Section 11.6

Substituting an Operator (Matrix) into a Function of a Numerical Argument

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Problem 1201

Find the kk-th power AkA^{k} of the Jordan block

A=(α100…000α10…0000α1…00…………………0000…α10000…0α) of order n A =\left(\begin{smallmatrix} \alpha & 1 & 0 & 0 & \ldots & 0 & 0 \\ 0 & \alpha & 1 & 0 & \ldots & 0 & 0 \\ 0 & 0 & \alpha & 1 & \ldots & 0 & 0 \\ \ldots & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots \\ 0 & 0 & 0 & 0 & \ldots & \alpha & 1 \\ 0 & 0 & 0 & 0 & \ldots & 0 & \alpha \end{smallmatrix}\right) \text{ of order } n
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Problem 1202

Prove that the value of the polynomial f(x)f(x) of the Jordan block (11.1) is given by the formula

f(A)=(f(α)f′(α)1!f′′(α)2!f′′′(α)3!…f(n−1)(α)(n−1)!0f(α)f′(α)1!f′′(α)2!…f(n−2)(α)(n−2)!……………………0000…f(α)) f(A) =\left(\begin{smallmatrix} f(\alpha ) & \frac{f^{\prime }(\alpha )}{1 !} & \frac{f^{\prime \prime }(\alpha )}{2 !} & \frac{f^{\prime \prime \prime }(\alpha )}{3 !} & \ldots & \frac{f^{(n-1)}(\alpha )}{(n-1) !} \\ 0 & f(\alpha ) & \frac{f^{\prime }(\alpha )}{1 !} & \frac{f^{\prime \prime }(\alpha )}{2 !} & \ldots & \frac{f^{(n-2)}(\alpha )}{(n-2) !} \\ \ldots \ldots & \ldots & \ldots \ldots \ldots \ldots \ldots \\ 0 & 0 & 0 & 0 & \ldots & f(\alpha ) \end{smallmatrix}\right)
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Problem 1203

Let μA(λ)=(λ−λ1)r1…(λ−λs)rs\mu_{\mathrm{A}}(\lambda ) = \left(\lambda -\lambda_{1}\right)^{r_{1}} \ldots \left(\lambda -\lambda_{s}\right)^{r_{s}} be the minimal polynomial of the operator A. If for each k=1,…,sk = 1, \ldots , s the function f(λ)f(\lambda ) is defined at the point λk\lambda_{k} and is differentiable there at least rk−1r_{k}-1 times, then f(λ)f(\lambda ) is said to be defined on the spectrum of the operator A, and the system of numbers

(f(j)(λk))1⩽k⩽s;1⩽j⩽rk−1 \left(f^{(j)}\left(\lambda _{k}\right)\right)_{1 \leqslant k \leqslant s ; 1 \leqslant j \leqslant r_{k}-1}

is called the system of values of the function f(λ)f(\lambda ) on the spectrum of the operator A. Prove that the values of the polynomials g(λ)g(\lambda ) and h(λ)h(\lambda ) of the operator A coincide if and only if the values of these polynomials on the spectrum of the operator A\mathbf{A} coincide.

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Problem 1204

Let the function f(λ)f(\lambda ) be defined on the spectrum of the operator A\mathbf{A} with minimal polynomial μA(λ)=(λ−λ1)r1…(λ−λs)rs\mu_{\mathrm{A}}(\lambda ) = \left(\lambda -\lambda_{1}\right)^{r_{1}} \ldots \left(\lambda -\lambda_{s}\right)^{r_{s}}. Consider the polynomial

r(λ)=∑k=1s(αk,1+αk,2(λ−λk)+…+αk,rs(λ−λk)rk−1)⋅μA(λ)(λ−λk)rk r(\lambda ) = \sum _{k = 1}^{s}\left(\alpha _{k, 1}+\alpha _{k, 2}\left(\lambda -\lambda _{k}\right)+\ldots +\alpha _{k, r_{s}}\left(\lambda -\lambda _{k}\right)^{r_{k}-1}\right) \cdot \frac{\mu _{\mathrm{A}}(\lambda )}{\left(\lambda -\lambda _{k}\right)^{r_{k}}}

where

αk,j=1(j−1)!((λ−λk)rkf(λ)μA(λ))λ=λk(j−1) \alpha _{k, j} = \frac{1}{(j-1) !}\left(\frac{\left(\lambda -\lambda _{k}\right)^{r_{k}} f(\lambda )}{\mu _{\mathrm{A}}(\lambda )}\right)_{\lambda = \lambda _{k}}^{(j-1)}

j=1,2,…,rk,k=1,2,…,sj = 1,2, \ldots , r_{k}, k = 1,2, \ldots , s. Prove that:

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(1)

the systems of values of the polynomial r(λ)r(\lambda ) and of the function f(λ)f(\lambda ) on the spectrum of the operator A coincide;

(2)

among all polynomials with this property, r(λ)r(\lambda ) has the smallest degree.

This polynomial r(λ)r(\lambda ) is called the interpolation polynomial of the function f(λ)f(\lambda ) on the spectrum of the operator A. The value f(A)f(\mathbf{A}) of the function ff of the operator A is by definition set equal to r(A)r(\mathbf{A}).

Problem 1205

Compute A100A^{100}, where

A=(10151327−20−2−10) A =\left(\begin{smallmatrix} 10 & 1 & 5 \\ 13 & 2 & 7 \\ -20 & -2 & -10 \end{smallmatrix}\right)
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Problem 1206

Compute A101A^{101}, where

A=(−2−3−4−100111) A =\left(\begin{smallmatrix} -2 & -3 & -4 \\ -1 & 0 & 0 \\ 1 & 1 & 1 \end{smallmatrix}\right)
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Problem 1207

Compute exp⁡A\exp A, where:

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(1)

A=(0−330)A = \left(\begin{array}{rr}0 & -3 \\ 3 & 0\end{array}\right);

(2)

A=(0−1−1−1−1−2011)A = \left(\begin{array}{rrr}0 & -1 & -1 \\ -1 & -1 & -2 \\ 0 & 1 & 1\end{array}\right);

(3)

A=(011−101−1−10)A = \left(\begin{array}{rrr}0 & 1 & 1 \\ -1 & 0 & 1 \\ -1 & -1 & 0\end{array}\right).

Problem 1208

For any square matrix AA prove the identity

det⁡exp⁡A=exp⁡tr⁡A \det \exp A = \exp \operatorname {tr} A
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Problem 1209

Find cos⁡A\cos A and sin⁡A\sin A, if A=(1−4−31−5−4−154)A = \left(\begin{array}{rrr}1 & -4 & -3 \\ 1 & -5 & -4 \\ -1 & 5 & 4\end{array}\right).

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Problem 1210

Find ln⁡A\ln A, if A=(2−4−31−4−4−155)A = \left(\begin{array}{rrr}2 & -4 & -3 \\ 1 & -4 & -4 \\ -1 & 5 & 5\end{array}\right).

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Problem 1211

Solve the equation X2=AX^{2} = A, if:

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(1)

A=(5−5−412−1−1−12)A = \left(\begin{array}{rrr}5 & -5 & -4 \\ 1 & 2 & -1 \\ -1 & -1 & 2\end{array}\right);

(2)

A=(2−4−31−4−4−155)A = \left(\begin{array}{rrr}2 & -4 & -3 \\ 1 & -4 & -4 \\ -1 & 5 & 5\end{array}\right).

Problem 1212

On the space Cn[t]\mathbb {C}_{n}[t] of polynomials of degree at most nn, consider the differentiation operator D\mathbf{D}:

(Df)(x)=f′(x) (\mathbf{D} f)(x) = f^{\prime }(x)

Compute expD.

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