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.

(a)

Consider the matrix .

The objective is to find the eigenvalues of A using geometric interpretation of this transformation.

The linear transformation corresponding to the matrix A is, $T\left(v\right)=Av$for $v∈R^2$.

A sketch of the transformation $T\left(v\right)=Av$ is shown below:

The matrix $A=\left[\begin{array}{cc}3& 4\\ 4& -3\end{array}\right]$can be written as, $A=5\left[\begin{array}{cc}3/5& 4/5\\ 4/5& -3/5\end{array}\right]$.

Note that the matrix on the right-hand side of the above equation represents a reflection about a line since it is of the form $\left[\begin{array}{cc}a& b\\ b& -a\end{array}\right]$with $a^2+b^2=1$.

Thus, the matrixrepresents a reflection about the line L followed by a scaling by a factor of 5.

If a vector$\stackrel{â‡\P Ä}{v}$is parallel to L , then$A\stackrel{â‡\P Ä}{v}=5\stackrel{â‡\P Ä}{v}$.

If $\stackrel{â‡\P Ä}{v}$is orthogonal to L , then $A\stackrel{â‡\P Ä}{v}=5\stackrel{â‡\P Ä}{v}$.

Therefore, the two eigenvalues of A are $+5,-5$.

(b)

Find the eigenspace corresponding to eigenvalue 5 using the equation$A\stackrel{â‡\P Ä}{v}=5\stackrel{â‡\P Ä}{v}$:

$\left[\begin{array}{cc}3& 4\\ 4& -3\end{array}\right]\left[\begin{array}{c}{v}_1\\ v_2\end{array}\right]=5\left[\begin{array}{c}{v}_1\\ v_2\end{array}\right]$

or equivalently

role="math" localid="1659529092342" $$\begin{gathered} 3v_1+4v_2=5v_1 \\ 4v_1+3v_2=5v_2 \end{gathered}$$

Multiply by 12 in $3v_1+4v_2=5v_1$and multiply by -12 in $4v_1-3v_2=5v_2$to find relation between $v_1$and $v_2$as follows:

$$\begin{gathered} 12v_1+16v_2=20v_1 \\ +\frac{-12v_1+9v_2=-15v_2}{25v_2=20v_1-15v_2} \end{gathered}$$

This implies, $v_1=2v_2$.

Now, role="math" localid="1659528880044" $$\begin{gathered} \left[\begin{array}{c}{v}_1\\ v_2\end{array}\right]=\left[\begin{array}{c}2v_2\\ v_2\end{array}\right] \\ =\left[\begin{array}{c}2\\ 1\end{array}\right] \end{gathered}$$

Next find the eigenspace corresponding to the eigenvalue -5 from the equation $A\stackrel{â‡\P Ä}{v}=-5\stackrel{â‡\P Ä}{v}$

role="math" localid="1659529222063" $\left[\begin{array}{cc}3& 4\\ 4& -3\end{array}\right]\left[\begin{array}{c}{v}_1\\ v_2\end{array}\right]=-5\left[\begin{array}{c}{v}_1\\ v_2\end{array}\right]$

Equivalently

$$\begin{gathered} 3v_1+4v_2=5v_1 \\ 4v_1+3v_2=5v_2 \end{gathered}$$

This implies,

$$\begin{gathered} 12v_1+16v_2=20v_1 \\ +\frac{-12v_1+9v_2=-15v_2}{25v_2=20v_1-15v_2} \end{gathered}$$

On solving, we get $v_1=-\frac{1}{2}{v}_2$.

Now,$$\begin{gathered} \left[\begin{array}{c}{v}_1\\ v_2\end{array}\right]=\left[\begin{array}{c}2v_2\\ v_2\end{array}\right] \\ =\left[\begin{array}{c}-1\\ 2\end{array}\right]\left[Take{v}_2=2\right] \end{gathered}$$

Thus, the eigenspace is the span of, role="math" localid="1659529330849" $\left[\begin{array}{c}-1\\ 2\end{array}\right]$.

Hence, the eigen basis are, $\left[\begin{array}{c}-1\\ 2\end{array}\right],\left[\begin{array}{c}2\\ 1\end{array}\right]$.

(c)

To diagonalize a square matrix A means to find an invertible matrix S and a diagonal matrix B such that .

Take $S=\left[\begin{array}{cc}-1& 2\\ 2& 1\end{array}\right]$(the eigen basis of matrix v ) and, $B=\left[\begin{array}{cc}5& 0\\ 0& -5\end{array}\right]$(eigenvalues of A on the diagonal entry).

Then,

$$\begin{gathered} S^{-1}AS={\left[\begin{array}{cc}-1& 2\\ 2& 1\end{array}\right]}^{-1}\left[\begin{array}{cc}3& 4\\ 4& -3\end{array}\right]\left[\begin{array}{cc}-1& 2\\ 2& 1\end{array}\right] \\ ={\left[\begin{array}{cc}-1& 2\\ 2& 1\end{array}\right]}^{-1}\left[\begin{array}{cc}5& 10\\ -10& 5\end{array}\right] \\ ={\left[\begin{array}{cc}\frac{-1}{5}& \frac{2}{5}\\ \frac{2}{5}& \frac{1}{5}\end{array}\right]}^{}\left[\begin{array}{cc}5& 10\\ -10& 5\end{array}\right] \\ =\left[\begin{array}{cc}5& 0\\ 0& -5\end{array}\right] \\ =B \end{gathered}$$

This implies, $S^{-1}AS=B$and hence, A is diagonalizable.