, we have $Cc_j = $ for all $j=1,ldots, n$. Now suppose by way of contradiction that $rank(C) > k$. Then since (by the rank-nullity theorem) $$nullity(C) = 2k+1-rank(C),$$ we have $$nullity(C) < 2k+1-k = k+1 Rightarrow nullity(C) le k.$$ But the columns of $C$ are all from the null space of $C$ and $rank(C) > k$. In view of the above, can you explain why this is impossible?
answered Mar 17 at 12:44
Minus One-TwelfthMinus One-Twelfth
2,823413
$endgroup$
$newcommandrankoperatornameranknewcommandnullityoperatornamenullitynewcommandcmathbfcnewcommandmathbf0$Yes. Hints: You want to show that $rank(C)le k$. Let the columns of $C$ be $c_1,ldots,c_n$. Then since $C^2 = O$, we have $Cc_j = $ for all $j=1,ldots, n$. Now suppose by way of contradiction that $rank(C) > k$. Then since (by the rank-nullity theorem) $$nullity(C) = 2k+1-rank(C),$$ we have $$nullity(C) < 2k+1-k = k+1 Rightarrow nullity(C) le k.$$ But the columns of $C$ are all from the null space of $C$ and $rank(C) > k$. In view of the above, can you explain why this is impossible?
answered Mar 17 at 12:44
Minus One-TwelfthMinus One-Twelfth
2,823413
edited Mar 17 at 12:50
answered Mar 17 at 12:44
Minus One-TwelfthMinus One-Twelfth
2,823413
answered Mar 17 at 12:44
Minus One-TwelfthMinus One-Twelfth
2,823413
answered Mar 17 at 12:44
Minus One-TwelfthMinus One-Twelfth
2,823413
2,823413
add a comment |
add a comment |
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$begingroup$
You need only assume $C^2 = O$, then in fact all eigenvalues are $0$, since $C$ is nilpotent.
$endgroup$
– Minus One-Twelfth
Mar 17 at 12:46
1
$begingroup$
Since $C^2=0$, then $operatornameIm(C)subseteqoperatornameKer(C)$. Then $n=operatornamedimoperatornameKer(C)+operatornamedimoperatornameIm(C)geq 2operatornamedimoperatornameIm(C)$. Since $n$ is odd and $operatornamedimoperatornameIm(C)$ a natural number, then $fracn-12geq operatornamedimoperatornameIm(C)$.
$endgroup$
– user647486
Mar 17 at 12:56