Prove a matrix equality












2












$begingroup$


Consider the square matrix



$$ J = begin{bmatrix}
lambda & 1 & 0 & cdots & 0 \
0 & lambda & 1 & cdots & 0 \
cdots & cdots & cdots & ddots & cdots \
0 & 0 & cdots & lambda & 1 \
0 & 0 & 0 & 0 & lambda \
end{bmatrix} $$



If $f(x)$ is a polynomial, prove that $f(J)$ is given by the matrix



$$ begin{bmatrix}
f(lambda) & dfrac{f'(lambda)}{1!} & dfrac{f''(lambda)}{2!} & cdots & dfrac{f^{(n-2)}(lambda)}{(n-2)!} & dfrac{f^{(n-1)}(lambda)}{(n-1)!}\[1em]
0 & f(lambda) & dfrac{f'(lambda)}{1!} & cdots & dfrac{f^{(n-3)}(lambda)}{(n-3)!} & dfrac{f^{(n-2)}(lambda)}{(n-2)!} \[1em]
vdots & vdots & vdots & ddots & vdots & vdots \
0 & 0 & 0 & cdots & f(lambda) & dfrac{f'(lambda)}{1!} \[1em]
0 & 0 & 0 & cdots & 0 & f(lambda) \
end{bmatrix} $$



Try



First, we can find the Taylor expansion of $f$ around $lambda$ and that is



$$ f(x) = sum_{k=0}^{infty} frac{ f^{(k)}( lambda) (x-lambda)^n }{k!} $$



Now, we have



$$ f(J) = sum frac{ f^{(k)} (lambda) ( J-lambda I)^k }{k!}$$



Now, notice that $J- lambda I $ is the matrix with ones in the upper diagonal and when we take powers of this matrix, the ones move to the right . That is,



$$ (J - lambda I)^k = begin{bmatrix} 0 & I_{n-k} \ 0 & 0 end{bmatrix} $$



as long as $k leq n-1 $ and $0$ otherwise, so we can write



$$ f(J) = sum_{k=0}^{n-1} frac{ f^{(k)}(lambda) I_{n-k} }{k!} $$



and I think this gives the result. Am I correct? Also, what is the motivation behind this exercise?






linear-algebra







share|cite|improve this question









$endgroup$












  • $begingroup$
    The last summation with different size matrices $I_{n-k}$ looks weird.
    $endgroup$
    – A.Γ.
    Dec 24 '18 at 9:42
















2












$begingroup$


Consider the square matrix



$$ J = begin{bmatrix}
lambda & 1 & 0 & cdots & 0 \
0 & lambda & 1 & cdots & 0 \
cdots & cdots & cdots & ddots & cdots \
0 & 0 & cdots & lambda & 1 \
0 & 0 & 0 & 0 & lambda \
end{bmatrix} $$



If $f(x)$ is a polynomial, prove that $f(J)$ is given by the matrix



$$ begin{bmatrix}
f(lambda) & dfrac{f'(lambda)}{1!} & dfrac{f''(lambda)}{2!} & cdots & dfrac{f^{(n-2)}(lambda)}{(n-2)!} & dfrac{f^{(n-1)}(lambda)}{(n-1)!}\[1em]
0 & f(lambda) & dfrac{f'(lambda)}{1!} & cdots & dfrac{f^{(n-3)}(lambda)}{(n-3)!} & dfrac{f^{(n-2)}(lambda)}{(n-2)!} \[1em]
vdots & vdots & vdots & ddots & vdots & vdots \
0 & 0 & 0 & cdots & f(lambda) & dfrac{f'(lambda)}{1!} \[1em]
0 & 0 & 0 & cdots & 0 & f(lambda) \
end{bmatrix} $$



Try



First, we can find the Taylor expansion of $f$ around $lambda$ and that is



$$ f(x) = sum_{k=0}^{infty} frac{ f^{(k)}( lambda) (x-lambda)^n }{k!} $$



Now, we have



$$ f(J) = sum frac{ f^{(k)} (lambda) ( J-lambda I)^k }{k!}$$



Now, notice that $J- lambda I $ is the matrix with ones in the upper diagonal and when we take powers of this matrix, the ones move to the right . That is,



$$ (J - lambda I)^k = begin{bmatrix} 0 & I_{n-k} \ 0 & 0 end{bmatrix} $$



as long as $k leq n-1 $ and $0$ otherwise, so we can write



$$ f(J) = sum_{k=0}^{n-1} frac{ f^{(k)}(lambda) I_{n-k} }{k!} $$



and I think this gives the result. Am I correct? Also, what is the motivation behind this exercise?






linear-algebra







share|cite|improve this question









$endgroup$












  • $begingroup$
    The last summation with different size matrices $I_{n-k}$ looks weird.
    $endgroup$
    – A.Γ.
    Dec 24 '18 at 9:42














2












2








2





$begingroup$


Consider the square matrix



$$ J = begin{bmatrix}
lambda & 1 & 0 & cdots & 0 \
0 & lambda & 1 & cdots & 0 \
cdots & cdots & cdots & ddots & cdots \
0 & 0 & cdots & lambda & 1 \
0 & 0 & 0 & 0 & lambda \
end{bmatrix} $$



If $f(x)$ is a polynomial, prove that $f(J)$ is given by the matrix



$$ begin{bmatrix}
f(lambda) & dfrac{f'(lambda)}{1!} & dfrac{f''(lambda)}{2!} & cdots & dfrac{f^{(n-2)}(lambda)}{(n-2)!} & dfrac{f^{(n-1)}(lambda)}{(n-1)!}\[1em]
0 & f(lambda) & dfrac{f'(lambda)}{1!} & cdots & dfrac{f^{(n-3)}(lambda)}{(n-3)!} & dfrac{f^{(n-2)}(lambda)}{(n-2)!} \[1em]
vdots & vdots & vdots & ddots & vdots & vdots \
0 & 0 & 0 & cdots & f(lambda) & dfrac{f'(lambda)}{1!} \[1em]
0 & 0 & 0 & cdots & 0 & f(lambda) \
end{bmatrix} $$



Try



First, we can find the Taylor expansion of $f$ around $lambda$ and that is



$$ f(x) = sum_{k=0}^{infty} frac{ f^{(k)}( lambda) (x-lambda)^n }{k!} $$



Now, we have



$$ f(J) = sum frac{ f^{(k)} (lambda) ( J-lambda I)^k }{k!}$$



Now, notice that $J- lambda I $ is the matrix with ones in the upper diagonal and when we take powers of this matrix, the ones move to the right . That is,



$$ (J - lambda I)^k = begin{bmatrix} 0 & I_{n-k} \ 0 & 0 end{bmatrix} $$



as long as $k leq n-1 $ and $0$ otherwise, so we can write



$$ f(J) = sum_{k=0}^{n-1} frac{ f^{(k)}(lambda) I_{n-k} }{k!} $$



and I think this gives the result. Am I correct? Also, what is the motivation behind this exercise?






linear-algebra







share|cite|improve this question









$endgroup$




Consider the square matrix



$$ J = begin{bmatrix}
lambda & 1 & 0 & cdots & 0 \
0 & lambda & 1 & cdots & 0 \
cdots & cdots & cdots & ddots & cdots \
0 & 0 & cdots & lambda & 1 \
0 & 0 & 0 & 0 & lambda \
end{bmatrix} $$



If $f(x)$ is a polynomial, prove that $f(J)$ is given by the matrix



$$ begin{bmatrix}
f(lambda) & dfrac{f'(lambda)}{1!} & dfrac{f''(lambda)}{2!} & cdots & dfrac{f^{(n-2)}(lambda)}{(n-2)!} & dfrac{f^{(n-1)}(lambda)}{(n-1)!}\[1em]
0 & f(lambda) & dfrac{f'(lambda)}{1!} & cdots & dfrac{f^{(n-3)}(lambda)}{(n-3)!} & dfrac{f^{(n-2)}(lambda)}{(n-2)!} \[1em]
vdots & vdots & vdots & ddots & vdots & vdots \
0 & 0 & 0 & cdots & f(lambda) & dfrac{f'(lambda)}{1!} \[1em]
0 & 0 & 0 & cdots & 0 & f(lambda) \
end{bmatrix} $$



Try



First, we can find the Taylor expansion of $f$ around $lambda$ and that is



$$ f(x) = sum_{k=0}^{infty} frac{ f^{(k)}( lambda) (x-lambda)^n }{k!} $$



Now, we have



$$ f(J) = sum frac{ f^{(k)} (lambda) ( J-lambda I)^k }{k!}$$



Now, notice that $J- lambda I $ is the matrix with ones in the upper diagonal and when we take powers of this matrix, the ones move to the right . That is,



$$ (J - lambda I)^k = begin{bmatrix} 0 & I_{n-k} \ 0 & 0 end{bmatrix} $$



as long as $k leq n-1 $ and $0$ otherwise, so we can write



$$ f(J) = sum_{k=0}^{n-1} frac{ f^{(k)}(lambda) I_{n-k} }{k!} $$



and I think this gives the result. Am I correct? Also, what is the motivation behind this exercise?






linear-algebra




linear-algebra






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 24 '18 at 7:44









JamesJames

2,638325




2,638325












  • $begingroup$
    The last summation with different size matrices $I_{n-k}$ looks weird.
    $endgroup$
    – A.Γ.
    Dec 24 '18 at 9:42


















  • $begingroup$
    The last summation with different size matrices $I_{n-k}$ looks weird.
    $endgroup$
    – A.Γ.
    Dec 24 '18 at 9:42
















$begingroup$
The last summation with different size matrices $I_{n-k}$ looks weird.
$endgroup$
– A.Γ.
Dec 24 '18 at 9:42




$begingroup$
The last summation with different size matrices $I_{n-k}$ looks weird.
$endgroup$
– A.Γ.
Dec 24 '18 at 9:42










1 Answer
1






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oldest

votes


















1












$begingroup$

Here's a another way you could do it:




  • write the matrix of the linear operator $D: mathbb{P}_{leq N} rightarrow mathbb{P}_{leq N}$, the derivative of a polynomial $p in mathbb{P}_{leq N}$ of degree at most $N$, with respect to the basis $mathcal{B}={1, x, x^2, ..., x^{N} }$

  • now, composition of linear operators can be viewed by multiplying representative matrices you can deduce (and prove by induction) the matrix for $D^K=D circ D circ ...circ D$ $K$ times.

  • now, deduce a general formula for $J^n$ and prove it via induction (use the binomial expression of $(1+lambda)^n$)

  • do the necessary algebra and you're done


A motivation behind this exercise could be found in Jordan's canonical form, which can help you, for example, to find analytical solutions to linear systems of ordinary differential equations.






share|cite|improve this answer











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    1












    $begingroup$

    Here's a another way you could do it:




    • write the matrix of the linear operator $D: mathbb{P}_{leq N} rightarrow mathbb{P}_{leq N}$, the derivative of a polynomial $p in mathbb{P}_{leq N}$ of degree at most $N$, with respect to the basis $mathcal{B}={1, x, x^2, ..., x^{N} }$

    • now, composition of linear operators can be viewed by multiplying representative matrices you can deduce (and prove by induction) the matrix for $D^K=D circ D circ ...circ D$ $K$ times.

    • now, deduce a general formula for $J^n$ and prove it via induction (use the binomial expression of $(1+lambda)^n$)

    • do the necessary algebra and you're done


    A motivation behind this exercise could be found in Jordan's canonical form, which can help you, for example, to find analytical solutions to linear systems of ordinary differential equations.






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      Here's a another way you could do it:




      • write the matrix of the linear operator $D: mathbb{P}_{leq N} rightarrow mathbb{P}_{leq N}$, the derivative of a polynomial $p in mathbb{P}_{leq N}$ of degree at most $N$, with respect to the basis $mathcal{B}={1, x, x^2, ..., x^{N} }$

      • now, composition of linear operators can be viewed by multiplying representative matrices you can deduce (and prove by induction) the matrix for $D^K=D circ D circ ...circ D$ $K$ times.

      • now, deduce a general formula for $J^n$ and prove it via induction (use the binomial expression of $(1+lambda)^n$)

      • do the necessary algebra and you're done


      A motivation behind this exercise could be found in Jordan's canonical form, which can help you, for example, to find analytical solutions to linear systems of ordinary differential equations.






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        Here's a another way you could do it:




        • write the matrix of the linear operator $D: mathbb{P}_{leq N} rightarrow mathbb{P}_{leq N}$, the derivative of a polynomial $p in mathbb{P}_{leq N}$ of degree at most $N$, with respect to the basis $mathcal{B}={1, x, x^2, ..., x^{N} }$

        • now, composition of linear operators can be viewed by multiplying representative matrices you can deduce (and prove by induction) the matrix for $D^K=D circ D circ ...circ D$ $K$ times.

        • now, deduce a general formula for $J^n$ and prove it via induction (use the binomial expression of $(1+lambda)^n$)

        • do the necessary algebra and you're done


        A motivation behind this exercise could be found in Jordan's canonical form, which can help you, for example, to find analytical solutions to linear systems of ordinary differential equations.






        share|cite|improve this answer











        $endgroup$



        Here's a another way you could do it:




        • write the matrix of the linear operator $D: mathbb{P}_{leq N} rightarrow mathbb{P}_{leq N}$, the derivative of a polynomial $p in mathbb{P}_{leq N}$ of degree at most $N$, with respect to the basis $mathcal{B}={1, x, x^2, ..., x^{N} }$

        • now, composition of linear operators can be viewed by multiplying representative matrices you can deduce (and prove by induction) the matrix for $D^K=D circ D circ ...circ D$ $K$ times.

        • now, deduce a general formula for $J^n$ and prove it via induction (use the binomial expression of $(1+lambda)^n$)

        • do the necessary algebra and you're done


        A motivation behind this exercise could be found in Jordan's canonical form, which can help you, for example, to find analytical solutions to linear systems of ordinary differential equations.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 25 '18 at 1:07

























        answered Dec 24 '18 at 10:27









        LeonardoLeonardo

        3339




        3339






























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