User:Egm6322.s09.bit.gk/r6

$$\underline J =\begin{bmatrix} a & b\\ c & d \end{bmatrix}$$ and  $$det \underline J= \left (ad-bc  \right )$$

$$\underline J^T =\begin{bmatrix} a & c\\ b & d \end{bmatrix}$$ and  $$det \underline J^T= \left (ad-bc  \right )$$

We see that $$det \underline J=det \underline J^t$$

The Eigen Value problem of $$\underline A=\begin{bmatrix} a & b\\ b & c \end{bmatrix}$$ is $$det\left ( \underline A-\lambda \underline I\right )=0$$ Solving for $$ \lambda$$ ,We have ,

$$det\left(\begin{bmatrix} a & b\\ b & c \end{bmatrix} -\lambda \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}\right) =0$$

$$\Rightarrow det\left( \begin{bmatrix} a-\lambda & b\\ b & c-\lambda \end{bmatrix}\right)=0 $$

$$\Rightarrow \left(a-\lambda\right)\left(c-\lambda\right)-b^2=0$$

$$\Rightarrow \lambda^2-\lambda(a+c)+(ac-b^2)=0$$    (1)

The two roots of $$\lambda$$ are

$$\lambda_1=\frac{(a+c )+\sqrt{(a+c)^2-4(ac-b^2)}}{2}$$

and

$$\lambda_2=\frac{(a+c )-\sqrt{(a+c)^2-4(ac-b^2)}}{2}$$

$$\therefore$$ from equation (1) we have the sum of roots as $$(a+c)$$ and the product of roots $$(ac-b^2)$$

$$\therefore$$ from the factor theorem we have equation (1) as

$$\left (\lambda-\lambda_1  \right )\left (\lambda-\lambda_2   \right )=0$$

$$\underline {Theorem}$$

Any real matrix ,order $$n\times n$$,is diagonizable ,i.e,n Eigen values which are real numbers $$(\lambda_1,...\lambda_n)$$.This matrix called $$\underline A$$,can be decomposed as:

$$\underline A=\underline V \underline \Lambda     \underline V^T$$

and for an orthogonal matrix

$$V^{-1}=V^T$$ and $$\Lambda=diag[\lambda_1,...,\lambda_n]$$ and $$VV^{-1}=VV^T$$

$$\therefore det(I)=det(VV^T)=(detV)^2$$

$$\Rightarrow detV=\pm1$$

$$detV=1$$ in a righthanded co-ordinate system

&

$$detV=-1$$ in a left handed co-ordinate system

$$\therefore det \underline A=det\left( \underline V \underline \lambda \underline V^T\right)$$

For example ;

To find the Eigen values and Eigen vector of

$$\underline A=\begin{bmatrix} 2 & 5\\ 5 & 7 \end{bmatrix}$$

Using MATLAB ,from the given syntax,

$$[V,D]=eig(\underline A)$$

where the Eigen Values D are

$$D= \begin{bmatrix} -1.0902 & 0\\ 0      & 10.0902 \end{bmatrix}$$

we have ,the Eigen vector in form of Modal Matrix

$$V= \begin{bmatrix} -0.8507 & 0.5257\\ 0.5257 & 0.8507 \end{bmatrix}$$

The Modal matrix M is represented generally as

$$M^{-1}AM=D$$ Where A is the given matrix and D is the diagonal matrix containing the

eigenvalues of A

We have from the given Eigenvalues,we can find the eigen vectors

$$(A-\lambda_1 I)V_1=0$$

$$(A-\lambda_2 I)V_2=0$$

$$\Rightarrow AV_1=V_1\lambda_1$$

and

$$\Rightarrow AV_2=V_2\lambda_2$$

Writing these simultaneous equations in matrix form ,we have

$$A\begin{bmatrix} V_1 &V_2 \end{bmatrix}=\begin{bmatrix} V_1 &V_2 \end{bmatrix}\lambda$$

Where $$\lambda=\begin{bmatrix} \lambda_1 & 0\\ 0&\lambda_2 \end{bmatrix} $$ which represents the eigenvalues. $$ \therefore AV=V\lambda$$

This can also be shown by the numerical example from above using MATLAB.

$$A=\begin{bmatrix} 2 &5 \\ 5& 7 \end{bmatrix}$$   ;    $$V= \begin{bmatrix} -0.8507 & 0.5257\\ 0.5257 & 0.8507 \end{bmatrix}$$   ;     $$D= \begin{bmatrix} -1.0902 & 0\\ 0      & 10.0902 \end{bmatrix}$$

$$\therefore$$ using MATLAB ,we have

$$AV=VD=\begin{bmatrix} 0.9271 & 5.3049 \\   -0.5736& 8.5834

\end{bmatrix} $$

We have the derivative of an integral in 1-D given as

$$\frac{\mathrm{d} }{\mathrm{d} x}\int_{\xi=A(x)}^{\xi=B(x)}F(x,\xi)d\xi=\int_{\xi=A(x)}^{\xi=B(x)}\frac{\partial  }{\partial x}F(x,\xi)d\xi+...+\frac{\mathrm{d} B(x)}{\mathrm{d} x}F[x,\xi=B(x)]-\frac{\mathrm{d} A(x)}{\mathrm{d} x}F[x,\xi=A(x)] $$

when extended to the entire domain ,we have the Leibniz rule as follows,

$$\frac{\mathrm{d} }{\mathrm{d} t}\iiint_\mathrm{R(t)} T_{i,j}(x_i,t)dV=\iiint_\mathrm{R}\frac{\partial T_{i,j...}}{\partial t}dV+\iint_\mathrm{S}n_kw_kT_{i,j}dS$$

where $$T_{i,j }$$ is any property ,be it a scalar ,vector or tensor,which is a function of space and time.If the property around the region R changes over time is equal to rate of change of property within the region and the whatever property that goes through the surface where $$n_k$$ is the unit vector and $$w_k$$ is the velocity of the changing surface.

The Reynold's Transport Theorem,which can be shown to be analogous to the Leibniz theorem as follows

$$\frac{\mathrm{d} N}{\mathrm{d} t}_\mathrm{system}=\frac{\partial }{\partial t}\int_\mathrm{cv}NdV +\int_\mathrm{cs}N\vec{v}d\vec{A}$$

Which can further explained as

$$\frac{\mathrm{d} }{\mathrm{d} t}\iiint_\mathrm{MR}F(x_i,t)dV=\frac{\partial }{\partial t}\iiint_\mathrm{AR}FdV+\iint_\mathrm{CS}F\vec{v}.d\vec{S}$$

Which states that the rate of change of property $$F(x_i)$$ in the material region is equal to rate of change of property in the arbitrary region ,also the control volume the region and the flux of the material through the surface.

If the function F is a function of time then the equation can written as follows ,which is the Leibniz rule.

$$\frac{\mathrm{d} }{\mathrm{d} t}\iiint_\mathrm{MR}F(x_i,t)dV=\iiint_\mathrm{AR}\frac{\partial }{\partial t}FdV+\iint_\mathrm{CS}F\vec{v}.d\vec{S}

$$