Determine whether or not the given matrix \(A\) is diagonalizable. If it is find a diagonalizing matrix \(P\) and a diagonal matrix \(D\) such that \(P^{-1}AP=D\)\[ A=\begin {bmatrix} 3 & 0\\ 8 & -1 \end {bmatrix} \] Solution
The first step is to find the eigenvalues. This is found by solving\begin {align*} \det \left (A-\lambda I\right ) & =0\\\begin {vmatrix} 3-\lambda & 0\\ 8 & -1-\lambda \end {vmatrix} & =0\\ \left (3-\lambda \right ) \left (-1-\lambda \right ) & =0 \end {align*}
Hence the eigenvalues are \(\lambda _{1}=3,\lambda _{2}=-1\). Since the eigenvalues are unique, then the matrix is diagonalizable. We need to determine the corresponding eigenvector in order to find \(P\)
\(\lambda _{1}=3\)
Solving\begin {align*} \begin {bmatrix} 3-\lambda _{1} & 0\\ 8 & -1-\lambda _{1}\end {bmatrix}\begin {bmatrix} v_{1}\\ v_{2}\end {bmatrix} & =\begin {bmatrix} 0\\ 0 \end {bmatrix} \\\begin {bmatrix} 3-3 & 0\\ 8 & -1-3 \end {bmatrix}\begin {bmatrix} v_{1}\\ v_{2}\end {bmatrix} & =\begin {bmatrix} 0\\ 0 \end {bmatrix} \\\begin {bmatrix} 0 & 0\\ 8 & -4 \end {bmatrix}\begin {bmatrix} v_{1}\\ v_{2}\end {bmatrix} & =\begin {bmatrix} 0\\ 0 \end {bmatrix} \end {align*}
Hence \(v_{1}=t\) is free variable and \(v_{2}\) is base variable. Second row gives \(8t-4v_{2}=0\) or \(v_{2}=2t\). Therefore the eigenvector is \[\begin {bmatrix} v_{1}\\ v_{2}\end {bmatrix} =\begin {bmatrix} t\\ 2t \end {bmatrix} =t\begin {bmatrix} 1\\ 2 \end {bmatrix} \] Let \(t=1\), then \[ \vec {v}_{1}=\begin {bmatrix} 1\\ 2 \end {bmatrix} \] \(\lambda _{2}=-1\)
Solving\begin {align*} \begin {bmatrix} 3-\lambda _{2} & 0\\ 8 & -1-\lambda _{2}\end {bmatrix}\begin {bmatrix} v_{1}\\ v_{2}\end {bmatrix} & =\begin {bmatrix} 0\\ 0 \end {bmatrix} \\\begin {bmatrix} 3+1 & 0\\ 8 & -1+1 \end {bmatrix}\begin {bmatrix} v_{1}\\ v_{2}\end {bmatrix} & =\begin {bmatrix} 0\\ 0 \end {bmatrix} \\\begin {bmatrix} 4 & 0\\ 8 & 0 \end {bmatrix}\begin {bmatrix} v_{1}\\ v_{2}\end {bmatrix} & =\begin {bmatrix} 0\\ 0 \end {bmatrix} \end {align*}
\(R_{2}=R_{2}-2R_{1}\) gives\[\begin {bmatrix} 4 & 0\\ 0 & 0 \end {bmatrix}\begin {bmatrix} v_{1}\\ v_{2}\end {bmatrix} =\begin {bmatrix} 0\\ 0 \end {bmatrix} \] Hence \(v_{2}=t\) is free variable and \(v_{1}\) is base variable. First row gives \(4v_{1}=0\) or \(v_{1}=0\). Therefore the eigenvector is \[\begin {bmatrix} v_{1}\\ v_{2}\end {bmatrix} =\begin {bmatrix} 0\\ t \end {bmatrix} =t\begin {bmatrix} 0\\ 1 \end {bmatrix} \] Let \(t=1\), then \[ \vec {v}_{2}=\begin {bmatrix} 0\\ 1 \end {bmatrix} \] Now that both eigenvectors are found, then \begin {align*} P & =\begin {bmatrix} \vec {v}_{1} & \vec {v}_{2}\end {bmatrix} \\ & =\begin {bmatrix} 1 & 0\\ 2 & 1 \end {bmatrix} \end {align*}
And \(D\) is the diagonal matrix of the eigenvalues arranged in same order as the corresponding eigenvectors. (Will verify below). Hence\[ D=\begin {bmatrix} 3 & 0\\ 0 & -1 \end {bmatrix} \] Therefore \begin {align} P^{-1}AP & =D\nonumber \\\begin {bmatrix} 1 & 0\\ 2 & 1 \end {bmatrix} ^{-1}\begin {bmatrix} 3 & 0\\ 8 & -1 \end {bmatrix}\begin {bmatrix} 1 & 0\\ 2 & 1 \end {bmatrix} & =\begin {bmatrix} 3 & 0\\ 0 & -1 \end {bmatrix} \tag {1} \end {align}
To verify the above, the LHS of (1) is evaluated directly, to confirm that \(D\) is indeed the result and it is diagonal of the eigenvalues. The first step is to find \(\begin {bmatrix} 1 & 0\\ 2 & 1 \end {bmatrix} ^{-1}\). Since this is \(2\times 2\) then\[\begin {bmatrix} 1 & 0\\ 2 & 1 \end {bmatrix} ^{-1}=\frac {1}{\det \relax (P) }\begin {bmatrix} 1 & 0\\ -2 & 1 \end {bmatrix} \] But \(\det \relax (P) =1\). The above becomes\[\begin {bmatrix} 1 & 0\\ 2 & 1 \end {bmatrix} ^{-1}=\begin {bmatrix} 1 & 0\\ -2 & 1 \end {bmatrix} \] Therefore the LHS of (1) becomes\begin {align*} \begin {bmatrix} 1 & 0\\ 2 & 1 \end {bmatrix} ^{-1}\begin {bmatrix} 3 & 0\\ 8 & -1 \end {bmatrix} & =\begin {bmatrix} 1 & 0\\ -2 & 1 \end {bmatrix}\begin {bmatrix} 3 & 0\\ 8 & -1 \end {bmatrix} \\ & =\begin {bmatrix} 1\times 3 & 0\\ -2\times 3+1\times 8 & -1 \end {bmatrix} \\ & =\begin {bmatrix} 3 & 0\\ -6+8 & -1 \end {bmatrix} \\ & =\begin {bmatrix} 3 & 0\\ 2 & -1 \end {bmatrix} \end {align*}
And now LHS of (1) becomes\begin {align*} \begin {bmatrix} 3 & 0\\ 2 & -1 \end {bmatrix}\begin {bmatrix} 1 & 0\\ 2 & 1 \end {bmatrix} & =\begin {bmatrix} 3 & 0\\ 2\times 1-1\times 2 & -1 \end {bmatrix} \\ & =\begin {bmatrix} 3 & 0\\ 0 & -1 \end {bmatrix} \end {align*}
Hence\[ D=\begin {bmatrix} 3 & 0\\ 0 & -1 \end {bmatrix} \] Which confirms \(D\) is the matrix whose diagonal elements are the eigenvalues of \(A\).
Find the general solution of the homogeneous differential equation \(y^{\prime \prime \prime }+y^{\prime }-10y=0\)
Solution
This is a linear 3rd order constant coefficient ODE. Hence the method of characteristic equation will be used. Let the solution be \(y=Ae^{\lambda x}\). Substituting this into the ODE gives\begin {align*} A\lambda ^{3}e^{\lambda x}+A\lambda e^{\lambda x}-10Ae^{\lambda x} & =0\\ Ae^{\lambda x}\left (\lambda ^{3}+\lambda -10\right ) & =0 \end {align*}
Which simplifies (for non-trivial \(y\)) to the characteristic equation which is a polynomial in \(\lambda \) \[ \lambda ^{3}+\lambda -10=0 \] By inspection, we see that \(\lambda =2\) is a root. Therefore a factor of the equation is \(\left (\lambda -2\right ) \). Now doing long division \(\frac {\lambda ^{3}+\lambda -10}{\left (\lambda -2\right ) }\) gives \(\lambda ^{2}+2\lambda +5\).
Hence the above polynomial can be written as\begin {equation} \left (\lambda -2\right ) \left (\lambda ^{2}+2\lambda +5\right ) =0\tag {1} \end {equation} Now the roots for \(\left (\lambda ^{2}+2\lambda +5\right ) =0\) are found using the quadratic formula. \begin {align*} \lambda & =-\frac {b}{2a}\pm \frac {1}{2a}\sqrt {b^{2}-4ac}\\ & =-\frac {2}{2}\pm \frac {1}{2}\sqrt {4-4\times 5}\\ & =-1\pm \frac {1}{2}\sqrt {-16}\\ & =-1\pm \frac {1}{2}\left (4i\right ) \\ & =-1\pm 2i \end {align*}
Hence the roots of the characteristic equation are\begin {align*} \lambda _{1} & =2\\ \lambda _{2} & =-1+2i\\ \lambda _{3} & =-1-2i \end {align*}
Therefore the basis solution are \(\left \{ e^{2x},e^{\left (-1+2i\right ) x},e^{\left (-1-2i\right ) x}\right \} \) and the general solution is a linear combination of these basis solutions which gives\[ y=Ae^{2x}+Be^{\left (-1+2i\right ) x}+Ce^{\left (-1-2i\right ) x}\] Which can be simplified to \begin {equation} y=Ae^{2x}+e^{-x}\left (Be^{2ix}+Ce^{-2ix}\right ) \tag {2} \end {equation} By using Euler formula, the above can be simplified further as follows\begin {align*} Be^{2ix}+Ce^{-2ix} & =B\left (\cos \left (2x\right ) +i\sin \left ( 2x\right ) \right ) +C\left (\cos \left (2x\right ) -i\sin \left (2x\right ) \right ) \\ & =\left (B+C\right ) \cos \left (2x\right ) +\sin \left (2x\right ) \left ( i\left (B-C\right ) \right ) \end {align*}
Let \(\left (B+C\right ) =B_{0}\) a new constant and let \(i\left (B-C\right ) =C_{0}\) a new constant, the above becomes\[ Be^{2ix}+Ce^{-2ix}=B_{0}\cos \left (2x\right ) +C_{0}\sin \left (2x\right ) \] Substituting the above back in (2) gives the general solution as\[ y=Ae^{2x}+e^{-x}\left (B_{0}\cos \left (2x\right ) +C_{0}\sin \left ( 2x\right ) \right ) \] The constants \(A,B_{0},C_{0}\) can be found from initial conditions if given.
Using the method of undermined coefficients, compute the general solution of the given equation \(y^{\prime \prime }+3y^{\prime }+2y=2\sin \relax (x) \)
Solution
The solution is \[ y=y_{h}+y_{p}\] Where \(y_{h}\) is the solution to the homogenous ode \(y^{\prime \prime }+3y^{\prime }+2y=0\) and \(y_{p}\) is a particular solution to the given ODE. The ode \(y^{\prime \prime }+3y^{\prime }+2y=0\) is linear second order constant coefficient ODE. Hence the method of characteristic equation will be used. Let the solution be \(y_{h}=Ae^{\lambda x}\). Substituting this into \(y^{\prime \prime }+3y^{\prime }+2y=0\)\begin {align*} A\lambda ^{2}e^{\lambda x}+A\lambda 3e^{\lambda x}+2Ae^{\lambda x} & =0\\ Ae^{\lambda x}\left (\lambda ^{2}+3\lambda +2\right ) & =0 \end {align*}
And for non trivial solution the above simplifies to\begin {align*} \lambda ^{2}+3\lambda +2 & =0\\ \left (\lambda +1\right ) \left (\lambda +2\right ) & =0 \end {align*}
Hence the roots are \(\lambda =-1,\lambda =-2\). Therefore the basis solutions for \(y_{h}\) are \(\left \{ e^{-x},e^{-2x}\right \} \) and \(y_{h}\) is linear combination of these basis. Therefore \begin {equation} y_{h}=c_{1}e^{-x}+c_{2}e^{-2x}\tag {1} \end {equation} Now \(y_{p}\) is found. Since the RHS is \(\sin \relax (x) \) then the trial solution is \begin {equation} y_{p}=A\cos \relax (x) +B\sin \relax (x) \tag {2} \end {equation} This shows that the basis for \(y_{p}\) are \(\left \{ \sin x,\cos x\right \} \). There are no duplication between these basis and the basis for \(y_{h}\), so no need to multiply by an extra \(x\). Using (2) gives\begin {align} y_{p}^{\prime } & =-A\sin \relax (x) +B\cos \relax (x) \tag {3}\\ y_{p}^{\prime \prime } & =-A\cos \relax (x) -B\sin \relax (x) \tag {4} \end {align}
Substituting (2,3,4) back into the given ODE gives\begin {align*} y_{p}^{\prime \prime }+3y_{p}^{\prime }+2y_{p} & =2\sin \relax (x) \\ \left (-A\cos \relax (x) -B\sin \relax (x) \right ) +3\left ( -A\sin \relax (x) +B\cos \relax (x) \right ) +2\left ( A\cos \relax (x) +B\sin \relax (x) \right ) & =2\sin \left ( x\right ) \\ \cos \relax (x) \left (-A+3B+2A\right ) +\sin \relax (x) \left ( -B-3A+2B\right ) & =2\sin \relax (x) \\ \cos \relax (x) \left (3B+A\right ) +\sin \relax (x) \left ( -3A+B\right ) & =2\sin \relax (x) \end {align*}
Comparing coefficients on both sides gives two equations to solve for \(A,B\) \begin {align*} 3B+A & =0\\ -3A+B & =2 \end {align*}
Multiplying the second equation by \(-3\) gives \begin {align*} 3B+A & =0\\ 9A-3B & =-6 \end {align*}
Adding the above two equations gives \(10A=-6\). Hence \(A=-\frac {3}{5}\) and therefore \(3B=\frac {3}{5}\) or \(B=\frac {1}{5}\). Substituting these values of \(A,B\,\ \)into (2) gives\[ y_{p}=-\frac {3}{5}\cos \relax (x) +\frac {1}{5}\sin \relax (x) \] Hence the solution becomes\begin {align*} y & =y_{h}+y_{p}\\ & =\left (c_{1}e^{-x}+c_{2}e^{-2x}\right ) +\left (-\frac {3}{5}\cos \left ( x\right ) +\frac {1}{5}\sin \relax (x) \right ) \\ & =c_{1}e^{-x}+c_{2}e^{-2x}-\frac {3}{5}\cos \relax (x) +\frac {1}{5}\sin \relax (x) \end {align*}
Show that the given vector functions are linearly independent \[ \vec {x}_{1}\relax (t) =\begin {bmatrix} e^{t}\\ 2e^{t}\end {bmatrix} \qquad \vec {x}_{2}\relax (t) =\begin {bmatrix} \sin t\\ \cos t \end {bmatrix} \] Solution
These functions are defined for all \(t\). Hence domain is \(t\in \left ( -\infty ,\infty \right ) \,\). The Wronskian of these vectors is\begin {align*} W\relax (t) & =\begin {vmatrix} e^{t} & \sin t\\ 2e^{t} & \cos t \end {vmatrix} \\ & =e^{t}\cos t-2e^{t}\sin t\\ & =e^{t}\left (\cos t-2\sin t\right ) \end {align*}
We just need find one value \(t_{0}\) where \(W\left (t_{0}\right ) \neq 0\) to show linearly independence. At \(t=0\) the above becomes\[ W\left (t=0\right ) =1 \] Therefore the given vector functions are linearly independent. An alternative method is to write \begin {align*} c_{1}\vec {x}_{1}\relax (t) +c_{2}\vec {x}_{2}\relax (t) & =\vec {0}\\ c_{1}\begin {bmatrix} e^{t}\\ 2e^{t}\end {bmatrix} +c_{2}\begin {bmatrix} \sin t\\ \cos t \end {bmatrix} & =\begin {bmatrix} 0\\ 0 \end {bmatrix} \end {align*}
If the above is true only for \(c_{1}=0,c_{2}=0\) then \(\vec {x}_{1}\left ( t\right ) ,\vec {x}_{2}\relax (t) \) are linearly independent. The above can be written as\[\begin {bmatrix} e^{t} & \sin t\\ 2e^{t} & \cos t \end {bmatrix}\begin {bmatrix} c_{1}\\ c_{2}\end {bmatrix} =\begin {bmatrix} 0\\ 0 \end {bmatrix} \] \(R_{2}=R_{2}-2R_{1}\)\[\begin {bmatrix} e^{t} & \sin t\\ 0 & \cos t-2\sin t \end {bmatrix}\begin {bmatrix} c_{1}\\ c_{2}\end {bmatrix} =\begin {bmatrix} 0\\ 0 \end {bmatrix} \] Row two gives \[ c_{2}\left (\cos t-2\sin t\right ) =0 \] For this to be true for any \(t\) in the interval \(t\in \left ( -\infty ,\infty \right ) \), then only solution is \(c_{2}=0\). First row now gives \[ c_{1}e^{t}=0 \] But \(e^{t}\) is never zero which means \(c_{1}=0\).
Since the only solution to \(c_{1}\vec {x}_{1}+c_{2}\vec {x}_{2}=\vec {0}\) is \(c_{1}=c_{2}=0\), then this shows that \(\vec {x}_{1},\vec {x}_{2}\) are linearly independent.