91Ó°ÊÓ

Eigenvectors are a nonzero vector that is mapped by a given linear transformation of a vector space onto a vector that is the product of a scalar multiplied by the original vector.

Consider the following linear space:

V = { A : A is$2×2$matrix and$\left[\begin{array}{c}1\\ -3\end{array}\right]$is an eigenvector of A}

The objective is to determine the basis of the above linear space and determine the dimension of V.

Assume that$A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$such that$A∈V$.

Let $v=\left[\begin{array}{c}1\\ -3\end{array}\right]$is an eigenvector of A then it is known that $\stackrel{→}{Av}=\mathrm{λ}\stackrel{→}{v}$, thus it can be

written as:

$$\begin{gathered} \left[\begin{array}{cc}\mathrm{a}& \mathrm{b}\\ \mathrm{c}& \mathrm{d}\end{array}\right]\left[\begin{array}{c}1\\ -3\end{array}\right]=\mathrm{λ}\left[\begin{array}{c}1\\ -3\end{array}\right]\left\{\begin{array}{l}\mathrm{Here},\mathrm{λ}\mathrm{is}\mathrm{an}\mathrm{eignevalue}\mathrm{corresponding}\\ \mathrm{tothe}\mathrm{eigenvector}\mathrm{v}.\end{array}\right. \\ \left[\begin{array}{c}\mathrm{a}-3\mathrm{b}\\ \mathrm{c}-3\mathrm{d}\end{array}\right]=\left[\begin{array}{c}\mathrm{λ}\\ -3\mathrm{λ}\end{array}\right] \end{gathered}$$

This implies,

$$\begin{gathered} \mathrm{λ}=a-3b.....\left(1\right) \\ -3\mathrm{λ}=c-3d.....\left(2\right) \end{gathered}$$

Now, substitute the value of $\mathrm{λ}$from equation (1) in the equation (2) as follows:

$$\begin{gathered} -3(a-3b)=c-3d \\ -3a+9b=c-3d \\ -3a+9b+3d=c \end{gathered}$$

Now, substitute the value of $c=-3a+9b+3d$in the matrix Aas shown below:

$$\begin{gathered} A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right] \\ =\left[\begin{array}{cc}a& b\\ -3a+9b+3d& d\end{array}\right] \\ =a\left[\begin{array}{cc}1& 0\\ -3& 0\end{array}\right]+b\left[\begin{array}{cc}0& 1\\ 9& 0\end{array}\right]+d\left[\begin{array}{cc}0& 0\\ 3& 1\end{array}\right] \end{gathered}$$

Thus, all the matrices in the linear space Vis of the following form:

$$\begin{gathered} V=\left\{=a\left[\begin{array}{cc}1& 0\\ -3& 0\end{array}\right]+b\left[\begin{array}{cc}0& 1\\ 9& 0\end{array}\right]+d\left[\begin{array}{cc}0& 0\\ 3& 1\end{array}\right]suchthata,b,d∈R\right\} \\ a=spam\left\{\left[\begin{array}{cc}1& 0\\ -3& 0\end{array}\right],\left[\begin{array}{cc}0& 1\\ 9& 0\end{array}\right],\left[\begin{array}{cc}0& 0\\ 3& 1\end{array}\right]\right\} \end{gathered}$$

Hence, the basis for the linear space Vis$\left\{\left[\begin{array}{cc}1& 0\\ -3& 0\end{array}\right],\left[\begin{array}{cc}0& 1\\ 9& 0\end{array}\right],\left[\begin{array}{cc}0& 0\\ 3& 1\end{array}\right]\right\}$and dim (V)=3because there are 3 elements in the basis set of V.