Lagrange Interpolation Coefficients/ Frobenius Covariant
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I am having problems trying to show that
$$ sum_{k=1}^{n} L_k(A)=I$$
where A is a $nxn$ matrix,$I$ is the $nxn$ identity, and $L_k$ is the Lagrange interpolation coefficient defined as: $L_k(A)=prod_{j=1}^{n} frac{A-lambda_{j}I}{lambda_{k}-lambda_{j}}$ with $j neq k$. It is supposed to be a continuation of the fact that $$sum_{k=1}^{n} L_k(lambda)=1 $$ where $L_k(lambda)=prod_{j=1}^{n} frac{lambda-lambda_{j}}{lambda_{k}-lambda_{j}}$ with $j neq k$.
I don't have any idea of how to do this. I did a research and found that the thing I have to prove is also named as "Frobenius Covariants" but when I look for more help, the difficulty of this grows exponentially so I suppose there is some easy way that I can not do. Thank you very much if you can help me
differential-equations lagrange-interpolation
asked Nov 16 at 0:11
J.Rodriguez
15310
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up vote
0
down vote
favorite
I am having problems trying to show that
$$ sum_{k=1}^{n} L_k(A)=I$$
where A is a $nxn$ matrix,$I$ is the $nxn$ identity, and $L_k$ is the Lagrange interpolation coefficient defined as: $L_k(A)=prod_{j=1}^{n} frac{A-lambda_{j}I}{lambda_{k}-lambda_{j}}$ with $j neq k$. It is supposed to be a continuation of the fact that $$sum_{k=1}^{n} L_k(lambda)=1 $$ where $L_k(lambda)=prod_{j=1}^{n} frac{lambda-lambda_{j}}{lambda_{k}-lambda_{j}}$ with $j neq k$.
I don't have any idea of how to do this. I did a research and found that the thing I have to prove is also named as "Frobenius Covariants" but when I look for more help, the difficulty of this grows exponentially so I suppose there is some easy way that I can not do. Thank you very much if you can help me
differential-equations lagrange-interpolation
asked Nov 16 at 0:11
J.Rodriguez
15310
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am having problems trying to show that
$$ sum_{k=1}^{n} L_k(A)=I$$
where A is a $nxn$ matrix,$I$ is the $nxn$ identity, and $L_k$ is the Lagrange interpolation coefficient defined as: $L_k(A)=prod_{j=1}^{n} frac{A-lambda_{j}I}{lambda_{k}-lambda_{j}}$ with $j neq k$. It is supposed to be a continuation of the fact that $$sum_{k=1}^{n} L_k(lambda)=1 $$ where $L_k(lambda)=prod_{j=1}^{n} frac{lambda-lambda_{j}}{lambda_{k}-lambda_{j}}$ with $j neq k$.
I don't have any idea of how to do this. I did a research and found that the thing I have to prove is also named as "Frobenius Covariants" but when I look for more help, the difficulty of this grows exponentially so I suppose there is some easy way that I can not do. Thank you very much if you can help me
differential-equations lagrange-interpolation
asked Nov 16 at 0:11
J.Rodriguez
15310
I am having problems trying to show that
$$ sum_{k=1}^{n} L_k(A)=I$$
where A is a $nxn$ matrix,$I$ is the $nxn$ identity, and $L_k$ is the Lagrange interpolation coefficient defined as: $L_k(A)=prod_{j=1}^{n} frac{A-lambda_{j}I}{lambda_{k}-lambda_{j}}$ with $j neq k$. It is supposed to be a continuation of the fact that $$sum_{k=1}^{n} L_k(lambda)=1 $$ where $L_k(lambda)=prod_{j=1}^{n} frac{lambda-lambda_{j}}{lambda_{k}-lambda_{j}}$ with $j neq k$.
I don't have any idea of how to do this. I did a research and found that the thing I have to prove is also named as "Frobenius Covariants" but when I look for more help, the difficulty of this grows exponentially so I suppose there is some easy way that I can not do. Thank you very much if you can help me
differential-equations lagrange-interpolation
differential-equations lagrange-interpolation
asked Nov 16 at 0:11
J.Rodriguez
15310
asked Nov 16 at 0:11
J.Rodriguez
15310
asked Nov 16 at 0:11
J.Rodriguez
15310
asked Nov 16 at 0:11
J.Rodriguez
15310
asked Nov 16 at 0:11
J.Rodriguez
15310
15310
add a comment |
add a comment |
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